项目
内容
シラバス项目
15. 複素数平面（数学Ｃ）：(1) 複素数平面（① 複素数平面 ② 複素数の極形式）、(2) ド・モアブルの定理、(3) 複素数と図形
コース
仅コース2
国内书对应
国内书第 17 章：17.3 节（复平面、模、共轭的几何意义）、17.5 节（复数的三角形式，选学）
本章新内容的分量
大部分是新的。复平面和模国内学过；极形式国内是选学、不考；棣莫弗定理、n n n
国内高考的复数题只要求“算一次除法、读实部虚部”，三角形式是选学，棣莫弗公式只在国内书第 17 章 17.5 节的推论里出现过一句。EJU 的コース2 却会拿复数平面出一整道大问（2021 年第 1 回コース2 的 Ⅱ 問2 就是复数平面），而且考的是“复数当作平面上的点”：旋转、角度、三角形的形状、方程表示的图形。丢分主要丢在三处：偏角的范围没按题目要求（0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π n n n
シラバス小项目（日文原文）
国内书位置
相同 / 讲法不同 / 国内没有
本章哪节讲
複素数の四則演算、共役複素数の定義（シラバス第 8 项「いろいろな式」）
国内书第 17 章 17.2、17.4 节
相同，见国内书第 17 章；复数与二次方程见本书第 10 章
不展开
(1) ① 複素数平面
国内书第 17 章 17.3 节（点、向量、模、共轭关于实轴对称、两点距离）
讲法不同：国内只用坐标算模；EJU 还要直接用共轭和模的运算性质（∣ z ∣ 2 = z z ‾ \lvert z \rvert^2 = z\overline{z} ∣ z ∣ 2 = z z
17.2
(1) ② 複素数の極形式
国内书第 17 章 17.5 节（三角形式，选学）
讲法不同：国内选学、不考；EJU 必考，偏角范围按题目给定
17.3
(2) ド・モアブルの定理
国内书第 17 章 17.5 节“常用推论”一句
国内没有（系统讲）
17.4、17.5
(3) 複素数と図形：回転、2 直線のなす角、共線・垂直、三角形の形状
国内书第 17 章 17.5 节例 3（绕原点旋转）；内分点、重心见国内书第 14 章
国内没有（绕任意点旋转、用比值判断形状）
17.6
(3) 複素数と図形：方程式の表す図形（円、垂直二等分線、軌跡）
国内书第 17 章 17.3 节（∣ z − z 0 ∣ = r \lvert z - z_0 \rvert = r ∣ z − z 0 ∣ = r
讲法不同：EJU 要求不拆成 x x x y y y w = f ( z ) w = f(z) w = f ( z )
17.7
日本现行学习指导要领（2018 年告示）把这部分放在数学Ｃ的内容「平面上の曲線と複素数平面」里，和本书第 18 章「式と曲線」同属一项；教科书一般分成两章，如数研出版『数学C』的第 3 章「複素数平面」、第 4 章「式と曲線」。EJU シラバス只列三个中项目，实际范围就是上表最后五行。
本章六个知识点分两条线：一条是代数线，复平面（17.2）→ 极形式（17.3）→ 棣莫弗定理（17.4）→ n n n
複素数平面 ：复数 z = a + b i z = a + bi z = a + bi ( a , b ) (a, b) ( a , b ) 複素数平面 （国内叫复平面），横轴叫実軸 ，纵轴叫虚軸 。这部分和国内书第 17 章 17.3 节完全相同。
EJU 和日本教科书有一个写法习惯：直接把复数当作点的名字。「点 z z z z z z A ( α ) \mathrm{A}(\alpha) A ( α ) α \alpha α A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) C ( γ ) \mathrm{C}(\gamma) C ( γ )
加法、减法、实数倍的几何意义 （和平面向量相同，见国内书第 14 章）：
α + β \alpha + \beta α + β 0 0 0 α \alpha α β \beta β β − α \beta - \alpha β − α α \alpha α β \beta β ∣ β − α ∣ \lvert \beta - \alpha \rvert ∣ β − α ∣ A B \mathrm{AB} AB k α k\alpha k α k k k 0 α 0\alpha 0 α ∣ k ∣ \lvert k \rvert ∣ k ∣
共役複素数の性質 ：z = a + b i z = a + bi z = a + bi z ‾ = a − b i \overline{z} = a - bi z = a − bi α \alpha α β \beta β
α + β ‾ = α ‾ + β ‾ , α − β ‾ = α ‾ − β ‾ , α β ‾ = α ‾ β ‾ , ( α β ) ‾ = α ‾ β ‾ ( β ≠ 0 ) . \overline{\alpha + \beta} = \overline{\alpha} + \overline{\beta}, \quad \overline{\alpha - \beta} = \overline{\alpha} - \overline{\beta}, \quad \overline{\alpha\beta} = \overline{\alpha}\,\overline{\beta}, \quad \overline{\left(\frac{\alpha}{\beta}\right)} = \frac{\overline{\alpha}}{\overline{\beta}} \ (\beta \ne 0). α + β = α + β , α − β = α − β , α β = α β , ( β α ) = β α ( β = 0 ) . 一句话：共轭可以“穿过”加减乘除 。由乘法一条还得 α n ‾ = α ‾ n \overline{\alpha^n} = \overline{\alpha}^{\,n} α n = α n
实数、纯虚数的判定 ：
z が実数 ⟺ z ‾ = z , z が純虚数 ⟺ z ‾ = − z かつ z ≠ 0. z \text{ が実数} \iff \overline{z} = z, \qquad z \text{ が純虚数} \iff \overline{z} = -z \text{ かつ } z \ne 0. z が実数 ⟺ z = z , z が純虚数 ⟺ z = − z かつ z = 0. 絶対値の性質 ：
∣ z ∣ 2 = z z ‾ , ∣ α β ∣ = ∣ α ∣ ∣ β ∣ , ∣ α β ∣ = ∣ α ∣ ∣ β ∣ , ∣ z n ∣ = ∣ z ∣ n . \lvert z \rvert^2 = z\overline{z}, \quad \lvert \alpha\beta \rvert = \lvert \alpha \rvert \lvert \beta \rvert, \quad \left\lvert \frac{\alpha}{\beta} \right\rvert = \frac{\lvert \alpha \rvert}{\lvert \beta \rvert}, \quad \lvert z^n \rvert = \lvert z \rvert^n. ∣ z ∣ 2 = z z , ∣ α β ∣ = ∣ α ∣ ∣ β ∣ , β α = ∣ β ∣ ∣ α ∣ , ∣ z n ∣ = ∣ z ∣ n . 特别地，∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z z ‾ = 1 z\overline{z} = 1 z z = 1
∣ z ∣ = 1 ⟺ z ‾ = 1 z . \lvert z \rvert = 1 \iff \overline{z} = \frac{1}{z}. ∣ z ∣ = 1 ⟺ z = z 1 . 这一条在 EJU 里用得极多：单位圆上的点，取共轭就是取倒数。
题目只给几个复数的模（如 ∣ α ∣ = 2 \lvert \alpha \rvert = 2 ∣ α ∣ = 2 ∣ β ∣ = 3 \lvert \beta \rvert = 3 ∣ β ∣ = 3 ∣ α − β ∣ = 7 \lvert \alpha - \beta \rvert = \sqrt{7} ∣ α − β ∣ = 7 α β ‾ + α ‾ β \alpha\overline{\beta} + \overline{\alpha}\beta α β + α β
题目要证明或判断某个式子是实数、纯虚数，而条件只有 ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z z z
题目要求积、商、幂的模，如 ∣ ( 1 + 2 i ) 3 ( 3 − 4 i ) 2 ( 2 − i ) 4 ∣ \left\lvert \dfrac{(1 + 2i)^3(3 - 4i)^2}{(2 - i)^4} \right\rvert ( 2 − i ) 4 ( 1 + 2 i ) 3 ( 3 − 4 i ) 2
∣ z ∣ 2 \lvert z \rvert^2 ∣ z ∣ 2 z 2 z^2 z 2 z z ‾ z\overline{z} z z 展开 ∣ α + β ∣ 2 \lvert \alpha + \beta \rvert^2 ∣ α + β ∣ 2 α β ‾ \alpha\overline{\beta} α β α ‾ β \overline{\alpha}\beta α β α β ‾ \alpha\overline{\beta} α β 2 α β 2\alpha\beta 2 α β
纯虚数的条件里要有 z ≠ 0 z \ne 0 z = 0 z ‾ = − z \overline{z} = -z z = − z
EJU 题干的说法：「z ‾ \overline{z} z z z z z z z z − 1 z + 1 \dfrac{z - 1}{z + 1} z + 1 z − 1
结论 1：α β ‾ = α ‾ β ‾ \overline{\alpha\beta} = \overline{\alpha}\,\overline{\beta} α β = α β 设 α = a + b i \alpha = a + bi α = a + bi β = c + d i \beta = c + di β = c + d i
α β = ( a c − b d ) + ( a d + b c ) i , α β ‾ = ( a c − b d ) − ( a d + b c ) i . \alpha\beta = (ac - bd) + (ad + bc)i, \qquad \overline{\alpha\beta} = (ac - bd) - (ad + bc)i. α β = ( a c − b d ) + ( a d + b c ) i , α β = ( a c − b d ) − ( a d + b c ) i . α ‾ β ‾ = ( a − b i ) ( c − d i ) = ( a c − b d ) − ( a d + b c ) i . \overline{\alpha}\,\overline{\beta} = (a - bi)(c - di) = (ac - bd) - (ad + bc)i. α β = ( a − bi ) ( c − d i ) = ( a c − b d ) − ( a d + b c ) i . 两者相同。（关键一步 ：( a − b i ) ( c − d i ) (a - bi)(c - di) ( a − bi ) ( c − d i ) ( − b ) ( − d ) i 2 = − b d (-b)(-d)i^2 = -bd ( − b ) ( − d ) i 2 = − b d α β \alpha\beta α β α β \alpha\beta α β γ = α β \gamma = \dfrac{\alpha}{\beta} γ = β α α = β γ \alpha = \beta\gamma α = β γ α ‾ = β ‾ γ ‾ \overline{\alpha} = \overline{\beta}\,\overline{\gamma} α = β γ γ ‾ = α ‾ β ‾ \overline{\gamma} = \dfrac{\overline{\alpha}}{\overline{\beta}} γ = β α
结论 2：实数、纯虚数的判定。 z = a + b i z = a + bi z = a + bi z ‾ = z \overline{z} = z z = z a − b i = a + b i a - bi = a + bi a − bi = a + bi b = 0 b = 0 b = 0 z ‾ = − z \overline{z} = -z z = − z a − b i = − a − b i a - bi = -a - bi a − bi = − a − bi a = 0 a = 0 a = 0 z ≠ 0 z \ne 0 z = 0 b ≠ 0 b \ne 0 b = 0
结论 3：∣ α β ∣ = ∣ α ∣ ∣ β ∣ \lvert \alpha\beta \rvert = \lvert \alpha \rvert \lvert \beta \rvert ∣ α β ∣ = ∣ α ∣ ∣ β ∣ 由 ∣ z ∣ 2 = z z ‾ \lvert z \rvert^2 = z\overline{z} ∣ z ∣ 2 = z z
∣ α β ∣ 2 = α β α β ‾ = α α ‾ ⋅ β β ‾ = ∣ α ∣ 2 ∣ β ∣ 2 . \lvert \alpha\beta \rvert^2 = \alpha\beta\,\overline{\alpha\beta} = \alpha\overline{\alpha} \cdot \beta\overline{\beta} = \lvert \alpha \rvert^2 \lvert \beta \rvert^2. ∣ α β ∣ 2 = α β α β = α α ⋅ β β = ∣ α ∣ 2 ∣ β ∣ 2 . 两边都是非负实数，开方即得。商和幂同理（幂是乘法反复用）。
例 1 （场景：只知道模，求别的模）
問題 複素数 α \alpha α β \beta β ∣ α ∣ = 2 \lvert \alpha \rvert = 2 ∣ α ∣ = 2 ∣ β ∣ = 3 \lvert \beta \rvert = 3 ∣ β ∣ = 3 ∣ α − β ∣ = 7 \lvert \alpha - \beta \rvert = \sqrt{7} ∣ α − β ∣ = 7 α β ‾ + α ‾ β \alpha\overline{\beta} + \overline{\alpha}\beta α β + α β ∣ α + β ∣ \lvert \alpha + \beta \rvert ∣ α + β ∣
解答 与えられた絶対値を 2 2 2 ∣ z ∣ 2 = z z ‾ \lvert z \rvert^2 = z\overline{z} ∣ z ∣ 2 = z z
∣ α − β ∣ 2 = ( α − β ) ( α ‾ − β ‾ ) = α α ‾ − α β ‾ − α ‾ β + β β ‾ = ∣ α ∣ 2 + ∣ β ∣ 2 − ( α β ‾ + α ‾ β ) . \lvert \alpha - \beta \rvert^2 = (\alpha - \beta)(\overline{\alpha} - \overline{\beta}) = \alpha\overline{\alpha} - \alpha\overline{\beta} - \overline{\alpha}\beta + \beta\overline{\beta} = \lvert \alpha \rvert^2 + \lvert \beta \rvert^2 - (\alpha\overline{\beta} + \overline{\alpha}\beta). ∣ α − β ∣ 2 = ( α − β ) ( α − β ) = α α − α β − α β + β β = ∣ α ∣ 2 + ∣ β ∣ 2 − ( α β + α β ) . 代入すると 7 = 4 + 9 − ( α β ‾ + α ‾ β ) 7 = 4 + 9 - (\alpha\overline{\beta} + \overline{\alpha}\beta) 7 = 4 + 9 − ( α β + α β ) α β ‾ + α ‾ β = 6 \alpha\overline{\beta} + \overline{\alpha}\beta = 6 α β + α β = 6
∣ α + β ∣ 2 = ∣ α ∣ 2 + ∣ β ∣ 2 + ( α β ‾ + α ‾ β ) = 4 + 9 + 6 = 19 , \lvert \alpha + \beta \rvert^2 = \lvert \alpha \rvert^2 + \lvert \beta \rvert^2 + (\alpha\overline{\beta} + \overline{\alpha}\beta) = 4 + 9 + 6 = 19, ∣ α + β ∣ 2 = ∣ α ∣ 2 + ∣ β ∣ 2 + ( α β + α β ) = 4 + 9 + 6 = 19 , よって ∣ α + β ∣ = 19 \lvert \alpha + \beta \rvert = \sqrt{19} ∣ α + β ∣ = 19
中文翻译 题目：复数 α \alpha α β \beta β ∣ α ∣ = 2 \lvert \alpha \rvert = 2 ∣ α ∣ = 2 ∣ β ∣ = 3 \lvert \beta \rvert = 3 ∣ β ∣ = 3 ∣ α − β ∣ = 7 \lvert \alpha - \beta \rvert = \sqrt{7} ∣ α − β ∣ = 7 α β ‾ + α ‾ β \alpha\overline{\beta} + \overline{\alpha}\beta α β + α β ∣ α + β ∣ \lvert \alpha + \beta \rvert ∣ α + β ∣ ∣ z ∣ 2 = z z ‾ \lvert z \rvert^2 = z\overline{z} ∣ z ∣ 2 = z z ∣ α − β ∣ 2 = ( α − β ) ( α ‾ − β ‾ ) = α α ‾ − α β ‾ − α ‾ β + β β ‾ = ∣ α ∣ 2 + ∣ β ∣ 2 − ( α β ‾ + α ‾ β ) \lvert \alpha - \beta \rvert^2 = (\alpha - \beta)(\overline{\alpha} - \overline{\beta}) = \alpha\overline{\alpha} - \alpha\overline{\beta} - \overline{\alpha}\beta + \beta\overline{\beta} = \lvert \alpha \rvert^2 + \lvert \beta \rvert^2 - (\alpha\overline{\beta} + \overline{\alpha}\beta) ∣ α − β ∣ 2 = ( α − β ) ( α − β ) = α α − α β − α β + β β = ∣ α ∣ 2 + ∣ β ∣ 2 − ( α β + α β ) 7 = 4 + 9 − ( α β ‾ + α ‾ β ) 7 = 4 + 9 - (\alpha\overline{\beta} + \overline{\alpha}\beta) 7 = 4 + 9 − ( α β + α β ) α β ‾ + α ‾ β = 6 \alpha\overline{\beta} + \overline{\alpha}\beta = 6 α β + α β = 6 ∣ α + β ∣ 2 = ∣ α ∣ 2 + ∣ β ∣ 2 + ( α β ‾ + α ‾ β ) = 4 + 9 + 6 = 19 \lvert \alpha + \beta \rvert^2 = \lvert \alpha \rvert^2 + \lvert \beta \rvert^2 + (\alpha\overline{\beta} + \overline{\alpha}\beta) = 4 + 9 + 6 = 19 ∣ α + β ∣ 2 = ∣ α ∣ 2 + ∣ β ∣ 2 + ( α β + α β ) = 4 + 9 + 6 = 19 ∣ α + β ∣ = 19 \lvert \alpha + \beta \rvert = \sqrt{19} ∣ α + β ∣ = 19
解说 容易卡在：第一步把 ( α − β ) ( α ‾ − β ‾ ) (\alpha - \beta)(\overline{\alpha} - \overline{\beta}) ( α − β ) ( α − β ) ( α − β ) 2 (\alpha - \beta)^2 ( α − β ) 2
例 2 （场景：判断是实数还是纯虚数）
問題 複素数 z z z ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z ≠ − 1 z \ne -1 z = − 1 w = z − 1 z + 1 w = \dfrac{z - 1}{z + 1} w = z + 1 z − 1 0 0 0
解答 ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z ‾ = 1 z \overline{z} = \dfrac{1}{z} z = z 1 w w w
w ‾ = z ‾ − 1 z ‾ + 1 = 1 z − 1 1 z + 1 = 1 − z 1 + z = − w . \overline{w} = \frac{\overline{z} - 1}{\overline{z} + 1} = \frac{\frac{1}{z} - 1}{\frac{1}{z} + 1} = \frac{1 - z}{1 + z} = -w. w = z + 1 z − 1 = z 1 + 1 z 1 − 1 = 1 + z 1 − z = − w . （3 番目の等号は分母と分子に z z z w ‾ = − w \overline{w} = -w w = − w w w w 0 0 0 z = 1 z = 1 z = 1 w = 0 w = 0 w = 0 z ≠ 1 z \ne 1 z = 1 w w w
中文翻译 题目：复数 z z z ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z ≠ − 1 z \ne -1 z = − 1 w = z − 1 z + 1 w = \dfrac{z - 1}{z + 1} w = z + 1 z − 1 ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z ‾ = 1 z \overline{z} = \dfrac{1}{z} z = z 1 w w w w ‾ = z ‾ − 1 z ‾ + 1 = 1 z − 1 1 z + 1 = 1 − z 1 + z = − w \overline{w} = \frac{\overline{z} - 1}{\overline{z} + 1} = \frac{\frac{1}{z} - 1}{\frac{1}{z} + 1} = \frac{1 - z}{1 + z} = -w w = z + 1 z − 1 = z 1 + 1 z 1 − 1 = 1 + z 1 − z = − w z z z w ‾ = − w \overline{w} = -w w = − w w w w z = 1 z = 1 z = 1 w = 0 w = 0 w = 0 z ≠ 1 z \ne 1 z = 1 w w w
解说 另一种解法 ：看图形。w = z − 1 z + 1 w = \dfrac{z - 1}{z + 1} w = z + 1 z − 1 − 1 → z -1 \to z − 1 → z 1 → z 1 \to z 1 → z z z z 1 1 1 − 1 -1 − 1 ± π 2 \pm\dfrac{\pi}{2} ± 2 π w w w
例 3 （场景：积、商、幂的模）
問題 ∣ ( 1 + 2 i ) 3 ( 3 − 4 i ) 2 ( 2 − i ) 4 ∣ \left\lvert \dfrac{(1 + 2i)^3(3 - 4i)^2}{(2 - i)^4} \right\rvert ( 2 − i ) 4 ( 1 + 2 i ) 3 ( 3 − 4 i ) 2
解答 ∣ 1 + 2 i ∣ = 5 \lvert 1 + 2i \rvert = \sqrt{5} ∣ 1 + 2 i ∣ = 5 ∣ 3 − 4 i ∣ = 5 \lvert 3 - 4i \rvert = 5 ∣ 3 − 4 i ∣ = 5 ∣ 2 − i ∣ = 5 \lvert 2 - i \rvert = \sqrt{5} ∣ 2 − i ∣ = 5
( 5 ) 3 ⋅ 5 2 ( 5 ) 4 = 5 5 ⋅ 25 25 = 5 5 . \frac{(\sqrt{5})^3 \cdot 5^2}{(\sqrt{5})^4} = \frac{5\sqrt{5} \cdot 25}{25} = 5\sqrt{5}. ( 5 ) 4 ( 5 ) 3 ⋅ 5 2 = 25 5 5 ⋅ 25 = 5 5 . 中文翻译 题目：求 ∣ ( 1 + 2 i ) 3 ( 3 − 4 i ) 2 ( 2 − i ) 4 ∣ \left\lvert \dfrac{(1 + 2i)^3(3 - 4i)^2}{(2 - i)^4} \right\rvert ( 2 − i ) 4 ( 1 + 2 i ) 3 ( 3 − 4 i ) 2 ∣ 1 + 2 i ∣ = 5 \lvert 1 + 2i \rvert = \sqrt{5} ∣ 1 + 2 i ∣ = 5 ∣ 3 − 4 i ∣ = 5 \lvert 3 - 4i \rvert = 5 ∣ 3 − 4 i ∣ = 5 ∣ 2 − i ∣ = 5 \lvert 2 - i \rvert = \sqrt{5} ∣ 2 − i ∣ = 5 ( 5 ) 3 ⋅ 5 2 ( 5 ) 4 = 5 5 ⋅ 25 25 = 5 5 \frac{(\sqrt{5})^3 \cdot 5^2}{(\sqrt{5})^4} = \frac{5\sqrt{5} \cdot 25}{25} = 5\sqrt{5} ( 5 ) 4 ( 5 ) 3 ⋅ 5 2 = 25 5 5 ⋅ 25 = 5 5
解说 积、商、幂的模可以逐层拆开，不必先把复数本身算出来。( 2 − i ) 4 (2 - i)^4 ( 2 − i ) 4 ( 5 ) 4 = 25 (\sqrt{5})^4 = 25 ( 5 ) 4 = 25 4 5 4\sqrt{5} 4 5 ( 1 + 2 i ) 3 (1 + 2i)^3 ( 1 + 2 i ) 3 ( 5 ) 3 = 5 5 (\sqrt{5})^3 = 5\sqrt{5} ( 5 ) 3 = 5 5 25 25 25 5 5 5\sqrt{5} 5 5
平行四边形の性質 ：∣ α + β ∣ 2 + ∣ α − β ∣ 2 = 2 ( ∣ α ∣ 2 + ∣ β ∣ 2 ) \lvert \alpha + \beta \rvert^2 + \lvert \alpha - \beta \rvert^2 = 2\left(\lvert \alpha \rvert^2 + \lvert \beta \rvert^2\right) ∣ α + β ∣ 2 + ∣ α − β ∣ 2 = 2 ( ∣ α ∣ 2 + ∣ β ∣ 2 ) ∣ α + β ∣ 2 = 2 ( 4 + 9 ) − 7 = 19 \lvert \alpha + \beta \rvert^2 = 2(4 + 9) - 7 = 19 ∣ α + β ∣ 2 = 2 ( 4 + 9 ) − 7 = 19 ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z + 1 z = z + z ‾ z + \dfrac{1}{z} = z + \overline{z} z + z 1 = z + z z z z 1 z = z ‾ \dfrac{1}{z} = \overline{z} z 1 = z z + z ‾ = 2 a z + \overline{z} = 2a z + z = 2 a 三角不等式 ：∣ α ∣ − ∣ β ∣ ≤ ∣ α + β ∣ ≤ ∣ α ∣ + ∣ β ∣ \lvert \alpha \rvert - \lvert \beta \rvert \le \lvert \alpha + \beta \rvert \le \lvert \alpha \rvert + \lvert \beta \rvert ∣ α ∣ − ∣ β ∣ ≤ ∣ α + β ∣ ≤ ∣ α ∣ + ∣ β ∣ 0 0 0 α \alpha α α + β \alpha + \beta α + β ∣ z − c ∣ \lvert z - c \rvert ∣ z − c ∣ 实系数多项式 ：f ( x ) f(x) f ( x ) f ( z ‾ ) = f ( z ) ‾ f(\overline{z}) = \overline{f(z)} f ( z ) = f ( z ) f ( z ) = 0 f(z) = 0 f ( z ) = 0 f ( z ‾ ) = 0 f(\overline{z}) = 0 f ( z ) = 0
