Complex Numbers: Key Topic Mastery | 复数考点精讲

📚 Complex Numbers: Key Topic Mastery | 复数考点精讲

Complex numbers extend the real number system by introducing the imaginary unit i, where i² = −1. In IB AQA Mathematics, mastering complex numbers is essential for solving polynomial equations, understanding transformations in the Argand plane, and applying De Moivre’s theorem. This guide covers the full spectrum of topics from basic algebra to advanced locus problems, with clear bilingual explanations and exam‑style strategies.

复数通过引入虚数单位 i（满足 i² = −1）扩展了实数系统。在 IB AQA 数学中，精通复数是求解多项式方程、理解阿冈平面上变换以及应用棣莫弗定理的关键。本指南涵盖从基础代数到高等轨迹问题的全谱考点，提供清晰的英汉双语解析和考试型策略。

1. Imaginary Unit and Standard Form | 虚数单位与标准形式

A complex number is written as z = a + bi, where a and b are real numbers. a is the real part Re(z), and b is the imaginary part Im(z). The imaginary unit i satisfies i² = −1.

复数写作 z = a + bi，其中 a 和 b 为实数，a 称为实部 Re(z)，b 称为虚部 Im(z)。虚数单位 i 满足 i² = −1。

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

当且仅当实部相等且虚部相等时，两个复数相等。

Pure imaginary numbers are of the form bi (a = 0). Real numbers can be seen as a + 0i.

纯虚数为 bi 形式（a = 0）；实数可看作 a + 0i。

Example: z = 3 − 4i ⇒ Re(z) = 3, Im(z) = −4

示例：z = 3 − 4i ⇒ Re(z) = 3, Im(z) = −4

2. Basic Operations: Addition, Subtraction and Multiplication | 基本运算：加、减与乘法

Add or subtract complex numbers by combining real parts and imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.

复数加减法分别合并实部与虚部：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。

Multiplication uses the distributive law and the rule i² = −1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i.

乘法利用分配律以及 i² = −1：(a + bi)(c + di) = (ac − bd) + (ad + bc)i。

Powers of i cycle every four: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, etc. For large exponents, divide the exponent by 4 and use the remainder.

i 的幂以四为周期循环：i¹ = i, i² = −1, i³ = −i, i⁴ = 1。对大指数，用指数除以 4 取余数即可。

Example: (2 + 3i)(1 − 2i) = 2 − 4i + 3i − 6i² = 2 − i + 6 = 8 − i

示例：(2 + 3i)(1 − 2i) = 2 − 4i + 3i − 6i² = 8 − i

3. Complex Conjugate and Division | 共轭复数与除法

The complex conjugate of z = a + bi is z̅ = a − bi. It reflects z across the real axis. Key properties: z · z̅ = a² + b² (always a non‑negative real number), and (z̅) = z.

z = a + bi 的共轭复数是 z̅ = a − bi，它是 z 关于实轴的反射。关键性质：z · z̅ = a² + b²（恒为非负实数），且 (z̅) = z。

To divide two complex numbers, multiply numerator and denominator by the conjugate of the denominator: (a + bi)/(c + di) = ((a + bi)(c − di))/(c² + d²).

两复数相除时，分子分母同乘以分母的共轭：(a + bi)/(c + di) = ((a + bi)(c − di))/(c² + d²)。

Conjugates distribute over addition, subtraction, multiplication and division: e.g., (z/w)̅ = z̅ / w̅.

共轭对加减乘除满足分配律，例如 (z/w)̅ = z̅ / w̅。

Example: (3 + 2i)/(1 − i) = ((3 + 2i)(1 + i))/(1² + 1²) = (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2 = ½ + (5/2)i

示例：(3 + 2i)/(1 − i) = ½ + (5/2)i

4. Modulus and Argument | 模与辐角

The modulus |z| is the distance from the origin to the point representing z in the Argand diagram: |z| = √(a² + b²). It is always a non‑negative real number.

模 |z| 是阿冈图中代表 z 的点到原点的距离：|z| = √(a² + b²)，恒为非负实数。

The argument arg(z) is the angle θ (in radians) that the line connecting the origin to z makes with the positive real axis. Principal value usually lies in (−π, π] or [0, 2π).

辐角 arg(z) 是原点与 z 的连线与正实轴所成的角 θ（弧度）。主值通常在 (−π, π] 或 [0, 2π)。

For a complex number z = a + bi, θ = arctan(b/a) adjusted to the correct quadrant.

