Complex Numbers: Key Exam Points for IB & WJEC | IB WJEC 数学：复数考点精讲

📚 Complex Numbers: Key Exam Points for IB & WJEC | IB WJEC 数学：复数考点精讲

Complex numbers extend the real number system by introducing the imaginary unit i, where i² = -1. They are essential for solving equations that have no real solutions, and they form a core part of the IB and WJEC further mathematics syllabi. Mastery of complex numbers opens the door to elegant proofs, geometric transformations, and a deeper understanding of polynomial equations.

复数通过引入虚数单位 i（满足 i² = -1）扩展了实数系统。它们是求解无实数解方程的关键，也是 IB 和 WJEC 进阶数学大纲的核心部分。掌握复数知识将为精巧的证明、几何变换以及对多项式方程的深入理解打开大门。

1. Definition and the Imaginary Unit | 定义与虚数单位

A complex number is any number of the form z = a + bi, where a and b are real numbers, and i is the imaginary unit defined by i² = -1. The real part is Re(z) = a, and the imaginary part is Im(z) = b. If b = 0, z is purely real; if a = 0 and b ≠ 0, z is purely imaginary.

复数是指形如 z = a + bi 的数，其中 a 和 b 是实数，i 是虚数单位，满足 i² = -1。实部为 Re(z) = a，虚部为 Im(z) = b。若 b = 0，则 z 为实数；若 a = 0 且 b ≠ 0，则 z 为纯虚数。

The set of all complex numbers is denoted by C. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This allows us to equate real and imaginary parts when solving equations.

全体复数构成的集合记为 C。两个复数相等当且仅当它们的实部相等且虚部相等。这一性质使我们能够在解方程时分别令实部和虚部相等。

2. Algebraic Form and the Complex Conjugate | 代数形式与共轭复数

Working with complex numbers in algebraic form a + bi is straightforward. Addition and subtraction are performed component-wise: (a+bi) ± (c+di) = (a±c) + (b±d)i. Multiplication uses the distributive law and the identity i² = -1, yielding (a+bi)(c+di) = (ac – bd) + (ad + bc)i.

用代数形式 a + bi 处理复数非常直接。加法和减法按分量进行：(a+bi) ± (c+di) = (a±c) + (b±d)i。乘法利用分配律和恒等式 i² = -1，得到 (a+bi)(c+di) = (ac – bd) + (ad + bc)i。

The complex conjugate of z = a + bi is z̄ = a – bi. Geometrically it is a reflection across the real axis. Conjugation preserves addition and subtraction, swaps sign in multiplication, and allows division: (a+bi)/(c+di) = ((a+bi)(c-di)) / (c²+d²). Important identities include z·z̄ = a²+b² = |z|², and z + z̄ = 2a, z – z̄ = 2bi.

z = a + bi 的共轭复数为 z̄ = a – bi。它在几何上表示关于实轴的反射。共轭运算保持加减法，在乘法中交换符号，并可用于除法：(a+bi)/(c+di) = ((a+bi)(c-di)) / (c²+d²)。重要的恒等式包括 z·z̄ = a²+b² = |z|²，以及 z + z̄ = 2a，z – z̄ = 2bi。

3. Modulus and Argument | 模与幅角

The modulus of a complex number z = a + bi is its distance from the origin in the complex plane: |z| = √(a² + b²). The modulus is always a non-negative real number, and it satisfies |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0).

复数 z = a + bi 的模是它在复平面上到原点的距离：|z| = √(a² + b²)。模总是非负实数，并且满足 |z₁z₂| = |z₁||z₂|，|z₁/z₂| = |z₁|/|z₂|（z₂ ≠ 0）。

The argument of z, denoted arg(z), is the angle θ that the vector representing z makes with the positive real axis, measured in radians. The principal argument, denoted Arg(z), is usually taken in the interval (-π, π] or [0, 2π) depending on the syllabus. To find θ, use tanθ = b/a, but always consider the quadrant in which z lies. The argument is not unique: arg(z) = Arg(z) + 2kπ for any integer k.