错在哪 ：由 ∣ z ∣ = 2 \lvert z \rvert = 2 ∣ z ∣ = 2 z 2 = 4 z^2 = 4 z 2 = 4 为什么错 ：z 2 z^2 z 2 z = 2 i z = 2i z = 2 i z 2 = − 4 z^2 = -4 z 2 = − 4 z z ‾ z\overline{z} z z ∣ z ∣ 2 \lvert z \rvert^2 ∣ z ∣ 2 正确做法 ：写 z z ‾ = 4 z\overline{z} = 4 z z = 4 z ‾ = 4 z \overline{z} = \dfrac{4}{z} z = z 4 错在哪 ：展开 ∣ α + β ∣ 2 = ∣ α ∣ 2 + 2 α β + ∣ β ∣ 2 \lvert \alpha + \beta \rvert^2 = \lvert \alpha \rvert^2 + 2\alpha\beta + \lvert \beta \rvert^2 ∣ α + β ∣ 2 = ∣ α ∣ 2 + 2 α β + ∣ β ∣ 2 为什么错 ：照搬了实数的完全平方公式。∣ α + β ∣ 2 = ( α + β ) ( α ‾ + β ‾ ) \lvert \alpha + \beta \rvert^2 = (\alpha + \beta)(\overline{\alpha} + \overline{\beta}) ∣ α + β ∣ 2 = ( α + β ) ( α + β ) α β ‾ + α ‾ β \alpha\overline{\beta} + \overline{\alpha}\beta α β + α β 正确做法 ：每次都先写成“乘共轭”再展开。错在哪 ：认为”z ‾ = − z \overline{z} = -z z = − z z z z 为什么错 ：z = 0 z = 0 z = 0 正确做法 ：纯虚数的条件写成”z ‾ = − z \overline{z} = -z z = − z z ≠ 0 z \ne 0 z = 0
共轭是“上下翻折”（关于实轴对称）。翻折不改变长度，也不改变加法和乘法的结构，所以翻折可以先做也可以后做，这就是“共轭穿过四则运算”。
∣ z ∣ 2 = z z ‾ \lvert z \rvert^2 = z\overline{z} ∣ z ∣ 2 = z z ∣ x ∣ 2 = x 2 \lvert x \rvert^2 = x^2 ∣ x ∣ 2 = x 2 z z z z ‾ \overline{z} z
前置：国内书第 17 章 17.2–17.4 节，复数的定义、共轭、四则运算、z z ‾ = ∣ z ∣ 2 z\overline{z} = \lvert z \rvert^2 z z = ∣ z ∣ 2
前置：国内书第 14 章，向量加法的平行四边形法则，复数加法和它一样。
后续：17.3 节极形式，“积的模 = 模的积”会换个角度再看一次。
后续：17.7 节，∣ z − α ∣ 2 = ( z − α ) ( z ‾ − α ‾ ) \lvert z - \alpha \rvert^2 = (z - \alpha)(\overline{z} - \overline{\alpha}) ∣ z − α ∣ 2 = ( z − α ) ( z − α ) z z ‾ z\overline{z} z z
本书第 10 章 いろいろな式：实系数方程虚根成对，用的是 f ( z ‾ ) = f ( z ) ‾ f(\overline{z}) = \overline{f(z)} f ( z ) = f ( z )
偏角と極形式 ：z ≠ 0 z \ne 0 z = 0 r = ∣ z ∣ r = \lvert z \rvert r = ∣ z ∣ 0 z 0z 0 z θ \theta θ z z z 偏角 （国内叫辐角），记作 arg z \arg z arg z
z = r ( cos θ + i sin θ ) ( r > 0 ) , z = r(\cos\theta + i\sin\theta) \quad (r > 0), z = r ( cos θ + i sin θ ) ( r > 0 ) , 这叫 z z z 極形式 （国内叫三角形式）。定义和国内书第 17 章 17.5 节相同，区别在于考法：
偏角有无数个，彼此相差 2 π 2\pi 2 π arg z = θ + 2 n π \arg z = \theta + 2n\pi arg z = θ + 2 nπ n n n EJU 题目会写明范围 ，最常见的是「ただし，偏角 θ \theta θ 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π − π < θ ≤ π -\pi < \theta \le \pi − π < θ ≤ π
极形式的格式固定：r > 0 r > 0 r > 0
積・商の極形式 ：设 z 1 = r 1 ( cos θ 1 + i sin θ 1 ) z_1 = r_1(\cos\theta_1 + i\sin\theta_1) z 1 = r 1 ( cos θ 1 + i sin θ 1 ) z 2 = r 2 ( cos θ 2 + i sin θ 2 ) z_2 = r_2(\cos\theta_2 + i\sin\theta_2) z 2 = r 2 ( cos θ 2 + i sin θ 2 )
z 1 z 2 = r 1 r 2 { cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) } , z 1 z 2 = r 1 r 2 { cos ( θ 1 − θ 2 ) + i sin ( θ 1 − θ 2 ) } . z_1 z_2 = r_1 r_2\left\{\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right\}, \qquad \frac{z_1}{z_2} = \frac{r_1}{r_2}\left\{\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right\}. z 1 z 2 = r 1 r 2 { cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) } , z 2 z 1 = r 2 r 1 { cos ( θ 1 − θ 2 ) + i sin ( θ 1 − θ 2 ) } . 写成模和偏角：
∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ , arg ( z 1 z 2 ) = arg z 1 + arg z 2 , arg z 1 z 2 = arg z 1 − arg z 2 . \lvert z_1 z_2 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert, \quad \arg(z_1 z_2) = \arg z_1 + \arg z_2, \quad \arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2. ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ , arg ( z 1 z 2 ) = arg z 1 + arg z 2 , arg z 2 z 1 = arg z 1 − arg z 2 . 偏角的等式都是“相差 2 π 2\pi 2 π
原点のまわりの回転 ：乘 cos θ + i sin θ \cos\theta + i\sin\theta cos θ + i sin θ θ \theta θ θ > 0 \theta > 0 θ > 0 r ( cos θ + i sin θ ) r(\cos\theta + i\sin\theta) r ( cos θ + i sin θ ) θ \theta θ r r r r r r
题目要求把复数写成极形式，而复数是 − 2 ( cos π 6 + i sin π 6 ) -2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) − 2 ( cos 6 π + i sin 6 π ) sin α + i cos α \sin\alpha + i\cos\alpha sin α + i cos α 1 + cos θ + i sin θ 1 + \cos\theta + i\sin\theta 1 + cos θ + i sin θ r > 0 r > 0 r > 0
题目让你用两个复数的积（或商）求 cos 5 π 12 \cos\dfrac{5\pi}{12} cos 12 5 π sin π 12 \sin\dfrac{\pi}{12} sin 12 π
题目问 α 2 β \dfrac{\alpha^2}{\beta} β α 2 α β ‾ \alpha\overline{\beta} α β
题目说「点 z z z θ \theta θ r r r r ( cos θ + i sin θ ) r(\cos\theta + i\sin\theta) r ( cos θ + i sin θ )
r r r − 2 ( cos π 6 + i sin π 6 ) -2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) − 2 ( cos 6 π + i sin 6 π ) − 1 = cos π + i sin π -1 = \cos\pi + i\sin\pi − 1 = cos π + i sin π 2 ( cos 7 π 6 + i sin 7 π 6 ) 2\left(\cos\dfrac{7\pi}{6} + i\sin\dfrac{7\pi}{6}\right) 2 ( cos 6 7 π + i sin 6 7 π ) 正弦在前的要换角：sin α + i cos α = cos ( π 2 − α ) + i sin ( π 2 − α ) \sin\alpha + i\cos\alpha = \cos\left(\dfrac{\pi}{2} - \alpha\right) + i\sin\left(\dfrac{\pi}{2} - \alpha\right) sin α + i cos α = cos ( 2 π − α ) + i sin ( 2 π − α )
减号的要换成负角：cos α − i sin α = cos ( − α ) + i sin ( − α ) \cos\alpha - i\sin\alpha = \cos(-\alpha) + i\sin(-\alpha) cos α − i sin α = cos ( − α ) + i sin ( − α )
定偏角要同时看 cos θ \cos\theta cos θ sin θ \sin\theta sin θ tan θ \tan\theta tan θ
EJU 的英文题写 polar form、argument，并注明 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π θ \theta θ
积的公式在国内书第 17 章 17.5 节已经证明（直接展开，用两角和公式）。这里补两点。
结论 1：arg ( z 1 z 2 ) = arg z 1 + arg z 2 \arg(z_1 z_2) = \arg z_1 + \arg z_2 arg ( z 1 z 2 ) = arg z 1 + arg z 2 由积的公式，θ 1 + θ 2 \theta_1 + \theta_2 θ 1 + θ 2 z 1 z 2 z_1 z_2 z 1 z 2 2 π 2\pi 2 π 2 π 2\pi 2 π 关键一步 ：θ 1 = 3 π 2 \theta_1 = \dfrac{3\pi}{2} θ 1 = 2 3 π θ 2 = π 2 \theta_2 = \dfrac{\pi}{2} θ 2 = 2 π θ 1 + θ 2 = 2 π \theta_1 + \theta_2 = 2\pi θ 1 + θ 2 = 2 π 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π
结论 2：1 + cos θ + i sin θ 1 + \cos\theta + i\sin\theta 1 + cos θ + i sin θ 用二倍角公式（国内书第 13 章 13.3 节）：1 + cos θ = 2 cos 2 θ 2 1 + \cos\theta = 2\cos^2\dfrac{\theta}{2} 1 + cos θ = 2 cos 2 2 θ sin θ = 2 sin θ 2 cos θ 2 \sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2} sin θ = 2 sin 2 θ cos 2 θ
1 + cos θ + i sin θ = 2 cos θ 2 ( cos θ 2 + i sin θ 2 ) . 1 + \cos\theta + i\sin\theta = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right). 1 + cos θ + i sin θ = 2 cos 2 θ ( cos 2 θ + i sin 2 θ ) . cos θ 2 > 0 \cos\dfrac{\theta}{2} > 0 cos 2 θ > 0 cos θ 2 < 0 \cos\dfrac{\theta}{2} < 0 cos 2 θ < 0 cos π + i sin π \cos\pi + i\sin\pi cos π + i sin π π \pi π
例 1 （场景：化“不标准”的式子为极形式）
問題 0 < θ < π 0 < \theta < \pi 0 < θ < π z = 1 + cos θ + i sin θ z = 1 + \cos\theta + i\sin\theta z = 1 + cos θ + i sin θ w = sin π 5 + i cos π 5 w = \sin\dfrac{\pi}{5} + i\cos\dfrac{\pi}{5} w = sin 5 π + i cos 5 π 0 ≦ arg < 2 π 0 \leqq \arg < 2\pi 0 ≦ arg < 2 π
解答 証明で示した結論 2 より
z = 2 cos θ 2 ( cos θ 2 + i sin θ 2 ) . z = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right). z = 2 cos 2 θ ( cos 2 θ + i sin 2 θ ) . 0 < θ < π 0 < \theta < \pi 0 < θ < π 0 < θ 2 < π 2 0 < \dfrac{\theta}{2} < \dfrac{\pi}{2} 0 < 2 θ < 2 π cos θ 2 > 0 \cos\dfrac{\theta}{2} > 0 cos 2 θ > 0 2 cos θ 2 2\cos\dfrac{\theta}{2} 2 cos 2 θ θ 2 \dfrac{\theta}{2} 2 θ
w w w sin π 5 = cos ( π 2 − π 5 ) = cos 3 π 10 \sin\dfrac{\pi}{5} = \cos\left(\dfrac{\pi}{2} - \dfrac{\pi}{5}\right) = \cos\dfrac{3\pi}{10} sin 5 π = cos ( 2 π − 5 π ) = cos 10 3 π cos π 5 = sin 3 π 10 \cos\dfrac{\pi}{5} = \sin\dfrac{3\pi}{10} cos 5 π = sin 10 3 π
w = cos 3 π 10 + i sin 3 π 10 . w = \cos\frac{3\pi}{10} + i\sin\frac{3\pi}{10}. w = cos 10 3 π + i sin 10 3 π . 中文翻译 题目：0 < θ < π 0 < \theta < \pi 0 < θ < π z = 1 + cos θ + i sin θ z = 1 + \cos\theta + i\sin\theta z = 1 + cos θ + i sin θ w = sin π 5 + i cos π 5 w = \sin\dfrac{\pi}{5} + i\cos\dfrac{\pi}{5} w = sin 5 π + i cos 5 π 0 ≤ arg < 2 π 0 \le \arg < 2\pi 0 ≤ arg < 2 π z = 2 cos θ 2 ( cos θ 2 + i sin θ 2 ) z = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right) z = 2 cos 2 θ ( cos 2 θ + i sin 2 θ ) 0 < θ < π 0 < \theta < \pi 0 < θ < π 0 < θ 2 < π 2 0 < \dfrac{\theta}{2} < \dfrac{\pi}{2} 0 < 2 θ < 2 π cos θ 2 > 0 \cos\dfrac{\theta}{2} > 0 cos 2 θ > 0 2 cos θ 2 2\cos\dfrac{\theta}{2} 2 cos 2 θ θ 2 \dfrac{\theta}{2} 2 θ w w w sin π 5 = cos ( π 2 − π 5 ) = cos 3 π 10 \sin\dfrac{\pi}{5} = \cos\left(\dfrac{\pi}{2} - \dfrac{\pi}{5}\right) = \cos\dfrac{3\pi}{10} sin 5 π = cos ( 2 π − 5 π ) = cos 10 3 π cos π 5 = sin 3 π 10 \cos\dfrac{\pi}{5} = \sin\dfrac{3\pi}{10} cos 5 π = sin 10 3 π w = cos 3 π 10 + i sin 3 π 10 w = \cos\frac{3\pi}{10} + i\sin\frac{3\pi}{10} w = cos 10 3 π + i sin 10 3 π
解说 容易卡在：w w w π 5 \dfrac{\pi}{5} 5 π π 2 \dfrac{\pi}{2} 2 π z z z 2 cos θ 2 2\cos\dfrac{\theta}{2} 2 cos 2 θ θ \theta θ 0 < θ 2 < π 2 0 < \dfrac{\theta}{2} < \dfrac{\pi}{2} 0 < 2 θ < 2 π cos θ 2 > 0 \cos\dfrac{\theta}{2} > 0 cos 2 θ > 0 r > 0 r > 0 r > 0
例 2 （场景：用积求非特殊角的三角函数值）
問題 ( 3 + i ) ( 1 + i ) (\sqrt{3} + i)(1 + i) ( 3 + i ) ( 1 + i ) cos 5 π 12 \cos\dfrac{5\pi}{12} cos 12 5 π sin 5 π 12 \sin\dfrac{5\pi}{12} sin 12 5 π
解答 代数形式で計算すると
( 3 + i ) ( 1 + i ) = 3 + 3 i + i + i 2 = ( 3 − 1 ) + ( 3 + 1 ) i . (\sqrt{3} + i)(1 + i) = \sqrt{3} + \sqrt{3}i + i + i^2 = (\sqrt{3} - 1) + (\sqrt{3} + 1)i. ( 3 + i ) ( 1 + i ) = 3 + 3 i + i + i 2 = ( 3 − 1 ) + ( 3 + 1 ) i . 極形式では，3 + i = 2 ( cos π 6 + i sin π 6 ) \sqrt{3} + i = 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) 3 + i = 2 ( cos 6 π + i sin 6 π ) 1 + i = 2 ( cos π 4 + i sin π 4 ) 1 + i = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 1 + i = 2 ( cos 4 π + i sin 4 π )
2 2 ( cos 5 π 12 + i sin 5 π 12 ) . 2\sqrt{2}\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right). 2 2 ( cos 12 5 π + i sin 12 5 π ) . 2 2 2
cos 5 π 12 = 3 − 1 2 2 = 6 − 2 4 , sin 5 π 12 = 3 + 1 2 2 = 6 + 2 4 . \cos\frac{5\pi}{12} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}, \qquad \sin\frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}. cos 12 5 π = 2 2 3 − 1 = 4 6 − 2 , sin 12 5 π = 2 2 3 + 1 = 4 6 + 2 . 中文翻译 题目：计算 ( 3 + i ) ( 1 + i ) (\sqrt{3} + i)(1 + i) ( 3 + i ) ( 1 + i ) cos 5 π 12 \cos\dfrac{5\pi}{12} cos 12 5 π sin 5 π 12 \sin\dfrac{5\pi}{12} sin 12 5 π ( 3 + i ) ( 1 + i ) = 3 + 3 i + i + i 2 = ( 3 − 1 ) + ( 3 + 1 ) i (\sqrt{3} + i)(1 + i) = \sqrt{3} + \sqrt{3}i + i + i^2 = (\sqrt{3} - 1) + (\sqrt{3} + 1)i ( 3 + i ) ( 1 + i ) = 3 + 3 i + i + i 2 = ( 3 − 1 ) + ( 3 + 1 ) i 3 + i = 2 ( cos π 6 + i sin π 6 ) \sqrt{3} + i = 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) 3 + i = 2 ( cos 6 π + i sin 6 π ) 1 + i = 2 ( cos π 4 + i sin π 4 ) 1 + i = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 1 + i = 2 ( cos 4 π + i sin 4 π ) 2 2 ( cos 5 π 12 + i sin 5 π 12 ) 2\sqrt{2}\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right) 2 2 ( cos 12 5 π + i sin 12 5 π ) cos 5 π 12 = 3 − 1 2 2 = 6 − 2 4 \cos\frac{5\pi}{12} = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4} cos 12 5 π = 2 2 3 − 1 = 4 6 − 2 sin 5 π 12 = 3 + 1 2 2 = 6 + 2 4 \sin\frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4} sin 12 5 π = 2 2 3 + 1 = 4 6 + 2
解说 最后一步分母有理化，EJU 填空时必须这样写：3 − 1 2 2 = ( 3 − 1 ) 2 4 = 6 − 2 4 \dfrac{\sqrt{3} - 1}{2\sqrt{2}} = \dfrac{(\sqrt{3} - 1)\sqrt{2}}{4} = \dfrac{\sqrt{6} - \sqrt{2}}{4} 2 2 3 − 1 = 4 ( 3 − 1 ) 2 = 4 6 − 2
例 3 （场景：求积商的偏角和模）
問題 α = 1 + i \alpha = 1 + i α = 1 + i β = 1 − 3 i \beta = 1 - \sqrt{3}i β = 1 − 3 i α 2 β \dfrac{\alpha^2}{\beta} β α 2 θ \theta θ 0 ≦ θ < 2 π 0 \leqq \theta < 2\pi 0 ≦ θ < 2 π a + b i a + bi a + bi
解答 α = 2 ( cos π 4 + i sin π 4 ) \alpha = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) α = 2 ( cos 4 π + i sin 4 π ) β \beta β 2 2 2 ( 1 , − 3 ) (1, -\sqrt{3}) ( 1 , − 3 ) 4 4 4 β = 2 ( cos 5 π 3 + i sin 5 π 3 ) \beta = 2\left(\cos\dfrac{5\pi}{3} + i\sin\dfrac{5\pi}{3}\right) β = 2 ( cos 3 5 π + i sin 3 5 π )
∣ α 2 β ∣ = ( 2 ) 2 2 = 1 , arg α 2 β = 2 ⋅ π 4 − 5 π 3 = − 7 π 6 . \left\lvert \frac{\alpha^2}{\beta} \right\rvert = \frac{(\sqrt{2})^2}{2} = 1, \qquad \arg\frac{\alpha^2}{\beta} = 2 \cdot \frac{\pi}{4} - \frac{5\pi}{3} = -\frac{7\pi}{6}. β α 2 = 2 ( 2 ) 2 = 1 , arg β α 2 = 2 ⋅ 4 π − 3 5 π = − 6 7 π . 2 π 2\pi 2 π θ = 5 π 6 \theta = \dfrac{5\pi}{6} θ = 6 5 π
α 2 β = cos 5 π 6 + i sin 5 π 6 = − 3 2 + 1 2 i . \frac{\alpha^2}{\beta} = \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i. β α 2 = cos 6 5 π + i sin 6 5 π = − 2 3 + 2 1 i . 中文翻译 题目：α = 1 + i \alpha = 1 + i α = 1 + i β = 1 − 3 i \beta = 1 - \sqrt{3}i β = 1 − 3 i α 2 β \dfrac{\alpha^2}{\beta} β α 2 θ \theta θ 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π a + b i a + bi a + bi α = 2 ( cos π 4 + i sin π 4 ) \alpha = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) α = 2 ( cos 4 π + i sin 4 π ) β \beta β ( 1 , − 3 ) (1, -\sqrt{3}) ( 1 , − 3 ) β = 2 ( cos 5 π 3 + i sin 5 π 3 ) \beta = 2\left(\cos\dfrac{5\pi}{3} + i\sin\dfrac{5\pi}{3}\right) β = 2 ( cos 3 5 π + i sin 3 5 π ) ∣ α 2 β ∣ = ( 2 ) 2 2 = 1 \left\lvert \frac{\alpha^2}{\beta} \right\rvert = \frac{(\sqrt{2})^2}{2} = 1 β α 2 = 2 ( 2 ) 2 = 1 arg α 2 β = 2 ⋅ π 4 − 5 π 3 = − 7 π 6 \arg\frac{\alpha^2}{\beta} = 2 \cdot \frac{\pi}{4} - \frac{5\pi}{3} = -\frac{7\pi}{6} arg β α 2 = 2 ⋅ 4 π − 3 5 π = − 6 7 π 2 π 2\pi 2 π θ = 5 π 6 \theta = \dfrac{5\pi}{6} θ = 6 5 π α 2 β = cos 5 π 6 + i sin 5 π 6 = − 3 2 + 1 2 i \frac{\alpha^2}{\beta} = \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i β α 2 = cos 6 5 π + i sin 6 5 π = − 2 3 + 2 1 i
