Complex Numbers and De Moivre’s Theorem: A Complete IB AA HL Guide — 复数与棣莫弗定理完全指南

一、什么是复数：从数轴的缺口到复平面 | What Are Complex Numbers: From a Gap on the Number Line to the Complex Plane

在 IB Mathematics Analysis and Approaches HL 课程中，学生第一次遇到一个看似矛盾的问题：x² + 1 = 0 这样的方程在实数范围内没有解，因为任何实数的平方都不可能等于负数。但这个方程在数学和物理学中太重要了，不能简单地宣布”无解”。

In IB Mathematics Analysis and Approaches HL, students first encounter a seemingly contradictory problem: an equation like x² + 1 = 0 has no solution within the real numbers, because the square of any real number can never be negative. Yet this equation is far too important in mathematics and physics to simply declare “no solution”.

数学家给出的答案是引入一个假想的数 i，规定 i² = -1。这个”虚数单位”（imaginary unit）把数系从一维的实数轴扩展到二维的复平面：任何一个复数都可以写成 z = a + bi 的形式，其中 a 是实部（real part），b 是虚部（imaginary part），a 和 b 都是实数。

Mathematicians resolved this by introducing an imaginary number i, defined by the rule i² = -1. This imaginary unit extends the number system from the one-dimensional real number line to the two-dimensional complex plane: every complex number can be written as z = a + bi, where a is the real part, b is the imaginary part, and both a and b are real numbers.

掌握复数的关键是转变思维：不要把它看成”奇怪的数”，而要看成一个二维向量或平面上的一个点。这样看待复数，代数运算和几何直觉就能互相印证，这也是本单元在 AA HL 中如此重要的原因。

The key to mastering complex numbers is a mental shift: do not treat them as “strange numbers”, but as two-dimensional vectors or points in the plane. Seen this way, algebraic operations and geometric intuition reinforce each other, which is exactly why this topic matters so much in AA HL.

二、代数形式与基本运算：实部、虚部、共轭与四则运算 | Algebraic Form and Basic Operations: Real Part, Imaginary Part, Conjugate and Arithmetic

复数的标准形式 z = a + bi 称为代数形式（Cartesian form 或 algebraic form）。两个复数相等当且仅当实部和虚部分别相等，这一条性质是很多方程求解题的基础。

The standard form z = a + bi is called the Cartesian or algebraic form. Two complex numbers are equal if and only if their real parts and imaginary parts are separately equal, a property that underpins many equation-solving problems.

加减法只需分别合并实部和虚部，例如 (3 + 2i) + (4 – 5i) = 7 – 3i。乘法按分配律展开，并利用 i² = -1 化简：(1 + 2i)(3 – i) = 3 – i + 6i – 2i² = 3 + 5i + 2 = 5 + 5i。注意 i² = -1，所以 -2i² 变成了 +2。

Addition and subtraction simply combine real and imaginary parts separately, for example (3 + 2i) + (4 – 5i) = 7 – 3i. Multiplication expands using the distributive law and simplifies with i² = -1: (1 + 2i)(3 – i) = 3 – i + 6i – 2i² = 3 + 5i + 2 = 5 + 5i. Note that i² = -1, so -2i² becomes +2.

除法是本章第一个难点。要计算 (2 + i)/(1 – i)，需要引入共轭（conjugate）概念：复数 z = a + bi 的共轭记作 z̄ = a – bi。任何复数乘以其共轭都得到实数 a² + b²，这个技巧称为”分母有理化”的复数版 – 分子分母同乘分母的共轭，就能把分母变成实数。

Division is the first real challenge of this chapter. To compute (2 + i)/(1 – i), you need the concept of the conjugate: the conjugate of z = a + bi is written z̄ = a – bi. Any complex number multiplied by its conjugate gives the real number a² + b². This trick, the complex analogue of rationalising the denominator, multiplies numerator and denominator by the conjugate of the denominator so the denominator becomes real.

完整过程如下：(2 + i)/(1 – i) = (2 + i)(1 + i)/((1 – i)(1 + i)) = (2 + 2i + i + i²)/(1 + 1) = (1 + 3i)/2 = 1/2 + (3/2)i。共轭还常用于求模长的平方：|z|² = z·z̄，这个公式在后面会反复出现。

The full process is: (2 + i)/(1 – i) = (2 + i)(1 + i)/((1 – i)(1 + i)) = (2 + 2i + i + i²)/(1 + 1) = (1 + 3i)/2 = 1/2 + (3/2)i. The conjugate is also used to compute squared moduli: |z|² = z·z̄, a formula that reappears constantly later in the course.