对于 z = a + bi，θ = arctan(b/a) 并根据象限校正。

Example: z = −1 − √3 i ⇒ |z| = √(1 + 3) = 2; arg(z) = −2π/3 (principal value).

示例：z = −1 − √3 i ⇒ |z| = 2; arg(z) = −2π/3（主值）。

5. Polar Form and Exponential Form | 极坐标形式与指数形式

A complex number can be expressed in polar form: z = r(cosθ + i sinθ), where r = |z| and θ = arg(z).

复数可表示为极坐标形式：z = r(cosθ + i sinθ)，其中 r = |z|，θ = arg(z)。

Using Euler’s formula, e^(iθ) = cosθ + i sinθ, we obtain the exponential form: z = r e^(iθ).

利用欧拉公式 e^(iθ) = cosθ + i sinθ，得到指数形式：z = r e^(iθ)。

These forms make multiplication and division straightforward: multiply moduli, add arguments; divide moduli, subtract arguments.

极坐标/指数形式使乘除法变得简单：模相乘，辐角相加；模相除，辐角相减。

If z₁ = r₁ e^(iθ₁) and z₂ = r₂ e^(iθ₂), then z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)).

若 z₁ = r₁ e^(iθ₁), z₂ = r₂ e^(iθ₂)，则 z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))。

6. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem states: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ) for any integer n.

棣莫弗定理：对任意整数 n，(cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)。

It is especially useful for finding powers and roots of complex numbers in polar form. Also applicable for rational n with care about multiple values.

该定理对于在极坐标下求复数的幂和方根极为有用，对有理数 n 也可用，但需注意多值性。

To compute zⁿ, write z in polar form r(cosθ + i sinθ), then zⁿ = rⁿ(cos nθ + i sin nθ).

计算 zⁿ 时，先将 z 化成极坐标形式 r(cosθ + i sinθ)，再得 zⁿ = rⁿ(cos nθ + i sin nθ)。

Example: (1 + i)⁸; 1+i = √2 (cos π/4 + i sin π/4) → (√2)⁸ (cos 2π + i sin 2π) = 16.

示例：(1 + i)⁸ = 16。

7. Roots of Complex Numbers | 复数的根

To find the n‑th roots of a complex number z ≠ 0, solve wⁿ = z. Write z = r(cosθ + i sinθ). The roots are given by wₖ = r^(1/n) [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, …, n−1.

求非零复数 z 的 n 次根，解 wⁿ = z。将 z 写作 r(cosθ + i sinθ)，根的 n 个表达式为 wₖ = r^(1/n) [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]，k = 0, 1, …, n−1。

These n roots are equally spaced on a circle of radius r^(1/n), separated by an angle of 2π/n.

这 n 个根均匀分布在半径为 r^(1/n) 的圆周上，角间距为 2π/n。

The sum of all n‑th roots of any complex number is zero (except for z = 0).

任何非零复数的所有 n 次根之和为零。

Example: Cube roots of 8 (real) are 2, −1 + i√3, −1 − i√3.

示例：8 的立方根为 2, −1 + i√3, −1 − i√3。

8. Polynomial Equations with Complex Roots | 具有复数根的多项式方程

Complex roots of real polynomial equations occur in conjugate pairs. If a + bi is a root, then a − bi is also a root.

实系数多项式的复数根成共轭对出现。若 a + bi 是根，则 a − bi 也是根。

For a quadratic, if the discriminant Δ < 0, the roots are complex conjugates given by (−b ± i√|Δ|)/(2a).

对二次方程，若判别式 Δ < 0，则根为共轭复数 (−b ± i√|Δ|)/(2a)。

Higher-degree polynomials can be factorised using complex roots after known real or complex factors.

已知实根或复根后，可将高次多项式因式分解，利用复根来找到其他因子。

Example: x³ − 4x² + 6x − 4 = 0 has root x = 2 and complex roots 1 ± i.

示例：x³ − 4x² + 6x − 4 = 0 有根 x = 2 和 1 ± i。

9. Loci in the Argand Diagram | 阿冈图中的轨迹

A locus is a set of points satisfying a given condition in the complex plane. Common loci include:

轨迹是满足给定条件的点在复平面上的集合。常见轨迹包括：

|z − a| = r represents a circle centered at a with radius r.

|z − a| = r 表示以 a 为圆心、r 为半径的圆。
|z − a| = |z − b| represents the perpendicular bisector of the segment joining a and b.

|z − a| = |z − b| 表示连接 a 和 b 线段的垂直平分线。
arg(z − a) = θ represents a ray from point a making an angle θ with the positive real direction.

arg(z − a) = θ 表示从点 a 出发、与正实轴成角 θ 的射线。
|z − a| = k|z − b| (k ≠ 1) describes a circle (Apollonius circle).

|z − a| = k|z − b| (k ≠ 1) 表示阿波罗尼奥斯圆。

When sketching loci, identify the type and mark key points with clear indication of included/excluded boundaries.