解说 另一种解法 ：直接算。α 2 = 2 i \alpha^2 = 2i α 2 = 2 i 2 i 1 − 3 i = 2 i ( 1 + 3 i ) 1 + 3 = 2 i − 2 3 4 = − 3 2 + 1 2 i \frac{2i}{1 - \sqrt{3}i} = \frac{2i(1 + \sqrt{3}i)}{1 + 3} = \frac{2i - 2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i 1 − 3 i 2 i = 1 + 3 2 i ( 1 + 3 i ) = 4 2 i − 2 3 = − 2 3 + 2 1 i 5 π 6 \dfrac{5\pi}{6} 6 5 π − 7 π 6 -\dfrac{7\pi}{6} − 6 7 π 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π 2 π 2\pi 2 π
例 4 （场景：绕原点旋转并放大）
問題 点 2 + 4 i 2 + 4i 2 + 4 i π 4 \dfrac{\pi}{4} 4 π 2 \sqrt{2} 2
解答 2 ( cos π 4 + i sin π 4 ) = 1 + i \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) = 1 + i 2 ( cos 4 π + i sin 4 π ) = 1 + i
( 2 + 4 i ) ( 1 + i ) = 2 + 2 i + 4 i + 4 i 2 = − 2 + 6 i . (2 + 4i)(1 + i) = 2 + 2i + 4i + 4i^2 = -2 + 6i. ( 2 + 4 i ) ( 1 + i ) = 2 + 2 i + 4 i + 4 i 2 = − 2 + 6 i . 中文翻译 题目：把点 2 + 4 i 2 + 4i 2 + 4 i π 4 \dfrac{\pi}{4} 4 π 2 \sqrt{2} 2 2 ( cos π 4 + i sin π 4 ) = 1 + i \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) = 1 + i 2 ( cos 4 π + i sin 4 π ) = 1 + i ( 2 + 4 i ) ( 1 + i ) = 2 + 2 i + 4 i + 4 i 2 = − 2 + 6 i (2 + 4i)(1 + i) = 2 + 2i + 4i + 4i^2 = -2 + 6i ( 2 + 4 i ) ( 1 + i ) = 2 + 2 i + 4 i + 4 i 2 = − 2 + 6 i
解说 “旋转 π 4 \dfrac{\pi}{4} 4 π 2 \sqrt{2} 2 2 ( cos π 4 + i sin π 4 ) = 1 + i \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) = 1 + i 2 ( cos 4 π + i sin 4 π ) = 1 + i cos ( − π 4 ) + i sin ( − π 4 ) \cos\left(-\dfrac{\pi}{4}\right) + i\sin\left(-\dfrac{\pi}{4}\right) cos ( − 4 π ) + i sin ( − 4 π ) 2 \sqrt{2} 2
z ‾ = r { cos ( − θ ) + i sin ( − θ ) } \overline{z} = r\left\{\cos(-\theta) + i\sin(-\theta)\right\} z = r { cos ( − θ ) + i sin ( − θ ) } 1 z = 1 r { cos ( − θ ) + i sin ( − θ ) } \dfrac{1}{z} = \dfrac{1}{r}\left\{\cos(-\theta) + i\sin(-\theta)\right\} z 1 = r 1 { cos ( − θ ) + i sin ( − θ ) } arg z ‾ = arg 1 z = − arg z \arg\overline{z} = \arg\dfrac{1}{z} = -\arg z arg z = arg z 1 = − arg z 乘 i i i π 2 \dfrac{\pi}{2} 2 π − i -i − i π 2 \dfrac{\pi}{2} 2 π − 1 -1 − 1 π \pi π i = cos π 2 + i sin π 2 i = \cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2} i = cos 2 π + i sin 2 π
乘 1 + i 1 + i 1 + i π 4 \dfrac{\pi}{4} 4 π 2 \sqrt{2} 2 1 + 3 i 2 \dfrac{1 + \sqrt{3}i}{2} 2 1 + 3 i π 3 \dfrac{\pi}{3} 3 π 3 + i 2 \dfrac{\sqrt{3} + i}{2} 2 3 + i π 6 \dfrac{\pi}{6} 6 π
arg β α \arg\dfrac{\beta}{\alpha} arg α β 0 α 0\alpha 0 α 0 β 0\beta 0 β
错在哪 ：说 − 2 ( cos π 6 + i sin π 6 ) -2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) − 2 ( cos 6 π + i sin 6 π ) − 2 -2 − 2 π 6 \dfrac{\pi}{6} 6 π 为什么错 ：极形式要求 r > 0 r > 0 r > 0 cos π + i sin π \cos\pi + i\sin\pi cos π + i sin π 正确做法 ：写成 2 ( cos 7 π 6 + i sin 7 π 6 ) 2\left(\cos\dfrac{7\pi}{6} + i\sin\dfrac{7\pi}{6}\right) 2 ( cos 6 7 π + i sin 6 7 π ) 7 π 6 \dfrac{7\pi}{6} 6 7 π 错在哪 ：例 3 写偏角 − 7 π 6 -\dfrac{7\pi}{6} − 6 7 π 为什么错 ：题目要求 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π − 7 π 6 -\dfrac{7\pi}{6} − 6 7 π 正确做法 ：偏角相加减之后，加减 2 π 2\pi 2 π 错在哪 ：sin α + i cos α \sin\alpha + i\cos\alpha sin α + i cos α α \alpha α 为什么错 ：极形式是余弦在前；正弦在前时实部是 sin α \sin\alpha sin α π 2 − α \dfrac{\pi}{2} - \alpha 2 π − α 正确做法 ：先用 sin α = cos ( π 2 − α ) \sin\alpha = \cos\left(\dfrac{\pi}{2} - \alpha\right) sin α = cos ( 2 π − α ) cos α = sin ( π 2 − α ) \cos\alpha = \sin\left(\dfrac{\pi}{2} - \alpha\right) cos α = sin ( 2 π − α )
极形式把复数说成“多长、朝哪”。乘法就是“长度相乘、方向相加”：乘一个模为 1 的数是纯旋转，乘一个正实数是纯放大。所以乘法在复平面上是一个“转一下再拉一下”的动作，这是复数能用来解几何题的根本原因。
例 2 的方法也说明了两角和公式从哪来：两个复数相乘时，实部是”cos cos − sin sin \cos\cos - \sin\sin cos cos − sin sin sin cos + cos sin \sin\cos + \cos\sin sin cos + cos sin
前置：国内书第 17 章 17.5 节，三角形式和积的公式。
前置：国内书第 13 章 13.2、13.3 节，两角和公式和二倍角公式。
后续：17.4 节，积的公式反复用就是棣莫弗定理。
后续：17.6 节，绕任意点的旋转和两直线的夹角。
本书第 18 章 式と曲線：极坐标 ( r , θ ) (r, \theta) ( r , θ )
ド・モアブルの定理 （棣莫弗定理）：对任意整数 n n n
( cos θ + i sin θ ) n = cos n θ + i sin n θ . (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta. ( cos θ + i sin θ ) n = cos n θ + i sin n θ . 于是 z = r ( cos θ + i sin θ ) z = r(\cos\theta + i\sin\theta) z = r ( cos θ + i sin θ ) n n n
z n = r n ( cos n θ + i sin n θ ) . z^n = r^n(\cos n\theta + i\sin n\theta). z n = r n ( cos n θ + i sin n θ ) . 模变成 n n n n n n n n n z 0 = 1 z^0 = 1 z 0 = 1 z − n = 1 z n z^{-n} = \dfrac{1}{z^n} z − n = z n 1
国内书第 17 章 17.5 节只把它当推论提了一句（n n n z n z^n z n z n = a z^n = a z n = a
题目要求 ( 1 + 3 i ) 7 (1 + \sqrt{3}i)^7 ( 1 + 3 i ) 7 ( 1 − i ) − 6 (1 - i)^{-6} ( 1 − i ) − 6 n n n n n n
题目问“使 z n z^n z n n n n n θ n\theta n θ π \pi π π 2 \dfrac{\pi}{2} 2 π 2 π 2\pi 2 π
题目给出 z + 1 z = 3 z + \dfrac{1}{z} = \sqrt{3} z + z 1 = 3 z n + 1 z n z^n + \dfrac{1}{z^n} z n + z n 1 z = cos θ + i sin θ z = \cos\theta + i\sin\theta z = cos θ + i sin θ z n + z − n = 2 cos n θ z^n + z^{-n} = 2\cos n\theta z n + z − n = 2 cos n θ
定理只对“标准极形式”成立。( sin θ + i cos θ ) n (\sin\theta + i\cos\theta)^n ( sin θ + i cos θ ) n sin n θ + i cos n θ \sin n\theta + i\cos n\theta sin n θ + i cos n θ cos ( π 2 − θ ) + i sin ( π 2 − θ ) \cos\left(\dfrac{\pi}{2} - \theta\right) + i\sin\left(\dfrac{\pi}{2} - \theta\right) cos ( 2 π − θ ) + i sin ( 2 π − θ )
模也要 n n n ( 1 + 3 i ) 7 (1 + \sqrt{3}i)^7 ( 1 + 3 i ) 7 2 7 = 128 2^7 = 128 2 7 = 128
偏角乘 n n n 2 π 2\pi 2 π 2 π 2\pi 2 π
EJU 题干：「ド・モアブルの定理を用いると」「z n z^n z n n n n z n z^n z n
n n n n = 1 n = 1 n = 1 n = k n = k n = k
( cos θ + i sin θ ) k + 1 = ( cos k θ + i sin k θ ) ( cos θ + i sin θ ) = cos ( k + 1 ) θ + i sin ( k + 1 ) θ . (\cos\theta + i\sin\theta)^{k + 1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) = \cos(k + 1)\theta + i\sin(k + 1)\theta. ( cos θ + i sin θ ) k + 1 = ( cos k θ + i sin k θ ) ( cos θ + i sin θ ) = cos ( k + 1 ) θ + i sin ( k + 1 ) θ . （关键一步 ：第二个等号用的是 17.3 节积的公式，模 1 × 1 = 1 1 \times 1 = 1 1 × 1 = 1 k θ + θ k\theta + \theta k θ + θ n = k + 1 n = k + 1 n = k + 1
n = 0 n = 0 n = 0
n n n n = − m n = -m n = − m m m m
( cos θ + i sin θ ) − m = 1 cos m θ + i sin m θ = cos ( − m θ ) + i sin ( − m θ ) = cos n θ + i sin n θ . (\cos\theta + i\sin\theta)^{-m} = \frac{1}{\cos m\theta + i\sin m\theta} = \cos(-m\theta) + i\sin(-m\theta) = \cos n\theta + i\sin n\theta. ( cos θ + i sin θ ) − m = cos m θ + i sin m θ 1 = cos ( − m θ ) + i sin ( − m θ ) = cos n θ + i sin n θ . 例 1 （场景：高次幂、负次幂）
問題 ( 1 + 3 i ) 7 (1 + \sqrt{3}i)^7 ( 1 + 3 i ) 7 ( 1 − i ) − 6 (1 - i)^{-6} ( 1 − i ) − 6
解答 1 + 3 i = 2 ( cos π 3 + i sin π 3 ) 1 + \sqrt{3}i = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right) 1 + 3 i = 2 ( cos 3 π + i sin 3 π )
( 1 + 3 i ) 7 = 2 7 ( cos 7 π 3 + i sin 7 π 3 ) = 128 ( cos π 3 + i sin π 3 ) = 64 + 64 3 i . (1 + \sqrt{3}i)^7 = 2^7\left(\cos\frac{7\pi}{3} + i\sin\frac{7\pi}{3}\right) = 128\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 64 + 64\sqrt{3}i. ( 1 + 3 i ) 7 = 2 7 ( cos 3 7 π + i sin 3 7 π ) = 128 ( cos 3 π + i sin 3 π ) = 64 + 64 3 i . ここで 7 π 3 = 2 π + π 3 \dfrac{7\pi}{3} = 2\pi + \dfrac{\pi}{3} 3 7 π = 2 π + 3 π 1 − i = 2 { cos ( − π 4 ) + i sin ( − π 4 ) } 1 - i = \sqrt{2}\left\{\cos\left(-\dfrac{\pi}{4}\right) + i\sin\left(-\dfrac{\pi}{4}\right)\right\} 1 − i = 2 { cos ( − 4 π ) + i sin ( − 4 π ) }
( 1 − i ) − 6 = ( 2 ) − 6 ( cos 3 π 2 + i sin 3 π 2 ) = 1 8 ( 0 − i ) = − 1 8 i . (1 - i)^{-6} = (\sqrt{2})^{-6}\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = \frac{1}{8}(0 - i) = -\frac{1}{8}i. ( 1 − i ) − 6 = ( 2 ) − 6 ( cos 2 3 π + i sin 2 3 π ) = 8 1 ( 0 − i ) = − 8 1 i . 中文翻译 题目：计算 ( 1 + 3 i ) 7 (1 + \sqrt{3}i)^7 ( 1 + 3 i ) 7 ( 1 − i ) − 6 (1 - i)^{-6} ( 1 − i ) − 6 1 + 3 i = 2 ( cos π 3 + i sin π 3 ) 1 + \sqrt{3}i = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right) 1 + 3 i = 2 ( cos 3 π + i sin 3 π ) ( 1 + 3 i ) 7 = 2 7 ( cos 7 π 3 + i sin 7 π 3 ) = 128 ( cos π 3 + i sin π 3 ) = 64 + 64 3 i (1 + \sqrt{3}i)^7 = 2^7\left(\cos\frac{7\pi}{3} + i\sin\frac{7\pi}{3}\right) = 128\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 64 + 64\sqrt{3}i ( 1 + 3 i ) 7 = 2 7 ( cos 3 7 π + i sin 3 7 π ) = 128 ( cos 3 π + i sin 3 π ) = 64 + 64 3 i 7 π 3 = 2 π + π 3 \dfrac{7\pi}{3} = 2\pi + \dfrac{\pi}{3} 3 7 π = 2 π + 3 π 1 − i = 2 { cos ( − π 4 ) + i sin ( − π 4 ) } 1 - i = \sqrt{2}\left\{\cos\left(-\dfrac{\pi}{4}\right) + i\sin\left(-\dfrac{\pi}{4}\right)\right\} 1 − i = 2 { cos ( − 4 π ) + i sin ( − 4 π ) } ( 1 − i ) − 6 = ( 2 ) − 6 ( cos 3 π 2 + i sin 3 π 2 ) = 1 8 ( 0 − i ) = − 1 8 i (1 - i)^{-6} = (\sqrt{2})^{-6}\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = \frac{1}{8}(0 - i) = -\frac{1}{8}i ( 1 − i ) − 6 = ( 2 ) − 6 ( cos 2 3 π + i sin 2 3 π ) = 8 1 ( 0 − i ) = − 8 1 i
解说 另一种解法 （第二问）：( 1 − i ) 2 = 1 − 2 i + i 2 = − 2 i (1 - i)^2 = 1 - 2i + i^2 = -2i ( 1 − i ) 2 = 1 − 2 i + i 2 = − 2 i ( 1 − i ) 6 = ( − 2 i ) 3 = − 8 i 3 = 8 i (1 - i)^6 = (-2i)^3 = -8i^3 = 8i ( 1 − i ) 6 = ( − 2 i ) 3 = − 8 i 3 = 8 i ( 1 − i ) − 6 = 1 8 i = − i 8 (1 - i)^{-6} = \dfrac{1}{8i} = -\dfrac{i}{8} ( 1 − i ) − 6 = 8 i 1 = − 8 i π 3 \dfrac{\pi}{3} 3 π π 6 \dfrac{\pi}{6} 6 π n n n ( 1 + 3 i ) 7 (1 + \sqrt{3}i)^7 ( 1 + 3 i ) 7 2 7 = 128 2^7 = 128 2 7 = 128
例 2 （场景：z n z^n z n n n n
問題 z = 3 + i 1 + i z = \dfrac{\sqrt{3} + i}{1 + i} z = 1 + i 3 + i z n z^n z n n n n z n z^n z n z n z^n z n n n n z n z^n z n
解答 分子は 2 ( cos π 6 + i sin π 6 ) 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) 2 ( cos 6 π + i sin 6 π ) 2 ( cos π 4 + i sin π 4 ) \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 2 ( cos 4 π + i sin 4 π )
z = 2 { cos ( − π 12 ) + i sin ( − π 12 ) } , z n = ( 2 ) n { cos ( − n π 12 ) + i sin ( − n π 12 ) } . z = \sqrt{2}\left\{\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right\}, \qquad z^n = (\sqrt{2})^n\left\{\cos\left(-\frac{n\pi}{12}\right) + i\sin\left(-\frac{n\pi}{12}\right)\right\}. z = 2 { cos ( − 12 π ) + i sin ( − 12 π ) } , z n = ( 2 ) n { cos ( − 12 nπ ) + i sin ( − 12 nπ ) } .
実数となるのは sin n π 12 = 0 \sin\dfrac{n\pi}{12} = 0 sin 12 nπ = 0 n π 12 \dfrac{n\pi}{12} 12 nπ π \pi π n n n 12 12 12 z 12 = ( 2 ) 12 ( cos ( − π ) + i sin ( − π ) ) = 64 × ( − 1 ) = − 64 z^{12} = (\sqrt{2})^{12}\left(\cos(-\pi) + i\sin(-\pi)\right) = 64 \times (-1) = -64 z 12 = ( 2 ) 12 ( cos ( − π ) + i sin ( − π ) ) = 64 × ( − 1 ) = − 64
純虚数となるのは cos n π 12 = 0 \cos\dfrac{n\pi}{12} = 0 cos 12 nπ = 0 n π 12 = π 2 + k π \dfrac{n\pi}{12} = \dfrac{\pi}{2} + k\pi 12 nπ = 2 π + k π n n n 6 6 6 z 6 = 8 { cos ( − π 2 ) + i sin ( − π 2 ) } = − 8 i z^6 = 8\left\{\cos\left(-\dfrac{\pi}{2}\right) + i\sin\left(-\dfrac{\pi}{2}\right)\right\} = -8i z 6 = 8 { cos ( − 2 π ) + i sin ( − 2 π ) } = − 8 i
中文翻译 题目：z = 3 + i 1 + i z = \dfrac{\sqrt{3} + i}{1 + i} z = 1 + i 3 + i z n z^n z n n n n z n z^n z n z n z^n z n n n n z n z^n z n 2 ( cos π 6 + i sin π 6 ) 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) 2 ( cos 6 π + i sin 6 π ) 2 ( cos π 4 + i sin π 4 ) \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 2 ( cos 4 π + i sin 4 π ) z = 2 { cos ( − π 12 ) + i sin ( − π 12 ) } z = \sqrt{2}\left\{\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right\} z = 2 { cos ( − 12 π ) + i sin ( − 12 π ) } z n = ( 2 ) n { cos ( − n π 12 ) + i sin ( − n π 12 ) } z^n = (\sqrt{2})^n\left\{\cos\left(-\frac{n\pi}{12}\right) + i\sin\left(-\frac{n\pi}{12}\right)\right\} z n = ( 2 ) n { cos ( − 12 nπ ) + i sin ( − 12 nπ ) } sin n π 12 = 0 \sin\dfrac{n\pi}{12} = 0 sin 12 nπ = 0 n π 12 \dfrac{n\pi}{12} 12 nπ π \pi π n = 12 n = 12 n = 12 z 12 = ( 2 ) 12 ( cos ( − π ) + i sin ( − π ) ) = 64 × ( − 1 ) = − 64 z^{12} = (\sqrt{2})^{12}\left(\cos(-\pi) + i\sin(-\pi)\right) = 64 \times (-1) = -64 z 12 = ( 2 ) 12 ( cos ( − π ) + i sin ( − π ) ) = 64 × ( − 1 ) = − 64 cos n π 12 = 0 \cos\dfrac{n\pi}{12} = 0 cos 12 nπ = 0 n π 12 = π 2 + k π \dfrac{n\pi}{12} = \dfrac{\pi}{2} + k\pi 12 nπ = 2 π + k π n = 6 n = 6 n = 6 z 6 = 8 { cos ( − π 2 ) + i sin ( − π 2 ) } = − 8 i z^6 = 8\left\{\cos\left(-\dfrac{\pi}{2}\right) + i\sin\left(-\dfrac{\pi}{2}\right)\right\} = -8i z 6 = 8 { cos ( − 2 π ) + i sin ( − 2 π ) } = − 8 i
解说 容易卡在：直接展开 z z z ( 3 + 1 ) + ( 1 − 3 ) i 2 \dfrac{(\sqrt{3} + 1) + (1 - \sqrt{3})i}{2} 2 ( 3 + 1 ) + ( 1 − 3 ) i z n z^n z n n π 12 = π \dfrac{n\pi}{12} = \pi 12 nπ = π z 12 = − 64 z^{12} = -64 z 12 = − 64 π \pi π 2 π 2\pi 2 π
例 3 （场景：z + 1 z z + \dfrac{1}{z} z + z 1
問題 複素数 z z z z + 1 z = 3 z + \dfrac{1}{z} = \sqrt{3} z + z 1 = 3 z 10 + 1 z 10 z^{10} + \dfrac{1}{z^{10}} z 10 + z 10 1