三、复平面上的几何：模长与幅角 | Geometry on the Argand Plane: Modulus and Argument

把复数 z = a + bi 画在复平面（Argand diagram）上，横轴是实轴，纵轴是虚轴，z 对应坐标为 (a, b) 的点。这个点也可以用极坐标描述：它到原点的距离叫模长（modulus），记作 |z|；它与原点的连线同正实轴之间的夹角叫幅角（argument），记作 arg(z)。

Plot the complex number z = a + bi on the Argand diagram with the real axis horizontal and the imaginary axis vertical; z corresponds to the point (a, b). This point can also be described in polar coordinates: its distance from the origin is the modulus, written |z|, and the angle between the line from the origin and the positive real axis is the argument, written arg(z).

由勾股定理，|z| = √(a² + b²)。幅角由 tan θ = b/a 决定，但必须根据点 (a, b) 所在的象限确定正确的角度，这一点最容易出错。例如 z = -1 + i 落在第二象限，所以 arg(z) = 3π/4，而不是 arctan(-1) 给出的 -π/4。

By Pythagoras’ theorem, |z| = √(a² + b²). The argument is determined by tan θ = b/a, but you must use the quadrant of the point (a, b) to find the correct angle, which is the most common source of error. For example z = -1 + i lies in the second quadrant, so arg(z) = 3π/4, not the -π/4 that arctan(-1) would suggest.

幅角有无限多个取值，彼此相差 2π 的整数倍。其中落在区间 (-π, π] 内的那个称为幅角主值（principal argument），记作 Arg(z)。IB 考试默认要求给出幅角主值，除非题目特别说明。

Arguments have infinitely many values, differing by integer multiples of 2π. The one lying in the interval (-π, π] is called the principal argument, written Arg(z). IB exams expect the principal argument by default unless the question states otherwise.

模长和幅角是连接代数与几何的桥梁：|z – w| 表示复平面上 z 与 w 两点之间的距离，|z – c| = r 表示以 c 为圆心、半径为 r 的圆。这类轨迹问题（locus problems）在 AA HL 考试中经常出现。

Modulus and argument are the bridge between algebra and geometry: |z – w| is the distance between the points z and w in the complex plane, and |z – c| = r represents a circle with centre c and radius r. Locus problems of this kind appear frequently in AA HL exams.

四、极坐标形式：从 (a, b) 到 r(cosθ + i sinθ) | Polar Form: From (a, b) to r(cos θ + i sin θ)

既然复平面上的点既可以用直角坐标 (a, b) 表示，也可以用极坐标 (r, θ) 表示，那么复数也就有两种写法。由 a = r cos θ 和 b = r sin θ，得到极坐标形式 z = r(cos θ + i sin θ)，有时简写为 z = r cis θ。

Since a point in the complex plane can be described by Cartesian coordinates (a, b) or polar coordinates (r, θ), complex numbers have two representations. From a = r cos θ and b = r sin θ we obtain the polar form z = r(cos θ + i sin θ), sometimes abbreviated as z = r cis θ.

两种形式的转换是基本技能。已知 (r, θ) 求 (a, b)：直接代入 a = r cos θ, b = r sin θ。已知 (a, b) 求 (r, θ)：r = √(a² + b²)，θ 由象限和 tan θ = b/a 共同确定。做题时建议画一个小的复平面草图，可以避免大多数符号错误。

Converting between the two forms is a basic skill. Given (r, θ), find (a, b) by substituting a = r cos θ and b = r sin θ. Given (a, b), find (r, θ) using r = √(a² + b²) with θ determined jointly by the quadrant and tan θ = b/a. Sketching a small Argand diagram before calculating avoids most sign errors.

例如 z = 2 – 2i：r = √(4 + 4) = 2√2，点 (2, -2) 在第四象限，θ = -π/4，所以 z = 2√2(cos(-π/4) + i sin(-π/4))。用幅角主值表示时，这就是它的标准极坐标形式。

For example z = 2 – 2i: r = √(4 + 4) = 2√2, the point (2, -2) lies in the fourth quadrant, θ = -π/4, so z = 2√2(cos(-π/4) + i sin(-π/4)). Using the principal argument, this is its standard polar form.

为什么要学极坐标形式？因为复数的乘除和幂运算在极坐标形式下变得极其简单，这正是下一节的内容。记住一条原则：加减用代数形式，乘除幂用极坐标形式，选对工具能让计算量减半。

Why learn polar form? Because multiplication, division and powers of complex numbers become dramatically simpler in polar form, which is the subject of the next section. Remember one principle: add and subtract in Cartesian form, multiply, divide and raise to powers in polar form. Choosing the right tool halves the work.