绘制轨迹时，需识别类型并标记关键点，明确包含或排除的边界。

Example: |z − i| = 2 → circle centre (0,1), radius 2.

示例：|z − i| = 2 → 圆心 (0,1)，半径 2 的圆。

10. Geometric Transformations | 几何变换

Multiplication by a complex number a + bi can be interpreted as a rotation and dilation. If z is multiplied by r e^(iθ), it is scaled by factor r and rotated by angle θ about the origin.

乘以复数 a + bi 可理解为旋转与伸缩。若 z 乘以 r e^(iθ)，则向量被伸缩 r 倍，并绕原点旋转 θ 角。

Addition z + c translates points by vector corresponding to c.

加法 z + c 将点按与 c 对应的向量平移。

Complex conjugation reflects points across the real axis. The transformation z → i z rotates anticlockwise by π/2.

共轭将点关于实轴反射。变换 z → i z 为逆时针旋转 π/2。

Combinations of these operations can model general transformations like reflections, rotations around arbitrary points, and spirals.

这些操作的组合可以模拟一般变换，如绕任意点的反射、旋转和螺旋。

11. Trigonometric Identities via De Moivre | 利用棣莫弗定理证明三角恒等式

De Moivre’s theorem can be used to derive multiple-angle formulas: cos nθ + i sin nθ = (cosθ + i sinθ)ⁿ. Expanding the right-hand side and equating real and imaginary parts yields identities for cos nθ and sin nθ.

棣莫弗定理可用于推导倍角公式：cos nθ + i sin nθ = (cosθ + i sinθ)ⁿ。展开右边并比较实部和虚部，可得 cos nθ 和 sin nθ 的表达式。

For example, using n = 2: (cosθ + i sinθ)² = cos²θ − sin²θ + i(2 sinθ cosθ), giving the double-angle formulas.

例如 n = 2：(cosθ + i sinθ)² = cos²θ − sin²θ + i(2 sinθ cosθ)，恰为二倍角公式。

Expressions like cos⁴θ or sin³θ can be expressed in terms of cosines and sines of multiple angles by writing cosθ = (e^(iθ) + e^(−iθ))/2, sinθ = (e^(iθ) − e^(−iθ))/(2i) and expanding.

利用 cosθ = (e^(iθ) + e^(−iθ))/2 和 sinθ = (e^(iθ) − e^(−iθ))/(2i) 展开，可将 cos⁴θ 或 sin³θ 表示为多倍角余弦/正弦的组合。

12. Binomial Expansions and Summation of Series | 二项式展开与级数求和

Complex numbers simplify the summation of trigonometric series. For instance, to sum Σ cos(kθ) (k = 0 to n), consider the real part of the geometric series Σ e^(ikθ).

复数可以简化三角级数求和。例如，求 Σ cos(kθ) (k = 0 到 n) 时，考虑几何级数 Σ e^(ikθ) 的实部。

The sum is a geometric progression with common ratio e^(iθ). Using the formula for the sum of a geometric series and then taking the real part yields a closed form.

该和为公比 e^(iθ) 的等比数列。利用等比级数求和公式再取实部，即可得到闭合形式。

This method also applies to series with binomial coefficients, where combining (1 + e^(iθ))ⁿ and taking real/imaginary parts helps evaluate sums like Σ C(n,k) cos(kθ) or Σ C(n,k) sin(kθ).

同样可用于含二项式系数的级数，结合 (1 + e^(iθ))ⁿ 并取实/虚部，可计算 Σ C(n,k) cos(kθ) 或 Σ C(n,k) sin(kθ)。

Example: Σ_{k=0}^{n} cos(kθ) = Re[ (1 − e^(i(n+1)θ))/(1 − e^(iθ)) ].

示例：Σ_{k=0}^{n} cos(kθ) = Re[ (1 − e^(i(n+1)θ))/(1 − e^(iθ)) ]。

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Complex Numbers: Key Topic Mastery | 复数 考点精讲