解答 両辺に z z z z 2 − 3 z + 1 = 0 z^2 - \sqrt{3}z + 1 = 0 z 2 − 3 z + 1 = 0 z = 3 ± i 2 = cos ( ± π 6 ) + i sin ( ± π 6 ) z = \dfrac{\sqrt{3} \pm i}{2} = \cos\left(\pm\dfrac{\pi}{6}\right) + i\sin\left(\pm\dfrac{\pi}{6}\right) z = 2 3 ± i = cos ( ± 6 π ) + i sin ( ± 6 π ) θ = ± π 6 \theta = \pm\dfrac{\pi}{6} θ = ± 6 π
z 10 + z − 10 = ( cos 10 θ + i sin 10 θ ) + ( cos 10 θ − i sin 10 θ ) = 2 cos 10 θ = 2 cos 5 π 3 = 1. z^{10} + z^{-10} = (\cos 10\theta + i\sin 10\theta) + (\cos 10\theta - i\sin 10\theta) = 2\cos 10\theta = 2\cos\frac{5\pi}{3} = 1. z 10 + z − 10 = ( cos 10 θ + i sin 10 θ ) + ( cos 10 θ − i sin 10 θ ) = 2 cos 10 θ = 2 cos 3 5 π = 1. 2 2 2 z z z cos \cos cos
中文翻译 题目：复数 z z z z + 1 z = 3 z + \dfrac{1}{z} = \sqrt{3} z + z 1 = 3 z 10 + 1 z 10 z^{10} + \dfrac{1}{z^{10}} z 10 + z 10 1 z z z z 2 − 3 z + 1 = 0 z^2 - \sqrt{3}z + 1 = 0 z 2 − 3 z + 1 = 0 z = 3 ± i 2 = cos ( ± π 6 ) + i sin ( ± π 6 ) z = \dfrac{\sqrt{3} \pm i}{2} = \cos\left(\pm\dfrac{\pi}{6}\right) + i\sin\left(\pm\dfrac{\pi}{6}\right) z = 2 3 ± i = cos ( ± 6 π ) + i sin ( ± 6 π ) θ = ± π 6 \theta = \pm\dfrac{\pi}{6} θ = ± 6 π z 10 + z − 10 = ( cos 10 θ + i sin 10 θ ) + ( cos 10 θ − i sin 10 θ ) = 2 cos 10 θ = 2 cos 5 π 3 = 1 z^{10} + z^{-10} = (\cos 10\theta + i\sin 10\theta) + (\cos 10\theta - i\sin 10\theta) = 2\cos 10\theta = 2\cos\frac{5\pi}{3} = 1 z 10 + z − 10 = ( cos 10 θ + i sin 10 θ ) + ( cos 10 θ − i sin 10 θ ) = 2 cos 10 θ = 2 cos 3 5 π = 1 z z z cos \cos cos
解说 z + 1 z = 3 z + \dfrac{1}{z} = \sqrt{3} z + z 1 = 3 z z z z z z z n + z − n = 2 cos n θ z^n + z^{-n} = 2\cos n\theta z n + z − n = 2 cos n θ z z z ± π 6 \pm\dfrac{\pi}{6} ± 6 π 2 cos 10 θ 2\cos 10\theta 2 cos 10 θ cos 10 θ = cos 5 π 3 = 1 2 \cos 10\theta = \cos\dfrac{5\pi}{3} = \dfrac{1}{2} cos 10 θ = cos 3 5 π = 2 1
3 倍角の公式 ：cos 3 θ = 4 cos 3 θ − 3 cos θ \cos 3\theta = 4\cos^3\theta - 3\cos\theta cos 3 θ = 4 cos 3 θ − 3 cos θ sin 3 θ = 3 sin θ − 4 sin 3 θ \sin 3\theta = 3\sin\theta - 4\sin^3\theta sin 3 θ = 3 sin θ − 4 sin 3 θ ( cos θ + i sin θ ) 3 (\cos\theta + i\sin\theta)^3 ( cos θ + i sin θ ) 3 cos 3 θ − 3 cos θ sin 2 θ \cos^3\theta - 3\cos\theta\sin^2\theta cos 3 θ − 3 cos θ sin 2 θ 3 cos 2 θ sin θ − sin 3 θ 3\cos^2\theta\sin\theta - \sin^3\theta 3 cos 2 θ sin θ − sin 3 θ cos 3 θ + i sin 3 θ \cos 3\theta + i\sin 3\theta cos 3 θ + i sin 3 θ sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 z = cos θ + i sin θ z = \cos\theta + i\sin\theta z = cos θ + i sin θ z n + 1 z n = 2 cos n θ z^n + \dfrac{1}{z^n} = 2\cos n\theta z n + z n 1 = 2 cos n θ z n − 1 z n = 2 i sin n θ z^n - \dfrac{1}{z^n} = 2i\sin n\theta z n − z n 1 = 2 i sin n θ z − n = cos n θ − i sin n θ z^{-n} = \cos n\theta - i\sin n\theta z − n = cos n θ − i sin n θ z n z^n z n z n z^n z n ⟺ n arg z \iff n\arg z ⟺ n arg z π \pi π z n z^n z n ⟺ n arg z \iff n\arg z ⟺ n arg z π 2 \dfrac{\pi}{2} 2 π z n = 1 ⟺ ∣ z ∣ = 1 z^n = 1 \iff \lvert z \rvert = 1 z n = 1 ⟺ ∣ z ∣ = 1 n arg z n\arg z n arg z 2 π 2\pi 2 π sin n θ \sin n\theta sin n θ cos n θ \cos n\theta cos n θ r n r^n r n
错在哪 ：( 1 + 3 i ) 7 = cos 7 π 3 + i sin 7 π 3 (1 + \sqrt{3}i)^7 = \cos\dfrac{7\pi}{3} + i\sin\dfrac{7\pi}{3} ( 1 + 3 i ) 7 = cos 3 7 π + i sin 3 7 π 为什么错 ：漏了模的 7 次方，∣ 1 + 3 i ∣ = 2 \lvert 1 + \sqrt{3}i \rvert = 2 ∣ 1 + 3 i ∣ = 2 正确做法 ：先写出 r r r z n z^n z n r n r^n r n 错在哪 ：( sin θ + i cos θ ) 2 = sin 2 θ + i cos 2 θ (\sin\theta + i\cos\theta)^2 = \sin 2\theta + i\cos 2\theta ( sin θ + i cos θ ) 2 = sin 2 θ + i cos 2 θ 为什么错 ：棣莫弗只对 cos θ + i sin θ \cos\theta + i\sin\theta cos θ + i sin θ θ = 0 \theta = 0 θ = 0 i 2 = − 1 i^2 = -1 i 2 = − 1 0 + i = i 0 + i = i 0 + i = i 正确做法 ：先写成 cos ( π 2 − θ ) + i sin ( π 2 − θ ) \cos\left(\dfrac{\pi}{2} - \theta\right) + i\sin\left(\dfrac{\pi}{2} - \theta\right) cos ( 2 π − θ ) + i sin ( 2 π − θ ) cos ( π − 2 θ ) + i sin ( π − 2 θ ) = − cos 2 θ + i sin 2 θ \cos(\pi - 2\theta) + i\sin(\pi - 2\theta) = -\cos 2\theta + i\sin 2\theta cos ( π − 2 θ ) + i sin ( π − 2 θ ) = − cos 2 θ + i sin 2 θ 错在哪 ：例 2 求”z n z^n z n z n z^n z n n = 24 n = 24 n = 24 为什么错 ：负实数也是实数，n π 12 = π \dfrac{n\pi}{12} = \pi 12 nπ = π z n = − 64 z^n = -64 z n = − 64 正确做法 ：实数的条件是 n θ = k π n\theta = k\pi n θ = k π k k k 2 k π 2k\pi 2 k π
z = cos θ + i sin θ z = \cos\theta + i\sin\theta z = cos θ + i sin θ z z z θ \theta θ z n z^n z n n n n n θ n\theta n θ r r r r > 1 r > 1 r > 1 r < 1 r < 1 r < 1
“z n z^n z n π 12 \dfrac{\pi}{12} 12 π
前置：17.3 节积的公式；国内书第 23 章 23.6 节数学归纳法。
前置：国内书第 34 章 34.5 节，二项式定理（推三倍角公式时展开三次方）。
后续：17.5 节，倒过来用棣莫弗定理解 z n = a z^n = a z n = a
本书第 13 章 三角関数：三倍角公式在 EJU 里常由棣莫弗推出，而不是直接给出。
本书第 14 章 数列：z n z^n z n z z z 1 + z + ⋯ + z n − 1 1 + z + \dots + z^{n - 1} 1 + z + ⋯ + z n − 1
方程式 z n = a z^n = a z n = a （a ≠ 0 a \ne 0 a = 0 n n n
把 a a a a = R ( cos φ + i sin φ ) a = R(\cos\varphi + i\sin\varphi) a = R ( cos φ + i sin φ )
设 z = r ( cos θ + i sin θ ) z = r(\cos\theta + i\sin\theta) z = r ( cos θ + i sin θ ) r > 0 r > 0 r > 0 z n = r n ( cos n θ + i sin n θ ) z^n = r^n(\cos n\theta + i\sin n\theta) z n = r n ( cos n θ + i sin n θ )
比较模：r n = R r^n = R r n = R r r r r = R n r = \sqrt[n]{R} r = n R
比较偏角：n θ = φ + 2 k π n\theta = \varphi + 2k\pi n θ = φ + 2 k π k k k θ = φ n + 2 k π n \theta = \dfrac{\varphi}{n} + \dfrac{2k\pi}{n} θ = n φ + n 2 k π
k = 0 , 1 , … , n − 1 k = 0, 1, \dots, n - 1 k = 0 , 1 , … , n − 1 n n n k = n k = n k = n k = 0 k = 0 k = 0
所以 z n = a z^n = a z n = a 恰好有 n n n ，它们都在以原点为中心、半径 R n \sqrt[n]{R} n R 2 π n \dfrac{2\pi}{n} n 2 π 正 n n n 的顶点。
1 の n n n ：a = 1 a = 1 a = 1 R = 1 R = 1 R = 1 φ = 0 \varphi = 0 φ = 0
z k = cos 2 k π n + i sin 2 k π n ( k = 0 , 1 , … , n − 1 ) . z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} \quad (k = 0, 1, \dots, n - 1). z k = cos n 2 k π + i sin n 2 k π ( k = 0 , 1 , … , n − 1 ) . 记 α = cos 2 π n + i sin 2 π n \alpha = \cos\dfrac{2\pi}{n} + i\sin\dfrac{2\pi}{n} α = cos n 2 π + i sin n 2 π z k = α k z_k = \alpha^k z k = α k n n n 1 , α , α 2 , … , α n − 1 1, \alpha, \alpha^2, \dots, \alpha^{n - 1} 1 , α , α 2 , … , α n − 1 n n n n = 3 n = 3 n = 3 1 , ω , ω 2 1, \omega, \omega^2 1 , ω , ω 2 ω \omega ω
1 の n n n （n ≥ 2 n \ge 2 n ≥ 2 α \alpha α
1 + α + α 2 + ⋯ + α n − 1 = 0 , z n − 1 = ( z − 1 ) ( z n − 1 + z n − 2 + ⋯ + z + 1 ) . 1 + \alpha + \alpha^2 + \dots + \alpha^{n - 1} = 0, \qquad z^n - 1 = (z - 1)(z^{n - 1} + z^{n - 2} + \dots + z + 1). 1 + α + α 2 + ⋯ + α n − 1 = 0 , z n − 1 = ( z − 1 ) ( z n − 1 + z n − 2 + ⋯ + z + 1 ) . α ≠ 1 \alpha \ne 1 α = 1 z n − 1 + ⋯ + z + 1 = 0 z^{n - 1} + \dots + z + 1 = 0 z n − 1 + ⋯ + z + 1 = 0
题目要求解 z 3 = 8 i z^3 = 8i z 3 = 8 i z 4 = − 8 + 8 3 i z^4 = -8 + 8\sqrt{3}i z 4 = − 8 + 8 3 i n n n
题目给出 α = cos 2 π 5 + i sin 2 π 5 \alpha = \cos\dfrac{2\pi}{5} + i\sin\dfrac{2\pi}{5} α = cos 5 2 π + i sin 5 2 π α 5 = 1 \alpha^5 = 1 α 5 = 1 α ≠ 1 \alpha \ne 1 α = 1 1 + α + ⋯ + α 4 1 + \alpha + \dots + \alpha^4 1 + α + ⋯ + α 4 ( 1 − α ) ( 1 − α 2 ) ⋯ (1 - \alpha)(1 - \alpha^2)\cdots ( 1 − α ) ( 1 − α 2 ) ⋯ cos 2 π 5 \cos\dfrac{2\pi}{5} cos 5 2 π α n = 1 \alpha^n = 1 α n = 1 1 + α + ⋯ + α n − 1 = 0 1 + \alpha + \dots + \alpha^{n - 1} = 0 1 + α + ⋯ + α n − 1 = 0
题目问方程的解在复平面上构成什么图形、面积多少：解在一个圆上构成正 n n n n n n
n n n n n n z 3 = 8 z^3 = 8 z 3 = 8 − 1 ± 3 i -1 \pm \sqrt{3}i − 1 ± 3 i 偏角的周期 2 k π 2k\pi 2 k π 除以 n n n 加上：θ = φ + 2 k π n \theta = \dfrac{\varphi + 2k\pi}{n} θ = n φ + 2 k π φ n + 2 k π \dfrac{\varphi}{n} + 2k\pi n φ + 2 k π
模 r r r r n = R r^n = R r n = R n n n − R n -\sqrt[n]{R} − n R
解写成 a + b i a + bi a + bi θ \theta θ z 5 = 1 z^5 = 1 z 5 = 1
EJU 题干：「方程式 z 3 = 8 i z^3 = 8i z 3 = 8 i α \alpha α
结论 1：恰好 n n n 由说明中的步骤，z z z r = R n r = \sqrt[n]{R} r = n R θ = φ + 2 k π n \theta = \dfrac{\varphi + 2k\pi}{n} θ = n φ + 2 k π k k k 关键一步 ：两个复数相等，当且仅当模相等、偏角相差 2 π 2\pi 2 π r n ( cos n θ + i sin n θ ) = R ( cos φ + i sin φ ) r^n(\cos n\theta + i\sin n\theta) = R(\cos\varphi + i\sin\varphi) r n ( cos n θ + i sin n θ ) = R ( cos φ + i sin φ ) r n = R r^n = R r n = R n θ − φ n\theta - \varphi n θ − φ 2 π 2\pi 2 π k = 0 , 1 , … , n − 1 k = 0, 1, \dots, n - 1 k = 0 , 1 , … , n − 1 θ \theta θ 2 π 2\pi 2 π [ φ n , φ n + 2 π ) \left[\dfrac{\varphi}{n}, \dfrac{\varphi}{n} + 2\pi\right) [ n φ , n φ + 2 π ) n n n k k k n n n θ \theta θ 2 π 2\pi 2 π
结论 2：1 + α + ⋯ + α n − 1 = 0 1 + \alpha + \dots + \alpha^{n - 1} = 0 1 + α + ⋯ + α n − 1 = 0 记左边为 S S S α S = α + α 2 + ⋯ + α n \alpha S = \alpha + \alpha^2 + \dots + \alpha^n α S = α + α 2 + ⋯ + α n
( α − 1 ) S = α n − 1 = 0. (\alpha - 1)S = \alpha^n - 1 = 0. ( α − 1 ) S = α n − 1 = 0. α ≠ 1 \alpha \ne 1 α = 1 S = 0 S = 0 S = 0
结论 3：z n − 1 = ( z − 1 ) ( z − α ) ( z − α 2 ) ⋯ ( z − α n − 1 ) z^n - 1 = (z - 1)(z - \alpha)(z - \alpha^2)\cdots(z - \alpha^{n - 1}) z n − 1 = ( z − 1 ) ( z − α ) ( z − α 2 ) ⋯ ( z − α n − 1 ) 左边是 n n n n n n 1 , α , … , α n − 1 1, \alpha, \dots, \alpha^{n - 1} 1 , α , … , α n − 1 z − α k z - \alpha^k z − α k
例 1 （场景：解 z n = a z^n = a z n = a
問題 方程式 z 3 = 8 i z^3 = 8i z 3 = 8 i
解答 8 i = 8 ( cos π 2 + i sin π 2 ) 8i = 8\left(\cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}\right) 8 i = 8 ( cos 2 π + i sin 2 π ) z = r ( cos θ + i sin θ ) z = r(\cos\theta + i\sin\theta) z = r ( cos θ + i sin θ ) r > 0 r > 0 r > 0
r 3 ( cos 3 θ + i sin 3 θ ) = 8 ( cos π 2 + i sin π 2 ) . r^3(\cos 3\theta + i\sin 3\theta) = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right). r 3 ( cos 3 θ + i sin 3 θ ) = 8 ( cos 2 π + i sin 2 π ) . 絶対値について r 3 = 8 r^3 = 8 r 3 = 8 r = 2 r = 2 r = 2 3 θ = π 2 + 2 k π 3\theta = \dfrac{\pi}{2} + 2k\pi 3 θ = 2 π + 2 k π θ = π 6 + 2 k π 3 \theta = \dfrac{\pi}{6} + \dfrac{2k\pi}{3} θ = 6 π + 3 2 k π k = 0 , 1 , 2 k = 0, 1, 2 k = 0 , 1 , 2 θ = π 6 , 5 π 6 , 3 π 2 \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2} θ = 6 π , 6 5 π , 2 3 π
z = 3 + i , − 3 + i , − 2 i . z = \sqrt{3} + i, \quad -\sqrt{3} + i, \quad -2i. z = 3 + i , − 3 + i , − 2 i . 中文翻译 题目：解方程 z 3 = 8 i z^3 = 8i z 3 = 8 i 8 i = 8 ( cos π 2 + i sin π 2 ) 8i = 8\left(\cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}\right) 8 i = 8 ( cos 2 π + i sin 2 π ) z = r ( cos θ + i sin θ ) z = r(\cos\theta + i\sin\theta) z = r ( cos θ + i sin θ ) r > 0 r > 0 r > 0 r 3 ( cos 3 θ + i sin 3 θ ) = 8 ( cos π 2 + i sin π 2 ) r^3(\cos 3\theta + i\sin 3\theta) = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) r 3 ( cos 3 θ + i sin 3 θ ) = 8 ( cos 2 π + i sin 2 π ) r 3 = 8 r^3 = 8 r 3 = 8 r = 2 r = 2 r = 2 3 θ = π 2 + 2 k π 3\theta = \dfrac{\pi}{2} + 2k\pi 3 θ = 2 π + 2 k π θ = π 6 + 2 k π 3 \theta = \dfrac{\pi}{6} + \dfrac{2k\pi}{3} θ = 6 π + 3 2 k π k = 0 , 1 , 2 k = 0, 1, 2 k = 0 , 1 , 2 θ = π 6 , 5 π 6 , 3 π 2 \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2} θ = 6 π , 6 5 π , 2 3 π z = 3 + i z = \sqrt{3} + i z = 3 + i − 3 + i -\sqrt{3} + i − 3 + i − 2 i -2i − 2 i
解说 验算：( − 2 i ) 3 = − 8 i 3 = 8 i (-2i)^3 = -8i^3 = 8i ( − 2 i ) 3 = − 8 i 3 = 8 i 另一种解法 ：先找一个解再旋转。看出 − 2 i -2i − 2 i ω = − 1 2 + 3 2 i \omega = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i ω = − 2 1 + 2 3 i ω 2 \omega^2 ω 2 ( z ω ) 3 = z 3 ω 3 = z 3 (z\omega)^3 = z^3\omega^3 = z^3 ( z ω ) 3 = z 3 ω 3 = z 3 − 2 i ⋅ ω = 3 + i -2i \cdot \omega = \sqrt{3} + i − 2 i ⋅ ω = 3 + i − 2 i ⋅ ω 2 = − 3 + i -2i \cdot \omega^2 = -\sqrt{3} + i − 2 i ⋅ ω 2 = − 3 + i
例 2 （场景：用 1 的 n n n
問題 α = cos 2 π 5 + i sin 2 π 5 \alpha = \cos\dfrac{2\pi}{5} + i\sin\dfrac{2\pi}{5} α = cos 5 2 π + i sin 5 2 π
(1) α 4 + α 3 + α 2 + α + 1 \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1 α 4 + α 3 + α 2 + α + 1
(2) ( 1 − α ) ( 1 − α 2 ) ( 1 − α 3 ) ( 1 − α 4 ) (1 - \alpha)(1 - \alpha^2)(1 - \alpha^3)(1 - \alpha^4) ( 1 − α ) ( 1 − α 2 ) ( 1 − α 3 ) ( 1 − α 4 )
(3) cos 2 π 5 \cos\dfrac{2\pi}{5} cos 5 2 π
解答 (1) α 5 = cos 2 π + i sin 2 π = 1 \alpha^5 = \cos 2\pi + i\sin 2\pi = 1 α 5 = cos 2 π + i sin 2 π = 1 α ≠ 1 \alpha \ne 1 α = 1 0 0 0
(2) 証明で示した結論 3 より z 5 − 1 = ( z − 1 ) ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) z^5 - 1 = (z - 1)(z - \alpha)(z - \alpha^2)(z - \alpha^3)(z - \alpha^4) z 5 − 1 = ( z − 1 ) ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) z − 1 z - 1 z − 1
z 4 + z 3 + z 2 + z + 1 = ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) . z^4 + z^3 + z^2 + z + 1 = (z - \alpha)(z - \alpha^2)(z - \alpha^3)(z - \alpha^4). z 4 + z 3 + z 2 + z + 1 = ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) . z = 1 z = 1 z = 1 1 + 1 + 1 + 1 + 1 = 5 1 + 1 + 1 + 1 + 1 = 5 1 + 1 + 1 + 1 + 1 = 5
(3) (1) の式の両辺を α 2 \alpha^2 α 2