五、欧拉公式与指数形式：e^(iθ) = cosθ + i sinθ | Euler’s Formula and Exponential Form: e^(iθ) = cos θ + i sin θ

AA HL 的独特之处在于引入欧拉公式：e^(iθ) = cos θ + i sin θ。这个公式把指数函数、三角函数和复数统一在一起，被许多数学家称为”最美的数学公式”的一部分（当 θ = π 时得到 e^(iπ) + 1 = 0）。

A distinctive feature of AA HL is Euler’s formula: e^(iθ) = cos θ + i sin θ. It unifies exponential functions, trigonometric functions and complex numbers, and forms part of what many mathematicians call the most beautiful formula in mathematics (at θ = π it gives e^(iπ) + 1 = 0).

由此复数可以写成指数形式 z = re^(iθ)。指数形式让乘除运算变成指数的加减：z₁z₂ = r₁r₂e^(i(θ₁+θ₂))，z₁/z₂ = (r₁/r₂)e^(i(θ₁-θ₂))。这等价于”模长相乘、幅角相加；模长相除、幅角相减”的口诀。

Complex numbers can then be written in exponential form z = re^(iθ). Exponential form turns multiplication and division into addition and subtraction of exponents: z₁z₂ = r₁r₂e^(i(θ₁+θ₂)) and z₁/z₂ = (r₁/r₂)e^(i(θ₁-θ₂)). This is the rule “multiply moduli and add arguments; divide moduli and subtract arguments”.

考试中欧拉公式的常见考法包括：用 e^(iθ) 快速计算幂，把三角表达式改写成指数形式后化简，以及证明三角恒等式。熟练运用 e^(iθ) 的周期性 e^(i(θ+2π)) = e^(iθ) 也能帮助化简复杂表达式。

Common exam uses of Euler’s formula include computing powers quickly via e^(iθ), rewriting trigonometric expressions in exponential form for simplification, and proving trigonometric identities. The periodicity e^(i(θ+2π)) = e^(iθ) is also useful for simplifying complicated expressions.

注意：e^(iθ) 中 θ 的单位是弧度（radian），不是度。IB 考试中所有复数问题的角度都默认用弧度制，把角度写成度数通常会被判定为错误。

Note that θ in e^(iθ) is measured in radians, not degrees. All angle work in complex number problems in IB exams defaults to radians; writing angles in degrees is usually marked wrong.

六、棣莫弗定理：把幂运算变成乘法 | De Moivre’s Theorem: Turning Powers into Multiplications

棣莫弗定理（De Moivre’s Theorem）是本章的核心定理：对任意整数 n 和任意实数 θ，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。用指数形式写就是 (e^(iθ))ⁿ = e^(inθ)，两者本质相同。

De Moivre’s Theorem is the central theorem of this chapter: for any integer n and any real θ, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). In exponential form this is (e^(iθ))ⁿ = e^(inθ); the two statements are equivalent.

它最直接的应用是计算高次幂。例如计算 (1 + i)⁸：先写极坐标形式 1 + i = √2(cos(π/4) + i sin(π/4))，由棣莫弗定理得 (1 + i)⁸ = (√2)⁸(cos(8·π/4) + i sin(8·π/4)) = 16(cos 2π + i sin 2π) = 16。一个原本需要八次相乘的运算，两行就完成了。

Its most direct application is computing high powers. For example, to compute (1 + i)⁸: first write the polar form 1 + i = √2(cos(π/4) + i sin(π/4)), then by De Moivre’s Theorem (1 + i)⁸ = (√2)⁸(cos(8·π/4) + i sin(8·π/4)) = 16(cos 2π + i sin 2π) = 16. An operation that would require eight multiplications takes just two lines.

定理对负整数同样成立。计算 (1 + i)⁻³ 时，只需把指数换成 -3：(1 + i)⁻³ = (√2)⁻³(cos(-3π/4) + i sin(-3π/4)) = (1/(2√2))(-√2/2 – i√2/2) = -1/4 – (1/4)i。也可以直接用共轭和模长验证这个结果。

The theorem also holds for negative integers. To compute (1 + i)⁻³, simply replace the exponent with -3: (1 + i)⁻³ = (√2)⁻³(cos(-3π/4) + i sin(-3π/4)) = (1/(2√2))(-√2/2 – i√2/2) = -1/4 – (1/4)i. You can verify this result using the conjugate and modulus directly.