z 的幅角，记作 arg(z)，是表示 z 的向量与正实轴之间的夹角 θ，以弧度为单位。主值通常记作 Arg(z)，根据大纲要求，一般取区间 (-π, π] 或 [0, 2π)。求 θ 时可用 tanθ = b/a，但务必考虑 z 所在的象限。幅角不唯一：arg(z) = Arg(z) + 2kπ，其中 k 为任意整数。

Quadrant	Sign of a, b	Arg(z) adjustment
I	a>0, b>0	θ = arctan(b/a)
II	a<0, b>0	θ = π – arctan(\|b/a\|)
III	a<0, b<0	θ = -π + arctan(\|b/a\|) or π + arctan(b/a) (adjust range)
IV	a>0, b<0	θ = -arctan(\|b/a\|) or 2π – arctan(\|b/a\|)

象限表有助于快速确定主幅角。务必遵循考试局要求的主值范围。

4. Polar and Exponential Forms | 极坐标形式与指数形式

Using modulus r and argument θ, a complex number can be expressed in polar form: z = r(cosθ + i sinθ). This representation is invaluable for multiplication, division, and exponentiation. Euler’s formula, e^(iθ) = cosθ + i sinθ, gives the compact exponential form: z = r e^(iθ).

使用模 r 和幅角 θ，复数可表示为极坐标形式：z = r(cosθ + i sinθ)。这种表示法对乘法、除法和乘方极为有用。欧拉公式 e^(iθ) = cosθ + i sinθ 则给出了紧凑的指数形式：z = r e^(iθ)。

Conversion between forms is standard: given a+bi, compute r = √(a²+b²) and θ accordingly. Given r and θ, a = r cosθ, b = r sinθ. The exponential form makes multiplication and division particularly elegant: r₁e^(iθ₁) · r₂e^(iθ₂) = r₁r₂ e^(i(θ₁+θ₂)), and similarly for division we divide the moduli and subtract the arguments.

不同形式之间的转换是常规操作：给定 a+bi，计算 r = √(a²+b²) 并确定相应的 θ。给定 r 和 θ，则 a = r cosθ，b = r sinθ。指数形式使乘除法特别优雅：r₁e^(iθ₁) · r₂e^(iθ₂) = r₁r₂ e^(i(θ₁+θ₂))，类似地，除法则是模相除、幅角相减。

A key identity to remember is e^(iπ) + 1 = 0, which elegantly links five fundamental mathematical constants.

需要牢记的关键恒等式是 e^(iπ) + 1 = 0，它优美地连接了五个基本数学常数。

5. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem states that for any integer n, (cosθ + i sinθ)^n = cos(nθ) + i sin(nθ). Combined with the polar form, it becomes (r(cosθ + i sinθ))^n = r^n (cos(nθ) + i sin(nθ)). This theorem is fundamental for computing powers of complex numbers and for deriving multiple-angle trigonometric formulas.

棣莫弗定理指出，对于任意整数 n，有 (cosθ + i sinθ)^n = cos(nθ) + i sin(nθ)。结合极坐标形式可得 (r(cosθ + i sinθ))^n = r^n (cos(nθ) + i sin(nθ))。该定理是计算复数乘方和推导倍角三角公式的基础。

In exponential form, the result is immediate: (re^(iθ))^n = r^n e^(inθ). The theorem can be extended to rational exponents (with care for multiple values) and is used to find roots of complex numbers.

在指数形式下，结果变得一目了然：(re^(iθ))^n = r^n e^(inθ)。该定理可推广到有理指数（需注意多值性），并用于求复数的方根。

When applying De Moivre’s theorem, ensure the argument is expressed in the correct range for the principal value if required, and remember that for negative powers you can still use the same rule with n being a negative integer.