α 2 + α + 1 + 1 α + 1 α 2 = 0. \alpha^2 + \alpha + 1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} = 0. α 2 + α + 1 + α 1 + α 2 1 = 0. t = α + 1 α t = \alpha + \dfrac{1}{\alpha} t = α + α 1 z + 1 z = 2 cos θ z + \dfrac{1}{z} = 2\cos\theta z + z 1 = 2 cos θ t = 2 cos 2 π 5 t = 2\cos\dfrac{2\pi}{5} t = 2 cos 5 2 π α 2 + 1 α 2 = t 2 − 2 \alpha^2 + \dfrac{1}{\alpha^2} = t^2 - 2 α 2 + α 2 1 = t 2 − 2 t 2 + t − 1 = 0 t^2 + t - 1 = 0 t 2 + t − 1 = 0 t = − 1 ± 5 2 t = \dfrac{-1 \pm \sqrt{5}}{2} t = 2 − 1 ± 5 0 < 2 π 5 < π 2 0 < \dfrac{2\pi}{5} < \dfrac{\pi}{2} 0 < 5 2 π < 2 π t > 0 t > 0 t > 0 t = − 1 + 5 2 t = \dfrac{-1 + \sqrt{5}}{2} t = 2 − 1 + 5
cos 2 π 5 = t 2 = − 1 + 5 4 . \cos\frac{2\pi}{5} = \frac{t}{2} = \frac{-1 + \sqrt{5}}{4}. cos 5 2 π = 2 t = 4 − 1 + 5 . 中文翻译 题目：α = cos 2 π 5 + i sin 2 π 5 \alpha = \cos\dfrac{2\pi}{5} + i\sin\dfrac{2\pi}{5} α = cos 5 2 π + i sin 5 2 π α 4 + α 3 + α 2 + α + 1 \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1 α 4 + α 3 + α 2 + α + 1 ( 1 − α ) ( 1 − α 2 ) ( 1 − α 3 ) ( 1 − α 4 ) (1 - \alpha)(1 - \alpha^2)(1 - \alpha^3)(1 - \alpha^4) ( 1 − α ) ( 1 − α 2 ) ( 1 − α 3 ) ( 1 − α 4 ) cos 2 π 5 \cos\dfrac{2\pi}{5} cos 5 2 π α 5 = cos 2 π + i sin 2 π = 1 \alpha^5 = \cos 2\pi + i\sin 2\pi = 1 α 5 = cos 2 π + i sin 2 π = 1 α ≠ 1 \alpha \ne 1 α = 1 z 5 − 1 = ( z − 1 ) ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) z^5 - 1 = (z - 1)(z - \alpha)(z - \alpha^2)(z - \alpha^3)(z - \alpha^4) z 5 − 1 = ( z − 1 ) ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) z − 1 z - 1 z − 1 z 4 + z 3 + z 2 + z + 1 = ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) z^4 + z^3 + z^2 + z + 1 = (z - \alpha)(z - \alpha^2)(z - \alpha^3)(z - \alpha^4) z 4 + z 3 + z 2 + z + 1 = ( z − α ) ( z − α 2 ) ( z − α 3 ) ( z − α 4 ) z = 1 z = 1 z = 1 = 1 + 1 + 1 + 1 + 1 = 5 = 1 + 1 + 1 + 1 + 1 = 5 = 1 + 1 + 1 + 1 + 1 = 5 α 2 \alpha^2 α 2 α 2 + α + 1 + 1 α + 1 α 2 = 0 \alpha^2 + \alpha + 1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} = 0 α 2 + α + 1 + α 1 + α 2 1 = 0 t = α + 1 α t = \alpha + \dfrac{1}{\alpha} t = α + α 1 z + 1 z = 2 cos θ z + \dfrac{1}{z} = 2\cos\theta z + z 1 = 2 cos θ t = 2 cos 2 π 5 t = 2\cos\dfrac{2\pi}{5} t = 2 cos 5 2 π α 2 + 1 α 2 = t 2 − 2 \alpha^2 + \dfrac{1}{\alpha^2} = t^2 - 2 α 2 + α 2 1 = t 2 − 2 t 2 + t − 1 = 0 t^2 + t - 1 = 0 t 2 + t − 1 = 0 t = − 1 ± 5 2 t = \dfrac{-1 \pm \sqrt{5}}{2} t = 2 − 1 ± 5 0 < 2 π 5 < π 2 0 < \dfrac{2\pi}{5} < \dfrac{\pi}{2} 0 < 5 2 π < 2 π t > 0 t > 0 t > 0 t = − 1 + 5 2 t = \dfrac{-1 + \sqrt{5}}{2} t = 2 − 1 + 5 cos 2 π 5 = t 2 = − 1 + 5 4 \cos\frac{2\pi}{5} = \frac{t}{2} = \frac{-1 + \sqrt{5}}{4} cos 5 2 π = 2 t = 4 − 1 + 5
解说 容易卡在：（3）不知道为什么要除以 α 2 \alpha^2 α 2 α + 1 α \alpha + \dfrac{1}{\alpha} α + α 1 2 cos 2 π 5 2\cos\dfrac{2\pi}{5} 2 cos 5 2 π cos 2 π 5 \cos\dfrac{2\pi}{5} cos 5 2 π cos 4 π 5 \cos\dfrac{4\pi}{5} cos 5 4 π α 2 \alpha^2 α 2 2 π 5 \dfrac{2\pi}{5} 5 2 π
例 3 （场景：解构成的图形）
問題 方程式 z 6 = − 64 z^6 = -64 z 6 = − 64
解答 − 64 = 64 ( cos π + i sin π ) -64 = 64(\cos\pi + i\sin\pi) − 64 = 64 ( cos π + i sin π ) r 6 = 64 r^6 = 64 r 6 = 64 r = 2 r = 2 r = 2 6 θ = π + 2 k π 6\theta = \pi + 2k\pi 6 θ = π + 2 k π θ = π 6 + k π 3 \theta = \dfrac{\pi}{6} + \dfrac{k\pi}{3} θ = 6 π + 3 k π k = 0 , 1 , … , 5 k = 0, 1, \dots, 5 k = 0 , 1 , … , 5
z = ± 3 ± i ( 4 個 ) , ± 2 i . z = \pm\sqrt{3} \pm i \ (\text{4 個}), \quad \pm 2i. z = ± 3 ± i ( 4 個 ) , ± 2 i . 6 6 6 2 2 2 π 3 \dfrac{\pi}{3} 3 π 1 1 1 2 2 2 6 6 6
6 × 3 4 × 2 2 = 6 3 . 6 \times \frac{\sqrt{3}}{4} \times 2^2 = 6\sqrt{3}. 6 × 4 3 × 2 2 = 6 3 . 中文翻译 题目：求方程 z 6 = − 64 z^6 = -64 z 6 = − 64 − 64 = 64 ( cos π + i sin π ) -64 = 64(\cos\pi + i\sin\pi) − 64 = 64 ( cos π + i sin π ) r 6 = 64 r^6 = 64 r 6 = 64 r = 2 r = 2 r = 2 6 θ = π + 2 k π 6\theta = \pi + 2k\pi 6 θ = π + 2 k π θ = π 6 + k π 3 \theta = \dfrac{\pi}{6} + \dfrac{k\pi}{3} θ = 6 π + 3 k π k = 0 , 1 , … , 5 k = 0, 1, \dots, 5 k = 0 , 1 , … , 5 z = ± 3 ± i z = \pm\sqrt{3} \pm i z = ± 3 ± i ± 2 i \pm 2i ± 2 i π 3 \dfrac{\pi}{3} 3 π 6 × 3 4 × 2 2 = 6 3 6 \times \frac{\sqrt{3}}{4} \times 2^2 = 6\sqrt{3} 6 × 4 3 × 2 2 = 6 3
解说 z n = a z^n = a z n = a ∣ a ∣ n \sqrt[n]{\lvert a \rvert} n ∣ a ∣ 2 π n \dfrac{2\pi}{n} n 2 π n = 6 n = 6 n = 6 2 2 2 ± 3 ± i \pm\sqrt{3} \pm i ± 3 ± i ± \pm ± ± 2 i \pm 2i ± 2 i
1 の 3 乗根 ：1 , ω , ω 2 1, \omega, \omega^2 1 , ω , ω 2 ω = − 1 + 3 i 2 \omega = \dfrac{-1 + \sqrt{3}i}{2} ω = 2 − 1 + 3 i ω 3 = 1 \omega^3 = 1 ω 3 = 1 ω 2 + ω + 1 = 0 \omega^2 + \omega + 1 = 0 ω 2 + ω + 1 = 0 ω ‾ = ω 2 \overline{\omega} = \omega^2 ω = ω 2 n = 3 n = 3 n = 3 z n = a z^n = a z n = a n n n n ≥ 2 n \ge 2 n ≥ 2 z 0 , z 0 α , … , z 0 α n − 1 z_0, z_0\alpha, \dots, z_0\alpha^{n - 1} z 0 , z 0 α , … , z 0 α n − 1 z 0 z_0 z 0 z 0 ( 1 + α + ⋯ + α n − 1 ) = 0 z_0(1 + \alpha + \dots + \alpha^{n - 1}) = 0 z 0 ( 1 + α + ⋯ + α n − 1 ) = 0 半径 r r r n n n = n 2 r 2 sin 2 π n = \dfrac{n}{2}r^2\sin\dfrac{2\pi}{n} = 2 n r 2 sin n 2 π n n n 2 π n \dfrac{2\pi}{n} n 2 π r r r 1 2 r 2 sin 2 π n \dfrac{1}{2}r^2\sin\dfrac{2\pi}{n} 2 1 r 2 sin n 2 π
α n = 1 \alpha^n = 1 α n = 1 α ‾ = 1 α = α n − 1 \overline{\alpha} = \dfrac{1}{\alpha} = \alpha^{n - 1} α = α 1 = α n − 1 ∣ α ∣ = 1 \lvert \alpha \rvert = 1 ∣ α ∣ = 1 z ‾ = 1 z \overline{z} = \dfrac{1}{z} z = z 1 α ⋅ α n − 1 = 1 \alpha \cdot \alpha^{n - 1} = 1 α ⋅ α n − 1 = 1 α \alpha α n n n α 5 = 1 \alpha^5 = 1 α 5 = 1 α 2026 = α 2025 α = α \alpha^{2026} = \alpha^{2025}\alpha = \alpha α 2026 = α 2025 α = α 2025 = 5 × 405 2025 = 5 \times 405 2025 = 5 × 405
错在哪 ：z 3 = 8 z^3 = 8 z 3 = 8 z = 2 z = 2 z = 2 为什么错 ：那是实数范围的答案。复数范围内三次方程有 3 个解。
正确做法 ：z = 2 , − 1 + 3 i , − 1 − 3 i z = 2, \ -1 + \sqrt{3}i, \ -1 - \sqrt{3}i z = 2 , − 1 + 3 i , − 1 − 3 i 1 , ω , ω 2 1, \omega, \omega^2 1 , ω , ω 2 错在哪 ：例 1 写 θ = π 6 + 2 k π \theta = \dfrac{\pi}{6} + 2k\pi θ = 6 π + 2 k π 为什么错 ：先除以 3 再加 2 k π 2k\pi 2 k π k k k 正确做法 ：3 θ = π 2 + 2 k π 3\theta = \dfrac{\pi}{2} + 2k\pi 3 θ = 2 π + 2 k π θ = π 6 + 2 k π 3 \theta = \dfrac{\pi}{6} + \dfrac{2k\pi}{3} θ = 6 π + 3 2 k π 错在哪 ：例 2（3）取了 t = − 1 − 5 2 t = \dfrac{-1 - \sqrt{5}}{2} t = 2 − 1 − 5 为什么错 ：二次方程的两个根对应 cos 2 π 5 \cos\dfrac{2\pi}{5} cos 5 2 π cos 4 π 5 \cos\dfrac{4\pi}{5} cos 5 4 π α 2 \alpha^2 α 2 正确做法 ：2 π 5 \dfrac{2\pi}{5} 5 2 π
z n = a z^n = a z n = a n n n 2 π n \dfrac{2\pi}{n} n 2 π n n n
1 的 n n n n n n n n n
前置：17.4 节棣莫弗定理。
前置：国内书第 23 章 23.3 节，等比数列求和（推 1 + α + ⋯ + α n − 1 = 0 1 + \alpha + \dots + \alpha^{n - 1} = 0 1 + α + ⋯ + α n − 1 = 0
前置：本书第 10 章 いろいろな式，因数定理和 ω \omega ω
后续：17.6 节，正多边形的顶点也可以用“绕中心旋转”得到。
本书第 9 章 図形の性質：正多边形的内角、对角线。
内分点・外分点・重心 ：和位置向量完全相同（国内书第 14 章、本书第 16 章）。A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) C ( γ ) \mathrm{C}(\gamma) C ( γ )
AB を m : n に内分する点 n α + m β m + n , 外分する点 − n α + m β m − n , △ A B C の重心 α + β + γ 3 . \text{AB を } m : n \text{ に内分する点} \ \frac{n\alpha + m\beta}{m + n}, \qquad \text{外分する点} \ \frac{-n\alpha + m\beta}{m - n}, \qquad \triangle\mathrm{ABC} \text{ の重心} \ \frac{\alpha + \beta + \gamma}{3}. AB を m : n に内分する点 m + n n α + m β , 外分する点 m − n − n α + m β , △ ABC の重心 3 α + β + γ . 点 α \alpha α ：点 β \beta β α \alpha α θ \theta θ γ \gamma γ
γ − α = ( cos θ + i sin θ ) ( β − α ) , 即 γ = α + ( cos θ + i sin θ ) ( β − α ) . \gamma - \alpha = (\cos\theta + i\sin\theta)(\beta - \alpha), \qquad \text{即} \ \gamma = \alpha + (\cos\theta + i\sin\theta)(\beta - \alpha). γ − α = ( cos θ + i sin θ ) ( β − α ) , 即 γ = α + ( cos θ + i sin θ ) ( β − α ) . 三步：先把旋转中心 α \alpha α β − α \beta - \alpha β − α cos θ + i sin θ \cos\theta + i\sin\theta cos θ + i sin θ α \alpha α
2 直線のなす角 ：三个不同的点 A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) C ( γ ) \mathrm{C}(\gamma) C ( γ )
arg γ − α β − α = 从半直线 AB 转到半直线 AC 的角 . \arg\frac{\gamma - \alpha}{\beta - \alpha} = \text{从半直线 AB 转到半直线 AC 的角}. arg β − α γ − α = 从半直线 AB 转到半直线 AC 的角 . ∠ B A C \angle\mathrm{BAC} ∠ BAC − π < θ ≤ π -\pi < \theta \le \pi − π < θ ≤ π
共線・垂直の条件 ：
3 点 A，B，C が一直線上にある ⟺ γ − α β − α が実数 , A B ⊥ A C ⟺ γ − α β − α が純虚数 . \text{3 点 A，B，C が一直線上にある} \iff \frac{\gamma - \alpha}{\beta - \alpha} \text{ が実数}, \qquad \mathrm{AB} \perp \mathrm{AC} \iff \frac{\gamma - \alpha}{\beta - \alpha} \text{ が純虚数}. 3 点 A ， B ， C が一直線上にある ⟺ β − α γ − α が実数 , AB ⊥ AC ⟺ β − α γ − α が純虚数 . 三角形の形状 ：把比值写成极形式 γ − α β − α = r ( cos θ + i sin θ ) \dfrac{\gamma - \alpha}{\beta - \alpha} = r(\cos\theta + i\sin\theta) β − α γ − α = r ( cos θ + i sin θ )
A C A B = r , ∠ B A C = ∣ θ ∣ . \frac{\mathrm{AC}}{\mathrm{AB}} = r, \qquad \angle\mathrm{BAC} = \lvert \theta \rvert. AB AC = r , ∠ BAC = ∣ θ ∣ .
比值 γ − α β − α \dfrac{\gamma - \alpha}{\beta - \alpha} β − α γ − α
三角形 ABC 的形状
cos ( ± π 3 ) + i sin ( ± π 3 ) = 1 ± 3 i 2 \cos\left(\pm\dfrac{\pi}{3}\right) + i\sin\left(\pm\dfrac{\pi}{3}\right) = \dfrac{1 \pm \sqrt{3}i}{2} cos ( ± 3 π ) + i sin ( ± 3 π ) = 2 1 ± 3 i 正三角形
± i \pm i ± i ∠ A = 90 ° \angle\mathrm{A} = 90° ∠ A = 90°
纯虚数 k i ki k i k ≠ ± 1 k \ne \pm 1 k = ± 1
∠ A = 90 ° \angle\mathrm{A} = 90° ∠ A = 90°
模为 1
A B = A C \mathrm{AB} = \mathrm{AC} AB = AC
题目给出两个顶点，要求正三角形、正方形的另一个顶点，或说「点 β \beta β α \alpha α θ \theta θ γ = α + ( cos θ + i sin θ ) ( β − α ) \gamma = \alpha + (\cos\theta + i\sin\theta)(\beta - \alpha) γ = α + ( cos θ + i sin θ ) ( β − α )
题目要求三点共线、两直线垂直时的参数，或求 ∠ B A C \angle\mathrm{BAC} ∠ BAC γ − α β − α \dfrac{\gamma - \alpha}{\beta - \alpha} β − α γ − α
题目给出 α \alpha α β \beta β γ \gamma γ 2 α 2 − 2 α β + β 2 = 0 2\alpha^2 - 2\alpha\beta + \beta^2 = 0 2 α 2 − 2 α β + β 2 = 0
旋转中心不是原点时，一定先减去中心。γ = ( cos θ + i sin θ ) β \gamma = (\cos\theta + i\sin\theta)\beta γ = ( cos θ + i sin θ ) β
比值的分子分母都从同一个顶点 A 出发：γ − α \gamma - \alpha γ − α β − α \beta - \alpha β − α
共线条件是“比值为实数”，正负都可以：正数表示 B、C 在 A 的同侧，负数表示 A 在 B、C 之间。
“逆时针”是日文「正の向き」，题目若不说方向（只说“旋转 π 3 \dfrac{\pi}{3} 3 π ± π 3 \pm\dfrac{\pi}{3} ± 3 π
EJU 题干：「点 β \beta β α \alpha α π 3 \dfrac{\pi}{3} 3 π ∠ B A C \angle\mathrm{BAC} ∠ BAC △ O A B \triangle\mathrm{OAB} △ OAB
结论 1：绕点 α \alpha α 平移整个平面使 α \alpha α z z z z − α z - \alpha z − α β \beta β α \alpha α θ \theta θ γ \gamma γ β − α \beta - \alpha β − α θ \theta θ γ − α \gamma - \alpha γ − α θ \theta θ cos θ + i sin θ \cos\theta + i\sin\theta cos θ + i sin θ γ − α = ( cos θ + i sin θ ) ( β − α ) \gamma - \alpha = (\cos\theta + i\sin\theta)(\beta - \alpha) γ − α = ( cos θ + i sin θ ) ( β − α )
结论 2：比值的偏角是夹角，模是边之比。 β − α \beta - \alpha β − α γ − α \gamma - \alpha γ − α
∣ γ − α β − α ∣ = ∣ γ − α ∣ ∣ β − α ∣ = A C A B , arg γ − α β − α = arg ( γ − α ) − arg ( β − α ) . \left\lvert \frac{\gamma - \alpha}{\beta - \alpha} \right\rvert = \frac{\lvert \gamma - \alpha \rvert}{\lvert \beta - \alpha \rvert} = \frac{\mathrm{AC}}{\mathrm{AB}}, \qquad \arg\frac{\gamma - \alpha}{\beta - \alpha} = \arg(\gamma - \alpha) - \arg(\beta - \alpha). β − α γ − α = ∣ β − α ∣ ∣ γ − α ∣ = AB AC , arg β − α γ − α = arg ( γ − α ) − arg ( β − α ) . （关键一步 ：偏角之差就是从 AB 方向转到 AC 方向的角。）共线就是这个角为 0 或 π \pi π ± π 2 \pm\dfrac{\pi}{2} ± 2 π γ ≠ α \gamma \ne \alpha γ = α
例 1 （场景：旋转求顶点）
問題 A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 3 + i ) \mathrm{B}(3 + i) B ( 3 + i ) △ A B C \triangle\mathrm{ABC} △ ABC C ( γ ) \mathrm{C}(\gamma) C ( γ )
解答 C \mathrm{C} C B \mathrm{B} B A \mathrm{A} A ± π 3 \pm\dfrac{\pi}{3} ± 3 π β − α = 2 \beta - \alpha = 2 β − α = 2
γ = ( 1 + i ) + ( 1 2 ± 3 2 i ) ⋅ 2 = 2 + ( 1 ± 3 ) i . \gamma = (1 + i) + \left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right) \cdot 2 = 2 + (1 \pm \sqrt{3})i. γ = ( 1 + i ) + ( 2 1 ± 2 3 i ) ⋅ 2 = 2 + ( 1 ± 3 ) i . よって γ = 2 + ( 1 + 3 ) i \gamma = 2 + (1 + \sqrt{3})i γ = 2 + ( 1 + 3 ) i γ = 2 + ( 1 − 3 ) i \gamma = 2 + (1 - \sqrt{3})i γ = 2 + ( 1 − 3 ) i
中文翻译 题目：A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 3 + i ) \mathrm{B}(3 + i) B ( 3 + i ) △ A B C \triangle\mathrm{ABC} △ ABC C ( γ ) \mathrm{C}(\gamma) C ( γ ) ± π 3 \pm\dfrac{\pi}{3} ± 3 π β − α = 2 \beta - \alpha = 2 β − α = 2 γ = ( 1 + i ) + ( 1 2 ± 3 2 i ) ⋅ 2 = 2 + ( 1 ± 3 ) i \gamma = (1 + i) + \left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right) \cdot 2 = 2 + (1 \pm \sqrt{3})i γ = ( 1 + i ) + ( 2 1 ± 2 3 i ) ⋅ 2 = 2 + ( 1 ± 3 ) i γ = 2 + ( 1 + 3 ) i \gamma = 2 + (1 + \sqrt{3})i γ = 2 + ( 1 + 3 ) i γ = 2 + ( 1 − 3 ) i \gamma = 2 + (1 - \sqrt{3})i γ = 2 + ( 1 − 3 ) i
解说 另一种解法 ：正三角形的第三个顶点在 AB 的垂直平分线上，到 AB 的距离是高 3 2 × 2 = 3 \dfrac{\sqrt{3}}{2} \times 2 = \sqrt{3} 2 3 × 2 = 3 2 + i 2 + i 2 + i 3 \sqrt{3} 3 2 + ( 1 ± 3 ) i 2 + (1 \pm \sqrt{3})i 2 + ( 1 ± 3 ) i ± π 3 \pm\dfrac{\pi}{3} ± 3 π
例 2 （场景：共线、垂直求参数）
問題 A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 3 + 2 i ) \mathrm{B}(3 + 2i) B ( 3 + 2 i ) C ( a + 4 i ) \mathrm{C}(a + 4i) C ( a + 4 i ) a a a 3 3 3 A \mathrm{A} A B \mathrm{B} B C \mathrm{C} C a a a A B ⊥ A C \mathrm{AB} \perp \mathrm{AC} AB ⊥ AC a a a
解答
γ − α β − α = ( a − 1 ) + 3 i 2 + i = { ( a − 1 ) + 3 i } ( 2 − i ) 5 = ( 2 a + 1 ) + ( 7 − a ) i 5 . \frac{\gamma - \alpha}{\beta - \alpha} = \frac{(a - 1) + 3i}{2 + i} = \frac{\{(a - 1) + 3i\}(2 - i)}{5} = \frac{(2a + 1) + (7 - a)i}{5}. β − α γ − α = 2 + i ( a − 1 ) + 3 i = 5 {( a − 1 ) + 3 i } ( 2 − i ) = 5 ( 2 a + 1 ) + ( 7 − a ) i .
一直線上にあるとき：虚部が 0 0 0 7 − a = 0 7 - a = 0 7 − a = 0 a = 7 a = 7 a = 7
A B ⊥ A C \mathrm{AB} \perp \mathrm{AC} AB ⊥ AC 0 0 0 7 − a ≠ 0 7 - a \ne 0 7 − a = 0 2 a + 1 = 0 2a + 1 = 0 2 a + 1 = 0 a = − 1 2 a = -\dfrac{1}{2} a = − 2 1
中文翻译 题目：A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 3 + 2 i ) \mathrm{B}(3 + 2i) B ( 3 + 2 i ) C ( a + 4 i ) \mathrm{C}(a + 4i) C ( a + 4 i ) a a a a a a A B ⊥ A C \mathrm{AB} \perp \mathrm{AC} AB ⊥ AC a a a γ − α β − α = ( a − 1 ) + 3 i 2 + i = { ( a − 1 ) + 3 i } ( 2 − i ) 5 = ( 2 a + 1 ) + ( 7 − a ) i 5 \frac{\gamma - \alpha}{\beta - \alpha} = \frac{(a - 1) + 3i}{2 + i} = \frac{\{(a - 1) + 3i\}(2 - i)}{5} = \frac{(2a + 1) + (7 - a)i}{5} β − α γ − α = 2 + i ( a − 1 ) + 3 i = 5 {( a − 1 ) + 3 i } ( 2 − i ) = 5 ( 2 a + 1 ) + ( 7 − a ) i 7 − a = 0 7 - a = 0 7 − a = 0 a = 7 a = 7 a = 7 7 − a ≠ 0 7 - a \ne 0 7 − a = 0 2 a + 1 = 0 2a + 1 = 0 2 a + 1 = 0 a = − 1 2 a = -\dfrac{1}{2} a = − 2 1
解说 比值的分子分母都从同一个顶点 A 出发：分子 γ − α \gamma - \alpha γ − α β − α \beta - \alpha β − α
例 3 （场景：由关系式判断形状）
問題 O \mathrm{O} O 0 0 0 α \alpha α β \beta β 2 α 2 − 2 α β + β 2 = 0 2\alpha^2 - 2\alpha\beta + \beta^2 = 0 2 α 2 − 2 α β + β 2 = 0 A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) △ O A B \triangle\mathrm{OAB} △ OAB
解答 両辺を α 2 \alpha^2 α 2 α ≠ 0 \alpha \ne 0 α = 0 w = β α w = \dfrac{\beta}{\alpha} w = α β
w 2 − 2 w + 2 = 0 , w = 1 ± i = 2 { cos ( ± π 4 ) + i sin ( ± π 4 ) } . w^2 - 2w + 2 = 0, \qquad w = 1 \pm i = \sqrt{2}\left\{\cos\left(\pm\frac{\pi}{4}\right) + i\sin\left(\pm\frac{\pi}{4}\right)\right\}. w 2 − 2 w + 2 = 0 , w = 1 ± i = 2 { cos ( ± 4 π ) + i sin ( ± 4 π ) } . よって O B O A = 2 \dfrac{\mathrm{OB}}{\mathrm{OA}} = \sqrt{2} OA OB = 2 ∠ A O B = π 4 \angle\mathrm{AOB} = \dfrac{\pi}{4} ∠ AOB = 4 π O A = k \mathrm{OA} = k OA = k
A B 2 = k 2 + 2 k 2 − 2 ⋅ k ⋅ 2 k cos π 4 = k 2 , \mathrm{AB}^2 = k^2 + 2k^2 - 2 \cdot k \cdot \sqrt{2}k \cos\frac{\pi}{4} = k^2, AB 2 = k 2 + 2 k 2 − 2 ⋅ k ⋅ 2 k cos 4 π = k 2 , よって A B = k = O A \mathrm{AB} = k = \mathrm{OA} AB = k = OA O A 2 + A B 2 = 2 k 2 = O B 2 \mathrm{OA}^2 + \mathrm{AB}^2 = 2k^2 = \mathrm{OB}^2 OA 2 + AB 2 = 2 k 2 = OB 2 △ O A B \triangle\mathrm{OAB} △ OAB ∠ A = 90 ° \angle\mathrm{A} = 90° ∠ A = 90°
中文翻译 题目：O 是原点，α \alpha α β \beta β 2 α 2 − 2 α β + β 2 = 0 2\alpha^2 - 2\alpha\beta + \beta^2 = 0 2 α 2 − 2 α β + β 2 = 0 A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) △ O A B \triangle\mathrm{OAB} △ OAB α 2 \alpha^2 α 2 α ≠ 0 \alpha \ne 0 α = 0 w = β α w = \dfrac{\beta}{\alpha} w = α β w 2 − 2 w + 2 = 0 w^2 - 2w + 2 = 0 w 2 − 2 w + 2 = 0 w = 1 ± i = 2 { cos ( ± π 4 ) + i sin ( ± π 4 ) } w = 1 \pm i = \sqrt{2}\left\{\cos\left(\pm\frac{\pi}{4}\right) + i\sin\left(\pm\frac{\pi}{4}\right)\right\} w = 1 ± i = 2 { cos ( ± 4 π ) + i sin ( ± 4 π ) } O B O A = 2 \dfrac{\mathrm{OB}}{\mathrm{OA}} = \sqrt{2} OA OB = 2 ∠ A O B = π 4 \angle\mathrm{AOB} = \dfrac{\pi}{4} ∠ AOB = 4 π O A = k \mathrm{OA} = k OA = k A B 2 = k 2 + 2 k 2 − 2 ⋅ k ⋅ 2 k cos π 4 = k 2 \mathrm{AB}^2 = k^2 + 2k^2 - 2 \cdot k \cdot \sqrt{2}k \cos\frac{\pi}{4} = k^2 AB 2 = k 2 + 2 k 2 − 2 ⋅ k ⋅ 2 k cos 4 π = k 2 A B = k = O A \mathrm{AB} = k = \mathrm{OA} AB = k = OA O A 2 + A B 2 = 2 k 2 = O B 2 \mathrm{OA}^2 + \mathrm{AB}^2 = 2k^2 = \mathrm{OB}^2 OA 2 + AB 2 = 2 k 2 = OB 2 △ O A B \triangle\mathrm{OAB} △ OAB ∠ A = 90 ° \angle\mathrm{A} = 90° ∠ A = 90°
解说 另一种解法 ：以 A 为顶点看比值。β − α 0 − α = 1 − β α = 1 − ( 1 ± i ) = ∓ i \dfrac{\beta - \alpha}{0 - \alpha} = 1 - \dfrac{\beta}{\alpha} = 1 - (1 \pm i) = \mp i 0 − α β − α = 1 − α β = 1 − ( 1 ± i ) = ∓ i ∓ π 2 \mp\dfrac{\pi}{2} ∓ 2 π A B = A O \mathrm{AB} = \mathrm{AO} AB = AO ∠ O A B = 90 ° \angle\mathrm{OAB} = 90° ∠ OAB = 90°
例 4 （场景：由三个点求角和形状）
問題 A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 4 + 2 i ) \mathrm{B}(4 + 2i) B ( 4 + 2 i ) C ( 2 + 3 i ) \mathrm{C}(2 + 3i) C ( 2 + 3 i ) ∠ B A C \angle\mathrm{BAC} ∠ BAC △ A B C \triangle\mathrm{ABC} △ ABC
解答
γ − α β − α = 1 + 2 i 3 + i = ( 1 + 2 i ) ( 3 − i ) 10 = 5 + 5 i 10 = 2 2 ( cos π 4 + i sin π 4 ) . \frac{\gamma - \alpha}{\beta - \alpha} = \frac{1 + 2i}{3 + i} = \frac{(1 + 2i)(3 - i)}{10} = \frac{5 + 5i}{10} = \frac{\sqrt{2}}{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right). β − α γ − α = 3 + i 1 + 2 i = 10 ( 1 + 2 i ) ( 3 − i ) = 10 5 + 5 i = 2 2 ( cos 4 π + i sin 4 π ) . よって ∠ B A C = π 4 \angle\mathrm{BAC} = \dfrac{\pi}{4} ∠ BAC = 4 π A C = 2 2 A B \mathrm{AC} = \dfrac{\sqrt{2}}{2}\mathrm{AB} AC = 2 2 AB A B = ∣ 3 + i ∣ = 10 \mathrm{AB} = \lvert 3 + i \rvert = \sqrt{10} AB = ∣ 3 + i ∣ = 10 A C = 5 \mathrm{AC} = \sqrt{5} AC = 5 B C = ∣ ( 2 + 3 i ) − ( 4 + 2 i ) ∣ = ∣ − 2 + i ∣ = 5 \mathrm{BC} = \lvert (2 + 3i) - (4 + 2i) \rvert = \lvert -2 + i \rvert = \sqrt{5} BC = ∣( 2 + 3 i ) − ( 4 + 2 i )∣ = ∣ − 2 + i ∣ = 5 A C = B C \mathrm{AC} = \mathrm{BC} AC = BC A C 2 + B C 2 = 10 = A B 2 \mathrm{AC}^2 + \mathrm{BC}^2 = 10 = \mathrm{AB}^2 AC 2 + BC 2 = 10 = AB 2 △ A B C \triangle\mathrm{ABC} △ ABC ∠ C = 90 ° \angle\mathrm{C} = 90° ∠ C = 90°
中文翻译 题目：A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 4 + 2 i ) \mathrm{B}(4 + 2i) B ( 4 + 2 i ) C ( 2 + 3 i ) \mathrm{C}(2 + 3i) C ( 2 + 3 i ) ∠ B A C \angle\mathrm{BAC} ∠ BAC △ A B C \triangle\mathrm{ABC} △ ABC γ − α β − α = 1 + 2 i 3 + i = ( 1 + 2 i ) ( 3 − i ) 10 = 5 + 5 i 10 = 2 2 ( cos π 4 + i sin π 4 ) \frac{\gamma - \alpha}{\beta - \alpha} = \frac{1 + 2i}{3 + i} = \frac{(1 + 2i)(3 - i)}{10} = \frac{5 + 5i}{10} = \frac{\sqrt{2}}{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) β − α γ − α = 3 + i 1 + 2 i = 10 ( 1 + 2 i ) ( 3 − i ) = 10 5 + 5 i = 2 2 ( cos 4 π + i sin 4 π ) ∠ B A C = π 4 \angle\mathrm{BAC} = \dfrac{\pi}{4} ∠ BAC = 4 π A C = 2 2 A B \mathrm{AC} = \dfrac{\sqrt{2}}{2}\mathrm{AB} AC = 2 2 AB A B = ∣ 3 + i ∣ = 10 \mathrm{AB} = \lvert 3 + i \rvert = \sqrt{10} AB = ∣ 3 + i ∣ = 10 A C = 5 \mathrm{AC} = \sqrt{5} AC = 5 B C = ∣ ( 2 + 3 i ) − ( 4 + 2 i ) ∣ = ∣ − 2 + i ∣ = 5 \mathrm{BC} = \lvert (2 + 3i) - (4 + 2i) \rvert = \lvert -2 + i \rvert = \sqrt{5} BC = ∣( 2 + 3 i ) − ( 4 + 2 i )∣ = ∣ − 2 + i ∣ = 5 A C = B C \mathrm{AC} = \mathrm{BC} AC = BC A C 2 + B C 2 = 10 = A B 2 \mathrm{AC}^2 + \mathrm{BC}^2 = 10 = \mathrm{AB}^2 AC 2 + BC 2 = 10 = AB 2 ∠ C = 90 ° \angle\mathrm{C} = 90° ∠ C = 90°
解说 比值的模和偏角同时给出“边之比”和“夹角”：模 2 2 \dfrac{\sqrt{2}}{2} 2 2 A C = 2 2 A B \mathrm{AC} = \dfrac{\sqrt{2}}{2}\mathrm{AB} AC = 2 2 AB π 4 \dfrac{\pi}{4} 4 π ∠ B A C = π 4 \angle\mathrm{BAC} = \dfrac{\pi}{4} ∠ BAC = 4 π B C = 5 \mathrm{BC} = \sqrt{5} BC = 5 A C \mathrm{AC} AC
正三角形の条件 ：△ A B C \triangle\mathrm{ABC} △ ABC ⟺ α 2 + β 2 + γ 2 = α β + β γ + γ α \iff \alpha^2 + \beta^2 + \gamma^2 = \alpha\beta + \beta\gamma + \gamma\alpha ⟺ α 2 + β 2 + γ 2 = α β + β γ + γ α ⟺ u = γ − α β − α = 1 ± 3 i 2 ⟺ u 2 − u + 1 = 0 \iff u = \dfrac{\gamma - \alpha}{\beta - \alpha} = \dfrac{1 \pm \sqrt{3}i}{2} \iff u^2 - u + 1 = 0 ⟺ u = β − α γ − α = 2 1 ± 3 i ⟺ u 2 − u + 1 = 0 u u u ( β − α ) 2 (\beta - \alpha)^2 ( β − α ) 2 ( γ − α ) 2 − ( γ − α ) ( β − α ) + ( β − α ) 2 = 0 (\gamma - \alpha)^2 - (\gamma - \alpha)(\beta - \alpha) + (\beta - \alpha)^2 = 0 ( γ − α ) 2 − ( γ − α ) ( β − α ) + ( β − α ) 2 = 0 正方形 ：A、B 是相邻顶点，ABCD 按逆时针排列时，δ = α + i ( β − α ) \delta = \alpha + i(\beta - \alpha) δ = α + i ( β − α ) γ = β + i ( β − α ) \gamma = \beta + i(\beta - \alpha) γ = β + i ( β − α ) π 2 \dfrac{\pi}{2} 2 π B C → = A D → \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AD}} BC = AD 原点を通る場合 ：O、A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) ⟺ β α \iff \dfrac{\beta}{\alpha} ⟺ α β ⟺ α β ‾ = α ‾ β \iff \alpha\overline{\beta} = \overline{\alpha}\beta ⟺ α β = α β O A ⊥ O B ⟺ α β ‾ + α ‾ β = 0 \mathrm{OA} \perp \mathrm{OB} \iff \alpha\overline{\beta} + \overline{\alpha}\beta = 0 OA ⊥ OB ⟺ α β + α β = 0 β α = β α ‾ ∣ α ∣ 2 \dfrac{\beta}{\alpha} = \dfrac{\beta\overline{\alpha}}{\lvert \alpha \rvert^2} α β = ∣ α ∣ 2 β α △ O A B \triangle\mathrm{OAB} △ OAB = 1 2 ∣ α ∣ ∣ β ∣ sin ∠ A O B = \dfrac{1}{2}\lvert \alpha \rvert \lvert \beta \rvert \sin\angle\mathrm{AOB} = 2 1 ∣ α ∣ ∣ β ∣ sin ∠ AOB ∠ A O B \angle\mathrm{AOB} ∠ AOB arg β α \arg\dfrac{\beta}{\alpha} arg α β
错在哪 ：B 绕 A ( α ) \mathrm{A}(\alpha) A ( α ) π 3 \dfrac{\pi}{3} 3 π γ = ( 1 2 + 3 2 i ) β \gamma = \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)\beta γ = ( 2 1 + 2 3 i ) β 为什么错 ：乘 cos θ + i sin θ \cos\theta + i\sin\theta cos θ + i sin θ 原点 转。
正确做法 ：γ = α + ( 1 2 + 3 2 i ) ( β − α ) \gamma = \alpha + \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)(\beta - \alpha) γ = α + ( 2 1 + 2 3 i ) ( β − α ) 错在哪 ：判断共线时要求比值是正实数，例 2 之类的题漏掉 A 在 B、C 之间的情形。
为什么错 ：比值为负表示 AC 与 AB 方向相反，三点仍在一条直线上。
正确做法 ：共线条件只看“虚部为 0”，不管实部正负。错在哪 ：正三角形只求出一个第三顶点。
为什么错 ：B 绕 A 可以逆时针转，也可以顺时针转，两边各有一个。
正确做法 ：题目没说方向或没说“逆时针排列”时，± π 3 \pm\dfrac{\pi}{3} ± 3 π
复数把“旋转 + 伸缩”压缩成一次乘法。所以一个比值 γ − α β − α \dfrac{\gamma - \alpha}{\beta - \alpha} β − α γ − α
这和向量的做法对照着看：向量判断垂直要算内积、判断夹角要算 cos \cos cos
前置：17.3 节，乘法是旋转加伸缩，商的偏角是偏角之差。
前置：国内书第 14 章 14.5 节共线向量定理、第 15 章 15.5 节垂直的判定，复数的条件是它们的另一种写法。
前置：国内书第 16 章，余弦定理和三角形面积。
后续：17.7 节，arg z − α z − β \arg\dfrac{z - \alpha}{z - \beta} arg z − β z − α
本书第 16 章 ベクトル：内分点、外分点、重心的公式两边一样。
∣ z − α ∣ \lvert z - \alpha \rvert ∣ z − α ∣ z z z α \alpha α
方程式
表す図形
∣ z − α ∣ = r \lvert z - \alpha \rvert = r ∣ z − α ∣ = r r > 0 r > 0 r > 0 点 α \alpha α r r r
∣ z − α ∣ = ∣ z − β ∣ \lvert z - \alpha \rvert = \lvert z - \beta \rvert ∣ z − α ∣ = ∣ z − β ∣ 2 点 α \alpha α β \beta β
∣ z − α ∣ = k ∣ z − β ∣ \lvert z - \alpha \rvert = k\lvert z - \beta \rvert ∣ z − α ∣ = k ∣ z − β ∣ k > 0 k > 0 k > 0 k ≠ 1 k \ne 1 k = 1 圆（アポロニウスの円）
第一行国内书第 17 章 17.3 节讲过，第二行在该节的常用推论里出现过；第三行国内书没有用复数讲。
アポロニウスの円 ：到两定点距离之比为 k : 1 k : 1 k : 1 k ≠ 1 k \ne 1 k = 1
中心 α − k 2 β 1 − k 2 , 半径 k ∣ α − β ∣ ∣ 1 − k 2 ∣ . \text{中心} \ \frac{\alpha - k^2\beta}{1 - k^2}, \qquad \text{半径} \ \frac{k\lvert \alpha - \beta \rvert}{\lvert 1 - k^2 \rvert}. 中心 1 − k 2 α − k 2 β , 半径 ∣ 1 − k 2 ∣ k ∣ α − β ∣ . 公式不必背，题目里用下面“平方、乘共轭、整理”的办法现场推。
z z ‾ z\overline{z} z z c c c
z z ‾ + α ‾ z + α z ‾ + c = 0 ⟺ ∣ z + α ∣ 2 = ∣ α ∣ 2 − c . z\overline{z} + \overline{\alpha}z + \alpha\overline{z} + c = 0 \iff \lvert z + \alpha \rvert^2 = \lvert \alpha \rvert^2 - c. z z + α z + α z + c = 0 ⟺ ∣ z + α ∣ 2 = ∣ α ∣ 2 − c . ∣ α ∣ 2 − c > 0 \lvert \alpha \rvert^2 - c > 0 ∣ α ∣ 2 − c > 0 − α -\alpha − α ∣ α ∣ 2 − c \sqrt{\lvert \alpha \rvert^2 - c} ∣ α ∣ 2 − c
軌跡（w = f ( z ) w = f(z) w = f ( z ) ：点 z z z w = f ( z ) w = f(z) w = f ( z ) w = f ( z ) w = f(z) w = f ( z ) z z z w w w z z z w w w
题目给出 ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ \lvert z + 2 \rvert = 2\lvert z - 1 \rvert ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ z z ‾ + ( 1 + i ) z + ( 1 − i ) z ‾ = 2 z\overline{z} + (1 + i)z + (1 - i)\overline{z} = 2 z z + ( 1 + i ) z + ( 1 − i ) z = 2 ∣ z − γ ∣ 2 = ρ 2 \lvert z - \gamma \rvert^2 = \rho^2 ∣ z − γ ∣ 2 = ρ 2
题目说「点 z z z w = 1 z w = \dfrac{1}{z} w = z 1 w = ( 1 + i ) z w = (1 + i)z w = ( 1 + i ) z z z z
题目说 z z z ∣ z − c ∣ \lvert z - c \rvert ∣ z − c ∣ arg z \arg z arg z
平方时两边都平方：∣ z + 2 ∣ = 2 ∣ z − 1 ∣ \lvert z + 2 \rvert = 2\lvert z - 1 \rvert ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ 4 ∣ z − 1 ∣ 2 4\lvert z - 1 \rvert^2 4 ∣ z − 1 ∣ 2
整理 z z ‾ z\overline{z} z z α ‾ z \overline{\alpha}z α z α z ‾ \alpha\overline{z} α z
求轨迹时 z z z z ≠ 0 z \ne 0 z = 0 w w w
拿不准时可以设 z = x + y i z = x + yi z = x + y i x x x y y y z z ‾ z\overline{z} z z
EJU 题干：「点 z z z w w w A \fbox{A} A B \fbox{B} B ∣ z ∣ \lvert z \rvert ∣ z ∣
结论 1：アポロニウスの円。 两边平方，写成乘共轭：
( z − α ) ( z ‾ − α ‾ ) = k 2 ( z − β ) ( z ‾ − β ‾ ) . (z - \alpha)(\overline{z} - \overline{\alpha}) = k^2(z - \beta)(\overline{z} - \overline{\beta}). ( z − α ) ( z − α ) = k 2 ( z − β ) ( z − β ) . 展开整理，
( 1 − k 2 ) z z ‾ − ( α ‾ − k 2 β ‾ ) z − ( α − k 2 β ) z ‾ + ∣ α ∣ 2 − k 2 ∣ β ∣ 2 = 0. (1 - k^2)z\overline{z} - (\overline{\alpha} - k^2\overline{\beta})z - (\alpha - k^2\beta)\overline{z} + \lvert \alpha \rvert^2 - k^2\lvert \beta \rvert^2 = 0. ( 1 − k 2 ) z z − ( α − k 2 β ) z − ( α − k 2 β ) z + ∣ α ∣ 2 − k 2 ∣ β ∣ 2 = 0. 除以 1 − k 2 1 - k^2 1 − k 2 关键一步 ：k ≠ 1 k \ne 1 k = 1 k = 1 k = 1 k = 1 z z ‾ z\overline{z} z z γ = α − k 2 β 1 − k 2 \gamma = \dfrac{\alpha - k^2\beta}{1 - k^2} γ = 1 − k 2 α − k 2 β
z z ‾ − γ ‾ z − γ z ‾ + ∣ α ∣ 2 − k 2 ∣ β ∣ 2 1 − k 2 = 0 , ∣ z − γ ∣ 2 = ∣ γ ∣ 2 − ∣ α ∣ 2 − k 2 ∣ β ∣ 2 1 − k 2 . z\overline{z} - \overline{\gamma}z - \gamma\overline{z} + \frac{\lvert \alpha \rvert^2 - k^2\lvert \beta \rvert^2}{1 - k^2} = 0, \qquad \lvert z - \gamma \rvert^2 = \lvert \gamma \rvert^2 - \frac{\lvert \alpha \rvert^2 - k^2\lvert \beta \rvert^2}{1 - k^2}. z z − γ z − γ z + 1 − k 2 ∣ α ∣ 2 − k 2 ∣ β ∣ 2 = 0 , ∣ z − γ ∣ 2 = ∣ γ ∣ 2 − 1 − k 2 ∣ α ∣ 2 − k 2 ∣ β ∣ 2 . 右边通分后分子是 ∣ α − k 2 β ∣ 2 − ( 1 − k 2 ) ( ∣ α ∣ 2 − k 2 ∣ β ∣ 2 ) = k 2 ∣ α − β ∣ 2 \lvert \alpha - k^2\beta \rvert^2 - (1 - k^2)(\lvert \alpha \rvert^2 - k^2\lvert \beta \rvert^2) = k^2\lvert \alpha - \beta \rvert^2 ∣ α − k 2 β ∣ 2 − ( 1 − k 2 ) (∣ α ∣ 2 − k 2 ∣ β ∣ 2 ) = k 2 ∣ α − β ∣ 2 − k 2 ( α β ‾ + α ‾ β ) -k^2(\alpha\overline{\beta} + \overline{\alpha}\beta) − k 2 ( α β + α β ) ∣ α ∣ 2 \lvert \alpha \rvert^2 ∣ α ∣ 2 ∣ β ∣ 2 \lvert \beta \rvert^2 ∣ β ∣ 2 k 2 ∣ α − β ∣ 2 k^2\lvert \alpha - \beta \rvert^2 k 2 ∣ α − β ∣ 2 = k 2 ∣ α − β ∣ 2 ( 1 − k 2 ) 2 > 0 = \dfrac{k^2\lvert \alpha - \beta \rvert^2}{(1 - k^2)^2} > 0 = ( 1 − k 2 ) 2 k 2 ∣ α − β ∣ 2 > 0
结论 2：z z ‾ z\overline{z} z z ∣ z + α ∣ 2 = ( z + α ) ( z ‾ + α ‾ ) = z z ‾ + α ‾ z + α z ‾ + ∣ α ∣ 2 \lvert z + \alpha \rvert^2 = (z + \alpha)(\overline{z} + \overline{\alpha}) = z\overline{z} + \overline{\alpha}z + \alpha\overline{z} + \lvert \alpha \rvert^2 ∣ z + α ∣ 2 = ( z + α ) ( z + α ) = z z + α z + α z + ∣ α ∣ 2 ∣ z + α ∣ 2 − ∣ α ∣ 2 \lvert z + \alpha \rvert^2 - \lvert \alpha \rvert^2 ∣ z + α ∣ 2 − ∣ α ∣ 2