应用棣莫弗定理时，如需主值，请确保幅角处于正确的范围内，并注意对于负整数指数，同样可以套用该法则，其中 n 为负整数。

6. Roots of Complex Numbers | 复数的方根

Finding the n-th roots of a complex number z = r(cosθ + i sinθ) leads to n distinct roots given by: z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], for k = 0, 1, 2, …, n-1. Geometrically these roots lie on a circle of radius r^(1/n) and are equally spaced by an angle of 2π/n radians.

求复数 z = r(cosθ + i sinθ) 的 n 次方根会得到 n 个不同的根：z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]，其中 k = 0, 1, 2, …, n-1。这些根在复平面上位于半径为 r^(1/n) 的圆上，且彼此间隔角度 2π/n。

The roots of unity are the special case when z = 1. The n-th roots of unity are cos(2kπ/n) + i sin(2kπ/n), and they sum to zero. They form a regular n-gon on the unit circle. This symmetric property is often tested in IB and WJEC exams.

单位根是当 z = 1 时的特例。1 的 n 次方根为 cos(2kπ/n) + i sin(2kπ/n)，它们的总和为零，并在单位圆上构成正 n 边形。这种对称性质在 IB 和 WJEC 考试中经常出现。

Example: The cube roots of 8 are found by writing 8 = 8(cos0 + i sin0). Then the roots are 2, 2(cos(2π/3) + i sin(2π/3)), 2(cos(4π/3) + i sin(4π/3)), which simplify to 2, -1 + i√3, -1 – i√3.

示例：求 8 的立方根，先将 8 写为 8(cos0 + i sin0)，则根为 2、2(cos(2π/3) + i sin(2π/3))、2(cos(4π/3) + i sin(4π/3))，化简得 2、-1 + i√3、-1 – i√3。

7. Solving Equations with Complex Numbers | 复数方程求解

Quadratic equations with real coefficients may have a negative discriminant, yielding complex conjugate solutions. For example, x² + 4x + 5 = 0 gives x = -2 ± i. The factorisation x² + 4x + 5 = (x+2)² + 1 illustrates completing the square.

实系数的二次方程当判别式为负时，会产生共轭复数解。例如 x² + 4x + 5 = 0 的解为 x = -2 ± i。通过配方法可得到因式分解 x² + 4x + 5 = (x+2)² + 1。

For equations with complex coefficients, separate real and imaginary parts to form two simultaneous real equations. Alternatively, treat the unknown as z = x + iy and equate corresponding parts. This technique is essential for solving equations like (1+i)z + (2-i)z̄ = 3.

对于含有复系数的方程，可分别令实部和虚部相等，转化为两个实数方程组。另一种思路是设未知数 z = x + iy，再令对应部分相等。这一技巧对于求解如 (1+i)z + (2-i)z̄ = 3 的方程至关重要。

Higher-degree polynomial equations can often be solved by noting that if a+bi is a root of a real polynomial, then its conjugate a-bi is also a root. This symmetry allows you to find the remaining roots or factor the polynomial into real quadratic factors.

更高次的多项式方程通常可以利用实系数多项式的性质：若 a+bi 是其根，则其共轭 a-bi 也是根。这种对称性可以帮助找到其余根，或将多项式分解为实二次因式。

8. The Complex Plane and Geometric Interpretations | 复平面与几何解释

The complex plane (Argand diagram) represents z = x + iy as the point (x, y). Addition of complex numbers corresponds to vector addition. Multiplication by a complex number can be seen as a rotation and a dilation. Specifically, multiplying by re^(iθ) scales by r and rotates by θ counterclockwise.

复平面（Argand 图）将 z = x + iy 表示为点 (x, y)。复数加法对应向量加法。乘以一个复数可以看作一次旋转和一次伸缩。具体来说，乘以 re^(iθ) 会将模伸缩为原来的 r 倍，并逆时针旋转 θ 角。

This geometric insight makes it easy to describe loci. For instance, |z – z₀| = r describes a circle centred at z₀ with radius r. The set |z – a| = |z – b| is the perpendicular bisector of the segment joining a and b. An argument condition arg(z – z₀) = α describes a ray from z₀ at angle α.