例 1 （场景：认出方程表示的图形）
問題 方程式 ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ \lvert z + 2 \rvert = 2\lvert z - 1 \rvert ∣ z + 2 ∣ = 2 ∣ z − 1 ∣
解答 両辺を 2 2 2 ( z + 2 ) ( z ‾ + 2 ) = 4 ( z − 1 ) ( z ‾ − 1 ) (z + 2)(\overline{z} + 2) = 4(z - 1)(\overline{z} - 1) ( z + 2 ) ( z + 2 ) = 4 ( z − 1 ) ( z − 1 )
z z ‾ + 2 z + 2 z ‾ + 4 = 4 z z ‾ − 4 z − 4 z ‾ + 4 , z\overline{z} + 2z + 2\overline{z} + 4 = 4z\overline{z} - 4z - 4\overline{z} + 4, z z + 2 z + 2 z + 4 = 4 z z − 4 z − 4 z + 4 , 3 z z ‾ − 6 z − 6 z ‾ = 0 , z z ‾ − 2 z − 2 z ‾ = 0 , ( z − 2 ) ( z ‾ − 2 ) = 4. 3z\overline{z} - 6z - 6\overline{z} = 0, \qquad z\overline{z} - 2z - 2\overline{z} = 0, \qquad (z - 2)(\overline{z} - 2) = 4. 3 z z − 6 z − 6 z = 0 , z z − 2 z − 2 z = 0 , ( z − 2 ) ( z − 2 ) = 4. よって ∣ z − 2 ∣ = 2 \lvert z - 2 \rvert = 2 ∣ z − 2 ∣ = 2 2 2 2 2 2 2
中文翻译 题目：方程 ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ \lvert z + 2 \rvert = 2\lvert z - 1 \rvert ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ ( z + 2 ) ( z ‾ + 2 ) = 4 ( z − 1 ) ( z ‾ − 1 ) (z + 2)(\overline{z} + 2) = 4(z - 1)(\overline{z} - 1) ( z + 2 ) ( z + 2 ) = 4 ( z − 1 ) ( z − 1 ) z z ‾ + 2 z + 2 z ‾ + 4 = 4 z z ‾ − 4 z − 4 z ‾ + 4 z\overline{z} + 2z + 2\overline{z} + 4 = 4z\overline{z} - 4z - 4\overline{z} + 4 z z + 2 z + 2 z + 4 = 4 z z − 4 z − 4 z + 4 3 z z ‾ − 6 z − 6 z ‾ = 0 3z\overline{z} - 6z - 6\overline{z} = 0 3 z z − 6 z − 6 z = 0 z z ‾ − 2 z − 2 z ‾ = 0 z\overline{z} - 2z - 2\overline{z} = 0 z z − 2 z − 2 z = 0 ( z − 2 ) ( z ‾ − 2 ) = 4 (z - 2)(\overline{z} - 2) = 4 ( z − 2 ) ( z − 2 ) = 4 ∣ z − 2 ∣ = 2 \lvert z - 2 \rvert = 2 ∣ z − 2 ∣ = 2
解说 另一种解法 ：阿波罗尼斯圆的直径两端是“把线段按 2 : 1 2 : 1 2 : 1 A ( − 2 ) \mathrm{A}(-2) A ( − 2 ) B ( 1 ) \mathrm{B}(1) B ( 1 ) 1 ⋅ ( − 2 ) + 2 ⋅ 1 3 = 0 \dfrac{1 \cdot (-2) + 2 \cdot 1}{3} = 0 3 1 ⋅ ( − 2 ) + 2 ⋅ 1 = 0 − 1 ⋅ ( − 2 ) + 2 ⋅ 1 2 − 1 = 4 \dfrac{-1 \cdot (-2) + 2 \cdot 1}{2 - 1} = 4 2 − 1 − 1 ⋅ ( − 2 ) + 2 ⋅ 1 = 4 4 ∣ z − 1 ∣ 2 4\lvert z - 1 \rvert^2 4 ∣ z − 1 ∣ 2
例 2 （场景：z z ‾ z\overline{z} z z
問題 方程式 z z ‾ + ( 1 + i ) z + ( 1 − i ) z ‾ = 2 z\overline{z} + (1 + i)z + (1 - i)\overline{z} = 2 z z + ( 1 + i ) z + ( 1 − i ) z = 2
解答 z z ‾ + α ‾ z + α z ‾ z\overline{z} + \overline{\alpha}z + \alpha\overline{z} z z + α z + α z α ‾ = 1 + i \overline{\alpha} = 1 + i α = 1 + i α = 1 − i \alpha = 1 - i α = 1 − i ∣ z + ( 1 − i ) ∣ 2 − ∣ 1 − i ∣ 2 = ∣ z + 1 − i ∣ 2 − 2 \lvert z + (1 - i) \rvert^2 - \lvert 1 - i \rvert^2 = \lvert z + 1 - i \rvert^2 - 2 ∣ z + ( 1 − i ) ∣ 2 − ∣ 1 − i ∣ 2 = ∣ z + 1 − i ∣ 2 − 2
∣ z − ( − 1 + i ) ∣ 2 = 4. \lvert z - (-1 + i) \rvert^2 = 4. ∣ z − ( − 1 + i ) ∣ 2 = 4. よって，点 − 1 + i -1 + i − 1 + i 2 2 2
中文翻译 题目：方程 z z ‾ + ( 1 + i ) z + ( 1 − i ) z ‾ = 2 z\overline{z} + (1 + i)z + (1 - i)\overline{z} = 2 z z + ( 1 + i ) z + ( 1 − i ) z = 2 z z ‾ + α ‾ z + α z ‾ z\overline{z} + \overline{\alpha}z + \alpha\overline{z} z z + α z + α z α ‾ = 1 + i \overline{\alpha} = 1 + i α = 1 + i α = 1 − i \alpha = 1 - i α = 1 − i = ∣ z + ( 1 − i ) ∣ 2 − ∣ 1 − i ∣ 2 = ∣ z + 1 − i ∣ 2 − 2 = \lvert z + (1 - i) \rvert^2 - \lvert 1 - i \rvert^2 = \lvert z + 1 - i \rvert^2 - 2 = ∣ z + ( 1 − i ) ∣ 2 − ∣ 1 − i ∣ 2 = ∣ z + 1 − i ∣ 2 − 2 ∣ z − ( − 1 + i ) ∣ 2 = 4 \lvert z - (-1 + i) \rvert^2 = 4 ∣ z − ( − 1 + i ) ∣ 2 = 4 − 1 + i -1 + i − 1 + i
解说 容易卡在：把中心写成 1 − i 1 - i 1 − i 1 + i 1 + i 1 + i ∣ z + α ∣ \lvert z + \alpha \rvert ∣ z + α ∣ − α -\alpha − α α = 1 − i \alpha = 1 - i α = 1 − i − ( 1 − i ) = − 1 + i -(1 - i) = -1 + i − ( 1 − i ) = − 1 + i
例 3 （场景：w = f ( z ) w = f(z) w = f ( z )
問題 点 z z z ∣ z − 1 ∣ = 1 \lvert z - 1 \rvert = 1 ∣ z − 1 ∣ = 1 z ≠ 0 z \ne 0 z = 0 w = 1 z w = \dfrac{1}{z} w = z 1
解答 z = 1 w z = \dfrac{1}{w} z = w 1 w ≠ 0 w \ne 0 w = 0
∣ 1 w − 1 ∣ = 1 , ∣ 1 − w ∣ ∣ w ∣ = 1 , ∣ w − 1 ∣ = ∣ w ∣ . \left\lvert \frac{1}{w} - 1 \right\rvert = 1, \qquad \frac{\lvert 1 - w \rvert}{\lvert w \rvert} = 1, \qquad \lvert w - 1 \rvert = \lvert w \rvert. w 1 − 1 = 1 , ∣ w ∣ ∣ 1 − w ∣ = 1 , ∣ w − 1 ∣ = ∣ w ∣ . w w w 1 1 1 0 0 0 0 0 0 1 1 1 R e w = 1 2 \mathrm{Re}\,w = \dfrac{1}{2} Re w = 2 1 1 2 \dfrac{1}{2} 2 1 w w w 0 0 0 z = 1 w z = \dfrac{1}{w} z = w 1 0 0 0
中文翻译 题目：点 z z z ∣ z − 1 ∣ = 1 \lvert z - 1 \rvert = 1 ∣ z − 1 ∣ = 1 z ≠ 0 z \ne 0 z = 0 w = 1 z w = \dfrac{1}{z} w = z 1 z = 1 w z = \dfrac{1}{w} z = w 1 w ≠ 0 w \ne 0 w = 0 ∣ 1 w − 1 ∣ = 1 \left\lvert \frac{1}{w} - 1 \right\rvert = 1 w 1 − 1 = 1 ∣ 1 − w ∣ ∣ w ∣ = 1 \frac{\lvert 1 - w \rvert}{\lvert w \rvert} = 1 ∣ w ∣ ∣ 1 − w ∣ = 1 ∣ w − 1 ∣ = ∣ w ∣ \lvert w - 1 \rvert = \lvert w \rvert ∣ w − 1 ∣ = ∣ w ∣ w w w w w w R e w = 1 2 \mathrm{Re}\,w = \dfrac{1}{2} Re w = 2 1 1 2 \dfrac{1}{2} 2 1 w w w z = 1 w z = \dfrac{1}{w} z = w 1
解说 容易卡在：把 w = 1 z w = \dfrac{1}{z} w = z 1 ∣ w − 1 ∣ = 1 \lvert w - 1 \rvert = 1 ∣ w − 1 ∣ = 1 w w w z z z w w w z z z z = 1 w z = \dfrac{1}{w} z = w 1 z z z z ≠ 0 z \ne 0 z = 0 w ≠ 0 w \ne 0 w = 0 R e w = 1 2 \mathrm{Re}\,w = \dfrac{1}{2} Re w = 2 1 w = 0 w = 0 w = 0
例 4 （场景：圆上的点的偏角的最值）
問題 点 z z z ∣ z − 2 i ∣ = 1 \lvert z - 2i \rvert = 1 ∣ z − 2 i ∣ = 1 arg z \arg z arg z 0 ≦ arg z < 2 π 0 \leqq \arg z < 2\pi 0 ≦ arg z < 2 π arg z \arg z arg z z z z
解答 円の中心を C ( 2 i ) \mathrm{C}(2i) C ( 2 i ) 1 1 1 O C = 2 \mathrm{OC} = 2 OC = 2 2 2 2 T \mathrm{T} T ∠ O T C = 90 ° \angle\mathrm{OTC} = 90° ∠ OTC = 90° sin ∠ C O T = 1 2 \sin\angle\mathrm{COT} = \dfrac{1}{2} sin ∠ COT = 2 1 ∠ C O T = π 6 \angle\mathrm{COT} = \dfrac{\pi}{6} ∠ COT = 6 π O C \mathrm{OC} OC π 2 \dfrac{\pi}{2} 2 π
π 3 ≦ arg z ≦ 2 π 3 . \frac{\pi}{3} \leqq \arg z \leqq \frac{2\pi}{3}. 3 π ≦ arg z ≦ 3 2 π . 最大となるとき，z z z O T = 2 2 − 1 2 = 3 \mathrm{OT} = \sqrt{2^2 - 1^2} = \sqrt{3} OT = 2 2 − 1 2 = 3
z = 3 ( cos 2 π 3 + i sin 2 π 3 ) = − 3 2 + 3 2 i . z = \sqrt{3}\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} + \frac{3}{2}i. z = 3 ( cos 3 2 π + i sin 3 2 π ) = − 2 3 + 2 3 i . 中文翻译 题目：点 z z z ∣ z − 2 i ∣ = 1 \lvert z - 2i \rvert = 1 ∣ z − 2 i ∣ = 1 arg z \arg z arg z 0 ≤ arg z < 2 π 0 \le \arg z < 2\pi 0 ≤ arg z < 2 π arg z \arg z arg z z z z C ( 2 i ) \mathrm{C}(2i) C ( 2 i ) O C = 2 \mathrm{OC} = 2 OC = 2 ∠ O T C = 90 ° \angle\mathrm{OTC} = 90° ∠ OTC = 90° sin ∠ C O T = 1 2 \sin\angle\mathrm{COT} = \dfrac{1}{2} sin ∠ COT = 2 1 ∠ C O T = π 6 \angle\mathrm{COT} = \dfrac{\pi}{6} ∠ COT = 6 π π 2 \dfrac{\pi}{2} 2 π π 3 ≤ arg z ≤ 2 π 3 \frac{\pi}{3} \le \arg z \le \frac{2\pi}{3} 3 π ≤ arg z ≤ 3 2 π z z z O T = 2 2 − 1 2 = 3 \mathrm{OT} = \sqrt{2^2 - 1^2} = \sqrt{3} OT = 2 2 − 1 2 = 3 z = 3 ( cos 2 π 3 + i sin 2 π 3 ) = − 3 2 + 3 2 i z = \sqrt{3}\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} + \frac{3}{2}i z = 3 ( cos 3 2 π + i sin 3 2 π ) = − 2 3 + 2 3 i
解说 容易卡在：认为 arg z \arg z arg z z z z − 1 + 2 i -1 + 2i − 1 + 2 i π − arctan 2 < 2 π 3 \pi - \arctan 2 < \dfrac{2\pi}{3} π − arctan 2 < 3 2 π sin ∠ C O T = r O C = 1 2 \sin\angle\mathrm{COT} = \dfrac{r}{\mathrm{OC}} = \dfrac{1}{2} sin ∠ COT = OC r = 2 1 π 2 ± π 6 \dfrac{\pi}{2} \pm \dfrac{\pi}{6} 2 π ± 6 π
アポロニウスの円の直径 ：满足 P A : P B = k : 1 \mathrm{PA} : \mathrm{PB} = k : 1 PA : PB = k : 1 k : 1 k : 1 k : 1 γ \gamma γ α \alpha α β \beta β 圆 ∣ z − α ∣ = r \lvert z - \alpha \rvert = r ∣ z − α ∣ = r ∣ z ∣ \lvert z \rvert ∣ z ∣ ∣ α ∣ + r \lvert \alpha \rvert + r ∣ α ∣ + r ∣ ∣ α ∣ − r ∣ \lvert \lvert \alpha \rvert - r \rvert ∣∣ α ∣ − r ∣ r − ∣ α ∣ r - \lvert \alpha \rvert r − ∣ α ∣
arg z − α z − β = θ \arg\dfrac{z - \alpha}{z - \beta} = \theta arg z − β z − α = θ 0 < ∣ θ ∣ < π 0 < \lvert \theta \rvert < \pi 0 < ∣ θ ∣ < π z z z θ \theta θ z z z z β z\beta z β z α z\alpha z α w = 1 z w = \dfrac{1}{z} w = z 1 w w ‾ w\overline{w} w w ∣ z − α ∣ + ∣ z − β ∣ = 2 a \lvert z - \alpha \rvert + \lvert z - \beta \rvert = 2a ∣ z − α ∣ + ∣ z − β ∣ = 2 a 2 a > ∣ α − β ∣ 2a > \lvert \alpha - \beta \rvert 2 a > ∣ α − β ∣ α \alpha α β \beta β
错在哪 ：∣ z + 2 ∣ = 2 ∣ z − 1 ∣ \lvert z + 2 \rvert = 2\lvert z - 1 \rvert ∣ z + 2 ∣ = 2 ∣ z − 1 ∣ ∣ z + 2 ∣ 2 = 2 ∣ z − 1 ∣ 2 \lvert z + 2 \rvert^2 = 2\lvert z - 1 \rvert^2 ∣ z + 2 ∣ 2 = 2 ∣ z − 1 ∣ 2 为什么错 ：右边的系数 2 也要平方。
正确做法 ：∣ z + 2 ∣ 2 = 4 ∣ z − 1 ∣ 2 \lvert z + 2 \rvert^2 = 4\lvert z - 1 \rvert^2 ∣ z + 2 ∣ 2 = 4 ∣ z − 1 ∣ 2 错在哪 ：例 3 把 w = 1 z w = \dfrac{1}{z} w = z 1 ∣ w − 1 ∣ = 1 \lvert w - 1 \rvert = 1 ∣ w − 1 ∣ = 1 w w w 为什么错 ：已知条件是关于 z z z w w w z z z z = 1 w z = \dfrac{1}{w} z = w 1 z z z 正确做法 ：∣ 1 w − 1 ∣ = 1 \left\lvert \dfrac{1}{w} - 1 \right\rvert = 1 w 1 − 1 = 1 ∣ w − 1 ∣ = ∣ w ∣ \lvert w - 1 \rvert = \lvert w \rvert ∣ w − 1 ∣ = ∣ w ∣ 错在哪 ：例 4 认为 arg z \arg z arg z z z z − 1 + 2 i -1 + 2i − 1 + 2 i 为什么错 ：最左点的偏角是 π − arctan 2 < 2 π 3 \pi - \arctan 2 < \dfrac{2\pi}{3} π − arctan 2 < 3 2 π 正确做法 ：偏角的最值看过原点的切线，距离的最值看过圆心的直线。
∣ z − α ∣ \lvert z - \alpha \rvert ∣ z − α ∣ z z z α \alpha α
w = 1 z w = \dfrac{1}{z} w = z 1 1 z = z ‾ ∣ z ∣ 2 \dfrac{1}{z} = \dfrac{\overline{z}}{\lvert z \rvert^2} z 1 = ∣ z ∣ 2 z
前置：17.2 节，∣ z ∣ 2 = z z ‾ \lvert z \rvert^2 = z\overline{z} ∣ z ∣ 2 = z z
前置：国内书第 29 章 29.2 节，圆的标准方程；国内书第 28 章，垂直平分线。
前置：17.6 节，比值的偏角是两直线的夹角。
本书第 11 章 図形と方程式：轨迹的求法（设动点、代入、去参数），アポロニウスの円的坐标做法。
本书第 18 章 式と曲線：椭圆、双曲线的定义也是“到两点距离”的条件。
2014–2021 年复数平面属于数学Ⅲ，以下出题位置据这段时间的公开过去问；2026 年新シラバス下还没有公开试题。EJU コース2 里复数平面通常是 Ⅱ 的問2（2015 年第 1、2 回，2016 年第 1 回，2017 年第 1 回，2019 年第 1 回，2020 年第 2 回，2021 年第 1 回都是），偶尔单独出成一整道大问（2016 年第 2 回的 Ⅲ，用偏角条件和直线方程求点 z）。一题只考一条线，誘導的骨架很固定：
代数线 ：(1) 化极形式，填模和偏角（偏角写成 B CD π \dfrac{\fbox{B}}{\fbox{CD}}\pi CD B π z n z^n z n z n z^n z n n n n z n = a z^n = a z n = a 1 + α + … 1 + \alpha + \dots 1 + α + … 形状线 ：(1) 由关系式求比值 β α \dfrac{\beta}{\alpha} α β β \beta β α \alpha α 轨迹线 ：(1) 把方程化成 ∣ z − γ ∣ = ρ \lvert z - \gamma \rvert = \rho ∣ z − γ ∣ = ρ ∣ z − c ∣ \lvert z - c \rvert ∣ z − c ∣ w = f ( z ) w = f(z) w = f ( z )
填涂提醒：偏角 5 12 π \dfrac{5}{12}\pi 12 5 π B CD π \dfrac{\fbox{B}}{\fbox{CD}}\pi CD B π − i -i − i AB i \fbox{AB}\,i AB i
問題 1 （極形式とド・モアブルの定理）
複素数 z = − 1 + 3 i 1 + i z = \dfrac{-1 + \sqrt{3}i}{1 + i} z = 1 + i − 1 + 3 i θ \theta θ 0 ≦ θ < 2 π 0 \leqq \theta < 2\pi 0 ≦ θ < 2 π
(1) z z z A \sqrt{\fbox{A}} A B CD π \dfrac{\fbox{B}}{\fbox{CD}}\pi CD B π
(2) z n z^n z n n n n n = EF n = \fbox{EF} n = EF z n = − GH z^n = -\fbox{GH} z n = − GH
(3) z n z^n z n n n n n = I n = \fbox{I} n = I z n = J i z^n = \fbox{J}\,i z n = J i
解答 (1) 分子は − 1 + 3 i = 2 ( cos 2 π 3 + i sin 2 π 3 ) -1 + \sqrt{3}i = 2\left(\cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}\right) − 1 + 3 i = 2 ( cos 3 2 π + i sin 3 2 π ) 1 + i = 2 ( cos π 4 + i sin π 4 ) 1 + i = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 1 + i = 2 ( cos 4 π + i sin 4 π ) 2 2 = 2 \dfrac{2}{\sqrt{2}} = \sqrt{2} 2 2 = 2 2 π 3 − π 4 = 5 π 12 \dfrac{2\pi}{3} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} 3 2 π − 4 π = 12 5 π A = 2 \fbox{A} = 2 A = 2 B = 5 \fbox{B} = 5 B = 5 CD \fbox{CD} CD 1 1 1 2 2 2
(2) z n = ( 2 ) n ( cos 5 n π 12 + i sin 5 n π 12 ) z^n = (\sqrt{2})^n\left(\cos\dfrac{5n\pi}{12} + i\sin\dfrac{5n\pi}{12}\right) z n = ( 2 ) n ( cos 12 5 nπ + i sin 12 5 nπ ) 5 n 12 \dfrac{5n}{12} 12 5 n 5 5 5 12 12 12 n n n 12 12 12 n = 12 n = 12 n = 12 EF \fbox{EF} EF 1 1 1 2 2 2 z 12 = 2 6 ( cos 5 π + i sin 5 π ) = − 64 z^{12} = 2^6(\cos 5\pi + i\sin 5\pi) = -64 z 12 = 2 6 ( cos 5 π + i sin 5 π ) = − 64 GH \fbox{GH} GH 6 6 6 4 4 4
(3) 純虚数となるのは 5 n π 12 = π 2 + k π \dfrac{5n\pi}{12} = \dfrac{\pi}{2} + k\pi 12 5 nπ = 2 π + k π 5 n = 6 + 12 k 5n = 6 + 12k 5 n = 6 + 12 k n = 1 , 2 , … n = 1, 2, \dots n = 1 , 2 , … 5 n 5n 5 n 12 12 12 6 6 6 n n n 6 6 6 5 × 6 = 30 = 6 + 24 5 \times 6 = 30 = 6 + 24 5 × 6 = 30 = 6 + 24 I = 6 \fbox{I} = 6 I = 6 z 6 = 8 ( cos 5 π 2 + i sin 5 π 2 ) = 8 i z^6 = 8\left(\cos\dfrac{5\pi}{2} + i\sin\dfrac{5\pi}{2}\right) = 8i z 6 = 8 ( cos 2 5 π + i sin 2 5 π ) = 8 i J = 8 \fbox{J} = 8 J = 8
中文翻译 题目：考虑复数 z = − 1 + 3 i 1 + i z = \dfrac{-1 + \sqrt{3}i}{1 + i} z = 1 + i − 1 + 3 i θ \theta θ 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π z z z A \sqrt{\fbox{A}} A B CD π \dfrac{\fbox{B}}{\fbox{CD}}\pi CD B π z n z^n z n n = EF n = \fbox{EF} n = EF z n = − GH z^n = -\fbox{GH} z n = − GH z n z^n z n n = I n = \fbox{I} n = I z n = J i z^n = \fbox{J}\,i z n = J i − 1 + 3 i = 2 ( cos 2 π 3 + i sin 2 π 3 ) -1 + \sqrt{3}i = 2\left(\cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}\right) − 1 + 3 i = 2 ( cos 3 2 π + i sin 3 2 π ) 1 + i = 2 ( cos π 4 + i sin π 4 ) 1 + i = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 1 + i = 2 ( cos 4 π + i sin 4 π ) 2 2 = 2 \dfrac{2}{\sqrt{2}} = \sqrt{2} 2 2 = 2 2 π 3 − π 4 = 5 π 12 \dfrac{2\pi}{3} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} 3 2 π − 4 π = 12 5 π A = 2 \fbox{A} = 2 A = 2 B = 5 \fbox{B} = 5 B = 5 CD \fbox{CD} CD z n = ( 2 ) n ( cos 5 n π 12 + i sin 5 n π 12 ) z^n = (\sqrt{2})^n\left(\cos\dfrac{5n\pi}{12} + i\sin\dfrac{5n\pi}{12}\right) z n = ( 2 ) n ( cos 12 5 nπ + i sin 12 5 nπ ) 5 n 12 \dfrac{5n}{12} 12 5 n n n n n = 12 n = 12 n = 12 EF \fbox{EF} EF z 12 = 2 6 ( cos 5 π + i sin 5 π ) = − 64 z^{12} = 2^6(\cos 5\pi + i\sin 5\pi) = -64 z 12 = 2 6 ( cos 5 π + i sin 5 π ) = − 64 GH \fbox{GH} GH 5 n π 12 = π 2 + k π \dfrac{5n\pi}{12} = \dfrac{\pi}{2} + k\pi 12 5 nπ = 2 π + k π 5 n = 6 + 12 k 5n = 6 + 12k 5 n = 6 + 12 k n = 1 , 2 , … n = 1, 2, \dots n = 1 , 2 , … 5 n 5n 5 n n n n 5 × 6 = 30 = 6 + 24 5 \times 6 = 30 = 6 + 24 5 × 6 = 30 = 6 + 24 I = 6 \fbox{I} = 6 I = 6 z 6 = 8 ( cos 5 π 2 + i sin 5 π 2 ) = 8 i z^6 = 8\left(\cos\dfrac{5\pi}{2} + i\sin\dfrac{5\pi}{2}\right) = 8i z 6 = 8 ( cos 2 5 π + i sin 2 5 π ) = 8 i J = 8 \fbox{J} = 8 J = 8
解说 两分钟做法：商先化极形式，模相除、偏角相减；再套棣莫弗，实数的条件看 sin \sin sin π \pi π cos \cos cos π 2 \dfrac{\pi}{2} 2 π A = 2 \fbox{A} = 2 A = 2 5 π 12 \dfrac{5\pi}{12} 12 5 π B CD π \dfrac{\fbox{B}}{\fbox{CD}}\pi CD B π GH \fbox{GH} GH J i \fbox{J}\,i J i
問題 2 （複素数と図形：三角形の形状）
O を原点とする複素数平面上で，0 でない複素数 α \alpha α β \beta β
α 2 − 2 α β + 4 β 2 = 0 , ∣ β ∣ = 2 \alpha^2 - 2\alpha\beta + 4\beta^2 = 0, \qquad \lvert \beta \rvert = 2 α 2 − 2 α β + 4 β 2 = 0 , ∣ β ∣ = 2 を満たしている。A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β )
(1) α β = A ± B i \dfrac{\alpha}{\beta} = \fbox{A} \pm \sqrt{\fbox{B}}\,i β α = A ± B i
(2) ∣ α ∣ = C \lvert \alpha \rvert = \fbox{C} ∣ α ∣ = C ∠ A O B = π D \angle\mathrm{AOB} = \dfrac{\pi}{\fbox{D}} ∠ AOB = D π A B = E F \mathrm{AB} = \fbox{E}\sqrt{\fbox{F}} AB = E F △ O A B \triangle\mathrm{OAB} △ OAB G \fbox{G} G G \fbox{G} G
⓪ 正三角形 ① ∠ O A B = 90 ° \angle\mathrm{OAB} = 90° ∠ OAB = 90° ∠ O B A = 90 ° \angle\mathrm{OBA} = 90° ∠ OBA = 90° O A = A B \mathrm{OA} = \mathrm{AB} OA = AB
(3) さらに β = 1 + 3 i \beta = 1 + \sqrt{3}i β = 1 + 3 i α \alpha α α = − H + I J i \alpha = -\fbox{H} + \fbox{I}\sqrt{\fbox{J}}\,i α = − H + I J i △ O A B \triangle\mathrm{OAB} △ OAB − K + L i -\fbox{K} + \sqrt{\fbox{L}}\,i − K + L i M \fbox{M} M
解答 (1) β 2 \beta^2 β 2 w = α β w = \dfrac{\alpha}{\beta} w = β α w 2 − 2 w + 4 = 0 w^2 - 2w + 4 = 0 w 2 − 2 w + 4 = 0 w = 1 ± 3 i w = 1 \pm \sqrt{3}i w = 1 ± 3 i A = 1 \fbox{A} = 1 A = 1 B = 3 \fbox{B} = 3 B = 3
(2) w = 2 { cos ( ± π 3 ) + i sin ( ± π 3 ) } w = 2\left\{\cos\left(\pm\dfrac{\pi}{3}\right) + i\sin\left(\pm\dfrac{\pi}{3}\right)\right\} w = 2 { cos ( ± 3 π ) + i sin ( ± 3 π ) } ∣ α ∣ = 2 ∣ β ∣ = 4 \lvert \alpha \rvert = 2\lvert \beta \rvert = 4 ∣ α ∣ = 2 ∣ β ∣ = 4 ∠ A O B = π 3 \angle\mathrm{AOB} = \dfrac{\pi}{3} ∠ AOB = 3 π C = 4 \fbox{C} = 4 C = 4 D = 3 \fbox{D} = 3 D = 3 A B 2 = 16 + 4 − 2 ⋅ 4 ⋅ 2 ⋅ 1 2 = 12 \mathrm{AB}^2 = 16 + 4 - 2 \cdot 4 \cdot 2 \cdot \dfrac{1}{2} = 12 AB 2 = 16 + 4 − 2 ⋅ 4 ⋅ 2 ⋅ 2 1 = 12 A B = 2 3 \mathrm{AB} = 2\sqrt{3} AB = 2 3 E = 2 \fbox{E} = 2 E = 2 F = 3 \fbox{F} = 3 F = 3 O B 2 + A B 2 = 4 + 12 = 16 = O A 2 \mathrm{OB}^2 + \mathrm{AB}^2 = 4 + 12 = 16 = \mathrm{OA}^2 OB 2 + AB 2 = 4 + 12 = 16 = OA 2 ∠ O B A = 90 ° \angle\mathrm{OBA} = 90° ∠ OBA = 90° G \fbox{G} G
(3) α = ( 1 ± 3 i ) β \alpha = (1 \pm \sqrt{3}i)\beta α = ( 1 ± 3 i ) β ( 1 + 3 i ) ( 1 + 3 i ) = 1 + 2 3 i − 3 = − 2 + 2 3 i (1 + \sqrt{3}i)(1 + \sqrt{3}i) = 1 + 2\sqrt{3}i - 3 = -2 + 2\sqrt{3}i ( 1 + 3 i ) ( 1 + 3 i ) = 1 + 2 3 i − 3 = − 2 + 2 3 i ( 1 − 3 i ) ( 1 + 3 i ) = 4 (1 - \sqrt{3}i)(1 + \sqrt{3}i) = 4 ( 1 − 3 i ) ( 1 + 3 i ) = 4 0 0 0 α = − 2 + 2 3 i \alpha = -2 + 2\sqrt{3}i α = − 2 + 2 3 i H = 2 \fbox{H} = 2 H = 2 I = 2 \fbox{I} = 2 I = 2 J = 3 \fbox{J} = 3 J = 3 O A \mathrm{OA} OA O A \mathrm{OA} OA α 2 = − 1 + 3 i \dfrac{\alpha}{2} = -1 + \sqrt{3}i 2 α = − 1 + 3 i O A 2 = 2 \dfrac{\mathrm{OA}}{2} = 2 2 OA = 2 K = 1 \fbox{K} = 1 K = 1 L = 3 \fbox{L} = 3 L = 3 M = 2 \fbox{M} = 2 M = 2
中文翻译 题目：在以 O 为原点的复平面上，非零复数 α \alpha α β \beta β α 2 − 2 α β + 4 β 2 = 0 \alpha^2 - 2\alpha\beta + 4\beta^2 = 0 α 2 − 2 α β + 4 β 2 = 0 ∣ β ∣ = 2 \lvert \beta \rvert = 2 ∣ β ∣ = 2 A ( α ) \mathrm{A}(\alpha) A ( α ) B ( β ) \mathrm{B}(\beta) B ( β ) α β = A ± B i \dfrac{\alpha}{\beta} = \fbox{A} \pm \sqrt{\fbox{B}}\,i β α = A ± B i ∣ α ∣ = C \lvert \alpha \rvert = \fbox{C} ∣ α ∣ = C ∠ A O B = π D \angle\mathrm{AOB} = \dfrac{\pi}{\fbox{D}} ∠ AOB = D π A B = E F \mathrm{AB} = \fbox{E}\sqrt{\fbox{F}} AB = E F △ O A B \triangle\mathrm{OAB} △ OAB G \fbox{G} G ∠ O A B = 90 ° \angle\mathrm{OAB} = 90° ∠ OAB = 90° ∠ O B A = 90 ° \angle\mathrm{OBA} = 90° ∠ OBA = 90° O A = A B \mathrm{OA} = \mathrm{AB} OA = AB β = 1 + 3 i \beta = 1 + \sqrt{3}i β = 1 + 3 i α \alpha α α = − H + I J i \alpha = -\fbox{H} + \fbox{I}\sqrt{\fbox{J}}\,i α = − H + I J i △ O A B \triangle\mathrm{OAB} △ OAB − K + L i -\fbox{K} + \sqrt{\fbox{L}}\,i − K + L i M \fbox{M} M β 2 \beta^2 β 2 w = α β w = \dfrac{\alpha}{\beta} w = β α w 2 − 2 w + 4 = 0 w^2 - 2w + 4 = 0 w 2 − 2 w + 4 = 0 w = 1 ± 3 i w = 1 \pm \sqrt{3}i w = 1 ± 3 i A = 1 \fbox{A} = 1 A = 1 B = 3 \fbox{B} = 3 B = 3 w = 2 { cos ( ± π 3 ) + i sin ( ± π 3 ) } w = 2\left\{\cos\left(\pm\dfrac{\pi}{3}\right) + i\sin\left(\pm\dfrac{\pi}{3}\right)\right\} w = 2 { cos ( ± 3 π ) + i sin ( ± 3 π ) } ∣ α ∣ = 2 ∣ β ∣ = 4 \lvert \alpha \rvert = 2\lvert \beta \rvert = 4 ∣ α ∣ = 2 ∣ β ∣ = 4 ∠ A O B = π 3 \angle\mathrm{AOB} = \dfrac{\pi}{3} ∠ AOB = 3 π C = 4 \fbox{C} = 4 C = 4 D = 3 \fbox{D} = 3 D = 3 A B 2 = 16 + 4 − 2 ⋅ 4 ⋅ 2 ⋅ 1 2 = 12 \mathrm{AB}^2 = 16 + 4 - 2 \cdot 4 \cdot 2 \cdot \dfrac{1}{2} = 12 AB 2 = 16 + 4 − 2 ⋅ 4 ⋅ 2 ⋅ 2 1 = 12 A B = 2 3 \mathrm{AB} = 2\sqrt{3} AB = 2 3 E = 2 \fbox{E} = 2 E = 2 F = 3 \fbox{F} = 3 F = 3 O B 2 + A B 2 = 4 + 12 = 16 = O A 2 \mathrm{OB}^2 + \mathrm{AB}^2 = 4 + 12 = 16 = \mathrm{OA}^2 OB 2 + AB 2 = 4 + 12 = 16 = OA 2 ∠ O B A = 90 ° \angle\mathrm{OBA} = 90° ∠ OBA = 90° G \fbox{G} G α = ( 1 ± 3 i ) β \alpha = (1 \pm \sqrt{3}i)\beta α = ( 1 ± 3 i ) β ( 1 + 3 i ) ( 1 + 3 i ) = 1 + 2 3 i − 3 = − 2 + 2 3 i (1 + \sqrt{3}i)(1 + \sqrt{3}i) = 1 + 2\sqrt{3}i - 3 = -2 + 2\sqrt{3}i ( 1 + 3 i ) ( 1 + 3 i ) = 1 + 2 3 i − 3 = − 2 + 2 3 i ( 1 − 3 i ) ( 1 + 3 i ) = 4 (1 - \sqrt{3}i)(1 + \sqrt{3}i) = 4 ( 1 − 3 i ) ( 1 + 3 i ) = 4 α = − 2 + 2 3 i \alpha = -2 + 2\sqrt{3}i α = − 2 + 2 3 i H = 2 \fbox{H} = 2 H = 2 I = 2 \fbox{I} = 2 I = 2 J = 3 \fbox{J} = 3 J = 3 O A \mathrm{OA} OA O A \mathrm{OA} OA α 2 = − 1 + 3 i \dfrac{\alpha}{2} = -1 + \sqrt{3}i 2 α = − 1 + 3 i O A 2 = 2 \dfrac{\mathrm{OA}}{2} = 2 2 OA = 2 K = 1 \fbox{K} = 1 K = 1 L = 3 \fbox{L} = 3 L = 3 M = 2 \fbox{M} = 2 M = 2
解说 两分钟做法：关系式里 α \alpha α β \beta β β 2 \beta^2 β 2 w = α β w = \dfrac{\alpha}{\beta} w = β α w w w w w w O A 2 = O B 2 + A B 2 \mathrm{OA}^2 = \mathrm{OB}^2 + \mathrm{AB}^2 OA 2 = OB 2 + AB 2 O A = A B \mathrm{OA} = \mathrm{AB} OA = AB O A = 4 \mathrm{OA} = 4 OA = 4 A B = 2 3 \mathrm{AB} = 2\sqrt{3} AB = 2 3 A \fbox{A} A B \fbox{B} B H \fbox{H} H K \fbox{K} K
問題 3 （方程式の表す図形と軌跡）
複素数平面上で，点 z z z ∣ z − 3 i ∣ = 2 ∣ z ∣ \lvert z - 3i \rvert = 2\lvert z \rvert ∣ z − 3 i ∣ = 2 ∣ z ∣
(1) 点 z z z AB i \fbox{AB}\,i AB i C \fbox{C} C
(2) ∣ z − 3 − 3 i ∣ \lvert z - 3 - 3i \rvert ∣ z − 3 − 3 i ∣ D \fbox{D} D E \fbox{E} E
(3) w = ( 1 + i ) z w = (1 + i)z w = ( 1 + i ) z w w w F − G i \fbox{F} - \fbox{G}\,i F − G i H I \fbox{H}\sqrt{\fbox{I}} H I
解答 (1) 両辺を 2 2 2 ( z − 3 i ) ( z ‾ + 3 i ) = 4 z z ‾ (z - 3i)(\overline{z} + 3i) = 4z\overline{z} ( z − 3 i ) ( z + 3 i ) = 4 z z z z ‾ + 3 i z − 3 i z ‾ + 9 = 4 z z ‾ z\overline{z} + 3iz - 3i\overline{z} + 9 = 4z\overline{z} z z + 3 i z − 3 i z + 9 = 4 z z
z z ‾ − i z + i z ‾ − 3 = 0. z\overline{z} - iz + i\overline{z} - 3 = 0. z z − i z + i z − 3 = 0. z z ‾ + α ‾ z + α z ‾ z\overline{z} + \overline{\alpha}z + \alpha\overline{z} z z + α z + α z α ‾ = − i \overline{\alpha} = -i α = − i α = i \alpha = i α = i ∣ z + i ∣ 2 − 1 − 3 = 0 \lvert z + i \rvert^2 - 1 - 3 = 0 ∣ z + i ∣ 2 − 1 − 3 = 0 ∣ z − ( − i ) ∣ = 2 \lvert z - (-i) \rvert = 2 ∣ z − ( − i )∣ = 2 − i -i − i 2 2 2 AB \fbox{AB} AB 1 1 1 C = 2 \fbox{C} = 2 C = 2
(2) ∣ z − ( 3 + 3 i ) ∣ \lvert z - (3 + 3i) \rvert ∣ z − ( 3 + 3 i )∣ z z z 3 + 3 i 3 + 3i 3 + 3 i − i -i − i 3 + 3 i 3 + 3i 3 + 3 i ∣ 3 + 4 i ∣ = 5 \lvert 3 + 4i \rvert = 5 ∣ 3 + 4 i ∣ = 5 2 2 2 3 + 3 i 3 + 3i 3 + 3 i 5 + 2 = 7 5 + 2 = 7 5 + 2 = 7 5 − 2 = 3 5 - 2 = 3 5 − 2 = 3 D = 7 \fbox{D} = 7 D = 7 E = 3 \fbox{E} = 3 E = 3
(3) z = w 1 + i z = \dfrac{w}{1 + i} z = 1 + i w ∣ z + i ∣ = 2 \lvert z + i \rvert = 2 ∣ z + i ∣ = 2 ∣ w 1 + i + i ∣ = 2 \left\lvert \dfrac{w}{1 + i} + i \right\rvert = 2 1 + i w + i = 2 ∣ 1 + i ∣ = 2 \lvert 1 + i \rvert = \sqrt{2} ∣ 1 + i ∣ = 2
∣ w + i ( 1 + i ) ∣ = 2 2 , ∣ w − ( 1 − i ) ∣ = 2 2 . \lvert w + i(1 + i) \rvert = 2\sqrt{2}, \qquad \lvert w - (1 - i) \rvert = 2\sqrt{2}. ∣ w + i ( 1 + i )∣ = 2 2 , ∣ w − ( 1 − i )∣ = 2 2 . よって中心は 1 − i 1 - i 1 − i 2 2 2\sqrt{2} 2 2 F = 1 \fbox{F} = 1 F = 1 G = 1 \fbox{G} = 1 G = 1 H = 2 \fbox{H} = 2 H = 2 I = 2 \fbox{I} = 2 I = 2
中文翻译 题目：复平面上点 z z z ∣ z − 3 i ∣ = 2 ∣ z ∣ \lvert z - 3i \rvert = 2\lvert z \rvert ∣ z − 3 i ∣ = 2 ∣ z ∣ z z z AB i \fbox{AB}\,i AB i C \fbox{C} C ∣ z − 3 − 3 i ∣ \lvert z - 3 - 3i \rvert ∣ z − 3 − 3 i ∣ D \fbox{D} D E \fbox{E} E w = ( 1 + i ) z w = (1 + i)z w = ( 1 + i ) z w w w F − G i \fbox{F} - \fbox{G}\,i F − G i H I \fbox{H}\sqrt{\fbox{I}} H I ( z − 3 i ) ( z ‾ + 3 i ) = 4 z z ‾ (z - 3i)(\overline{z} + 3i) = 4z\overline{z} ( z − 3 i ) ( z + 3 i ) = 4 z z z z ‾ + 3 i z − 3 i z ‾ + 9 = 4 z z ‾ z\overline{z} + 3iz - 3i\overline{z} + 9 = 4z\overline{z} z z + 3 i z − 3 i z + 9 = 4 z z z z ‾ − i z + i z ‾ − 3 = 0 z\overline{z} - iz + i\overline{z} - 3 = 0 z z − i z + i z − 3 = 0 z z ‾ + α ‾ z + α z ‾ z\overline{z} + \overline{\alpha}z + \alpha\overline{z} z z + α z + α z α ‾ = − i \overline{\alpha} = -i α = − i α = i \alpha = i α = i ∣ z + i ∣ 2 − 1 − 3 = 0 \lvert z + i \rvert^2 - 1 - 3 = 0 ∣ z + i ∣ 2 − 1 − 3 = 0 ∣ z − ( − i ) ∣ = 2 \lvert z - (-i) \rvert = 2 ∣ z − ( − i )∣ = 2 − i -i − i AB \fbox{AB} AB C = 2 \fbox{C} = 2 C = 2 ∣ z − ( 3 + 3 i ) ∣ \lvert z - (3 + 3i) \rvert ∣ z − ( 3 + 3 i )∣ 3 + 3 i 3 + 3i 3 + 3 i − i -i − i 3 + 3 i 3 + 3i 3 + 3 i ∣ 3 + 4 i ∣ = 5 \lvert 3 + 4i \rvert = 5 ∣ 3 + 4 i ∣ = 5 5 + 2 = 7 5 + 2 = 7 5 + 2 = 7 5 − 2 = 3 5 - 2 = 3 5 − 2 = 3 D = 7 \fbox{D} = 7 D = 7 E = 3 \fbox{E} = 3 E = 3 z = w 1 + i z = \dfrac{w}{1 + i} z = 1 + i w ∣ z + i ∣ = 2 \lvert z + i \rvert = 2 ∣ z + i ∣ = 2 ∣ w 1 + i + i ∣ = 2 \left\lvert \dfrac{w}{1 + i} + i \right\rvert = 2 1 + i w + i = 2 ∣ 1 + i ∣ = 2 \lvert 1 + i \rvert = \sqrt{2} ∣ 1 + i ∣ = 2 ∣ w + i ( 1 + i ) ∣ = 2 2 \lvert w + i(1 + i) \rvert = 2\sqrt{2} ∣ w + i ( 1 + i )∣ = 2 2 ∣ w − ( 1 − i ) ∣ = 2 2 \lvert w - (1 - i) \rvert = 2\sqrt{2} ∣ w − ( 1 − i )∣ = 2 2 1 − i 1 - i 1 − i 2 2 2\sqrt{2} 2 2 F = 1 \fbox{F} = 1 F = 1 G = 1 \fbox{G} = 1 G = 1 H = 2 \fbox{H} = 2 H = 2 I = 2 \fbox{I} = 2 I = 2
解说 两分钟做法：含模的方程先两边平方、乘共轭，整理成 z z ‾ + α ‾ z + α z ‾ + c = 0 z\overline{z} + \overline{\alpha}z + \alpha\overline{z} + c = 0 z z + α z + α z + c = 0 ∣ z − c ∣ \lvert z - c \rvert ∣ z − c ∣ c c c 1 + i 1 + i 1 + i π 4 \dfrac{\pi}{4} 4 π 2 \sqrt{2} 2 − i -i − i ( 1 + i ) ( − i ) = 1 − i (1 + i)(-i) = 1 - i ( 1 + i ) ( − i ) = 1 − i 2 2 2\sqrt{2} 2 2 AB \fbox{AB} AB − i -i − i G \fbox{G} G H I \fbox{H}\sqrt{\fbox{I}} H I
日文
读音
中文
例句或说明
複素数平面
ふくそすうへいめん
复平面
「複素数平面上の点 z z z
実軸
じつじく
实轴
虚軸
きょじく
虚轴
共役複素数
きょうやくふくそすう
共轭复数
「z ‾ \overline{z} z z z z
絶対値
ぜったいち
模（绝对值）
「z z z ∣ z ∣ \lvert z \rvert ∣ z ∣
実数
じっすう
实数
「z n z^n z n
純虚数
じゅんきょすう
纯虚数
「z − 1 z + 1 \dfrac{z - 1}{z + 1} z + 1 z − 1
実部
じつぶ
实部
虚部
きょぶ
虚部
「α \alpha α
極形式
きょくけいしき
极形式（三角形式）
「z z z
偏角
へんかく
辐角
「偏角 θ \theta θ 0 ≦ θ < 2 π 0 \leqq \theta < 2\pi 0 ≦ θ < 2 π
ド・モアブルの定理
ど・もあぶるのていり
棣莫弗定理
「ド・モアブルの定理により」
1 の n n n
いちのえぬじょうこん
1 的 n n n
「1 の 5 乗根のうち 1 でないもの」
n n n えぬじょうこん
n n n 「8 i 8i 8 i
回転
かいてん
旋转
「原点のまわりに θ \theta θ
〜を中心として
〜をちゅうしんとして
以……为中心
「点 α \alpha α π 3 \dfrac{\pi}{3} 3 π
正の向き
せいのむき
正方向（逆时针）
「正の向きに π 2 \dfrac{\pi}{2} 2 π
負の向き
ふのむき
负方向（顺时针）
拡大
かくだい
放大
「原点からの距離を 2 倍に拡大した点」
内分点
ないぶんてん
内分点
「線分 AB を 2 : 1 2 : 1 2 : 1
外分点
がいぶんてん
外分点
重心
じゅうしん
重心
外接円
がいせつえん
外接圆
「△ O A B \triangle\mathrm{OAB} △ OAB
一直線上にある
いっちょくせんじょうにある
共线
「3 点 A，B，C が一直線上にある」
垂直
すいちょく
垂直
「A B ⊥ A C \mathrm{AB} \perp \mathrm{AC} AB ⊥ AC
2 直線のなす角
にちょくせんのなすかく
两直线的夹角
「∠ B A C \angle\mathrm{BAC} ∠ BAC
正三角形
せいさんかくけい
正三角形
直角二等辺三角形
ちょっかくにとうへんさんかくけい
等腰直角三角形
中文语序相反
二等辺三角形
にとうへんさんかくけい
等腰三角形
形状
けいじょう
形状
「△ O A B \triangle\mathrm{OAB} △ OAB
垂直二等分線
すいちょくにとうぶんせん
垂直平分线
「線分 AB の垂直二等分線」
アポロニウスの円
あぽろにうすのえん
阿波罗尼斯圆
距離の比が一定の点の軌跡
軌跡
きせき
轨迹
〜を満たしながら動く
〜をみたしながらうごく
满足……而运动
「点 z z z ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1
〜が描く図形
〜がえがくずけい
……描出的图形
「点 w w w
〜上を動く
〜じょうをうごく
在……上运动
「点 z z z ∣ z − 1 ∣ = 1 \lvert z - 1 \rvert = 1 ∣ z − 1 ∣ = 1
最小の自然数
さいしょうのしぜんすう
最小的自然数
「z n z^n z n n n n
複号同順
ふくごうどうじゅん
正负号按顺序对应
± \pm ± ∓ \mp ∓
∣ α ∣ = 1 \lvert \alpha \rvert = 1 ∣ α ∣ = 1 ∣ β ∣ = 2 \lvert \beta \rvert = 2 ∣ β ∣ = 2 ∣ α + β ∣ = 3 \lvert \alpha + \beta \rvert = \sqrt{3} ∣ α + β ∣ = 3 ∣ α − β ∣ \lvert \alpha - \beta \rvert ∣ α − β ∣ ∣ z ∣ = 2 \lvert z \rvert = 2 ∣ z ∣ = 2 z z z a a a z + 4 z z + \dfrac{4}{z} z + z 4 a a a 把 − 1 − 3 i -1 - \sqrt{3}i − 1 − 3 i 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π
α = 2 ( cos π 12 + i sin π 12 ) \alpha = 2\left(\cos\dfrac{\pi}{12} + i\sin\dfrac{\pi}{12}\right) α = 2 ( cos 12 π + i sin 12 π ) β = 2 ( cos π 3 + i sin π 3 ) \beta = \sqrt{2}\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right) β = 2 ( cos 3 π + i sin 3 π ) α β \dfrac{\alpha}{\beta} β α a + b i a + bi a + bi 计算 ( 3 − i ) − 4 (\sqrt{3} - i)^{-4} ( 3 − i ) − 4
z = 1 + i 3 + i z = \dfrac{1 + i}{\sqrt{3} + i} z = 3 + i 1 + i z n z^n z n n n n z n z^n z n 解方程 z 4 = − 8 + 8 3 i z^4 = -8 + 8\sqrt{3}i z 4 = − 8 + 8 3 i
α = cos 2 π 7 + i sin 2 π 7 \alpha = \cos\dfrac{2\pi}{7} + i\sin\dfrac{2\pi}{7} α = cos 7 2 π + i sin 7 2 π α + α 2 + α 3 + α 4 + α 5 + α 6 \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 α + α 2 + α 3 + α 4 + α 5 + α 6 正方形 ABCD 的顶点按逆时针排列，A ( 1 + i ) \mathrm{A}(1 + i) A ( 1 + i ) B ( 4 + 2 i ) \mathrm{B}(4 + 2i) B ( 4 + 2 i )
三点 0 0 0 1 + 2 i 1 + 2i 1 + 2 i a + 6 i a + 6i a + 6 i a a a
方程 ∣ z − 2 ∣ = 2 ∣ z + 1 ∣ \lvert z - 2 \rvert = 2\lvert z + 1 \rvert ∣ z − 2 ∣ = 2 ∣ z + 1 ∣
点 z z z ∣ z ∣ = 1 \lvert z \rvert = 1 ∣ z ∣ = 1 w = ( 1 − i ) z + 2 i w = (1 - i)z + 2i w = ( 1 − i ) z + 2 i
答案 ：1. 7 \sqrt{7} 7 2 a 2a 2 a 2 ( cos 4 π 3 + i sin 4 π 3 ) 2\left(\cos\dfrac{4\pi}{3} + i\sin\dfrac{4\pi}{3}\right) 2 ( cos 3 4 π + i sin 3 4 π ) 1 − i 1 - i 1 − i − 1 32 + 3 32 i -\dfrac{1}{32} + \dfrac{\sqrt{3}}{32}i − 32 1 + 32 3 i n = 12 n = 12 n = 12 z 12 = − 1 64 z^{12} = -\dfrac{1}{64} z 12 = − 64 1 3 + i \sqrt{3} + i 3 + i − 1 + 3 i -1 + \sqrt{3}i − 1 + 3 i − 3 − i -\sqrt{3} - i − 3 − i 1 − 3 i 1 - \sqrt{3}i 1 − 3 i − 1 -1 − 1 C ( 3 + 5 i ) \mathrm{C}(3 + 5i) C ( 3 + 5 i ) D ( 4 i ) \mathrm{D}(4i) D ( 4 i ) a = 3 a = 3 a = 3 − 2 -2 − 2 2 i 2i 2 i 2 \sqrt{2} 2