这种几何直观使得描述轨迹变得简单。例如 |z – z₀| = r 表示以 z₀ 为圆心、r 为半径的圆；|z – a| = |z – b| 表示连接 a 与 b 的线段的垂直平分线；幅角条件 arg(z – z₀) = α 描述一条从 z₀ 出发、与正实轴成 α 角的射线。

Transformations such as translation, scaling, and rotation can be expressed compactly with complex functions: adding w translates by w; multiplying by k (real) scales by k; multiplying by e^(iθ) rotates by θ.

诸如平移、缩放和旋转等变换可以用复变函数简洁表示：加上 w 即平移 w；乘以实数 k 即缩放 k 倍；乘以 e^(iθ) 即旋转 θ 角。

9. Complex Roots of Real Polynomials | 实系数多项式的复数根

For any polynomial with real coefficients, non-real complex roots always occur in conjugate pairs. That is, if p+qi is a root, then p-qi is also a root. This property is a direct consequence of the conjugate symmetry of the real coefficients.

对于任何实系数多项式，非实的复数根总是成对共轭出现。也就是说，若 p+qi 是一个根，则 p-qi 也是一个根。这一性质是实系数共轭对称性的直接结果。

Consequently, any real polynomial of odd degree must have at least one real root. You can use known complex roots to construct real quadratic factors, simplifying polynomial long division or partial fractions. For instance, if a cubic has one real root and two complex conjugate roots, it can be factored into a linear factor and an irreducible quadratic.

因此，任何奇数次实系数多项式至少有一个实根。你可以利用已知的复根构造实二次因式，从而简化多项式长除或部分分式。例如，一个三次多项式若有一个实根和两个共轭复根，则可分解为一个一次因式和一个不可约二次因式的乘积。

In exam problems, you may be asked to find all roots of a polynomial given one complex root. Use the conjugate root theorem to identify the second complex root, then multiply the corresponding factors to obtain a real quadratic divisor, and finally divide the polynomial to find remaining roots.

在考试中，可能会给出一个复根，要求找出多项式的所有根。利用共轭根定理确定第二个复根，然后将对应的因式相乘得到实二次因式，最后通过多项式除法求出其余根。

10. Euler’s Formula and Further Applications | 欧拉公式及其应用

Euler’s formula, e^(iθ) = cosθ + i sinθ, is a cornerstone that links trigonometry with complex exponentials. It enables derivations of trigonometric identities such as cosⁿθ and sinⁿθ expressed in multiple angles. For example, using (e^(iθ))^n = e^(inθ) and expanding via binomial theorem yields expressions for cos nθ and sin nθ.

欧拉公式 e^(iθ) = cosθ + i sinθ 是连接三角学与复指数的基石。它可以用来推导三角恒等式，例如将 cosⁿθ 和 sinⁿθ 表示为倍角形式。例如，利用 (e^(iθ))^n = e^(inθ) 并结合二项式定理展开，即可得到 cos nθ 和 sin nθ 的表达式。

The exponential form also makes evaluating integrals like ∫ e^(ax) cos(bx) dx simpler by noting that the integrand is the real part of e^((a+ib)x). This technique is not always examined in IB but is part of further methods; however, the core idea of representing trigonometric functions via complex exponentials is fundamental.

指数形式还能使诸如 ∫ e^(ax) cos(bx) dx 的积分计算变得简单，只需将被积函数视为 e^((a+ib)x) 的实部。虽然这一技巧不一定在 IB 中考查，但它属于进阶方法；而通过复指数表示三角函数的核心思想则是基础且重要的。

Overall, Euler’s formula enriches the understanding of the interplay between algebra, geometry, and trigonometry, and it provides a powerful toolkit for solving a wide range of problems in the complex number topic.

总的来说，欧拉公式加深了对代数、几何和三角之间相互作用的理解，并提供了解决复数主题中各类问题的有力工具。

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Complex Numbers: Key Exam Points for IB & WJEC | IB WJEC 数学：复数 考点精讲