解题的标准流程是四步：一，把底数写成极坐标形式，确定 r 和 θ；二，套用 (r cis θ)ⁿ = rⁿ cis(nθ)；三，把 nθ 化简到基本角度（必要时加减 2π 的倍数）；四，若题目要求代数形式，再展开 cos(nθ) 和 sin(nθ) 并化简。

The standard four-step procedure is: first, write the base in polar form, identifying r and θ; second, apply (r cis θ)ⁿ = rⁿ cis(nθ); third, reduce nθ to a basic angle, adding or subtracting multiples of 2π if needed; fourth, if the question asks for Cartesian form, expand cos(nθ) and sin(nθ) and simplify.

七、极坐标下的乘除法：旋转与伸缩的几何意义 | Multiplication and Division in Polar Form: Rotation and Scaling

乘法 z₁z₂ 的几何意义是：把 z₁ 的模长放大 r₂ 倍，再绕原点逆时针旋转 θ₂。也就是说，乘以一个模长为 1 的复数相当于纯旋转，乘以一个正实数相当于纯伸缩。这个视角在轨迹题和图形变换题中非常有用。

The geometric meaning of multiplication z₁z₂ is: scale z₁ by a factor of r₂, then rotate it anticlockwise about the origin by θ₂. Multiplying by a complex number of modulus 1 is a pure rotation, while multiplying by a positive real number is a pure scaling. This viewpoint is very useful in locus and transformation problems.

例如，乘以 i = cis(π/2) 就是把图形逆时针旋转 90°。因此 z 和 iz 在复平面上互相垂直，|z| = |iz|。这类”旋转算子”的题目常以选择题或短答题形式出现，掌握几何意义可以几秒内给出答案。

For example, multiplying by i = cis(π/2) rotates a figure anticlockwise by 90°. Hence z and iz are perpendicular in the complex plane with |z| = |iz|. Questions about such rotation operators often appear as multiple-choice or short-answer items, and understanding the geometry lets you answer them in seconds.

除法是乘法的逆运算：z₁/z₂ 表示把 z₁ 的模长缩小 r₂ 倍，再顺时针旋转 θ₂。特别是 1/z，它把模长取倒数、幅角变号：1/z = (1/r)cis(-θ)。这个性质在证明对称性和简化表达式时经常用到。

Division is the inverse of multiplication: z₁/z₂ scales z₁ down by r₂ and rotates it clockwise by θ₂. In particular, 1/z takes the reciprocal of the modulus and negates the argument: 1/z = (1/r)cis(-θ). This property is often used when proving symmetries and simplifying expressions.

结合几何意义可以快速判断答案的合理性：如果题目说”z 乘以 2i”，那么答案的模长应是原来的 2 倍、方向应旋转 90°，任何不符合这两点的选项都可以直接排除。

Combined with geometric intuition you can quickly sanity-check answers: if a question multiplies z by 2i, the answer should have twice the modulus and be rotated by 90°, so any option violating either condition can be eliminated immediately.

八、单位根：解 z^n = w 的完整方法 | Roots of Unity: Solving z^n = w Completely

方程 zⁿ = w（w ≠ 0）恰好有 n 个不同的复数解，这是代数学基本定理的直接推论。求解步骤分三步：把 w 写成极坐标形式 w = r cis φ；设 z = ρ cis θ，则 ρⁿ = r，nθ = φ + 2kπ；最后令 k = 0, 1, …, n-1 得到全部解。

The equation zⁿ = w (w ≠ 0) has exactly n distinct complex solutions, a direct consequence of the Fundamental Theorem of Algebra. The solution procedure has three steps: write w in polar form w = r cis φ; let z = ρ cis θ, so ρⁿ = r and nθ = φ + 2kπ; finally set k = 0, 1, …, n-1 to obtain all solutions.

以 z³ = 8 为例：8 = 8 cis 0，所以 ρ = 2，θ = 2kπ/3。k = 0 给出 z = 2；k = 1 给出 z = 2 cis(2π/3) = -1 + i√3；k = 2 给出 z = 2 cis(4π/3) = -1 – i√3。三个根均匀分布在以原点为圆心、半径为 2 的圆上，相隔 120°。

Take z³ = 8 as an example: 8 = 8 cis 0, so ρ = 2 and θ = 2kπ/3. For k = 0 we get z = 2; for k = 1, z = 2 cis(2π/3) = -1 + i√3; for k = 2, z = 2 cis(4π/3) = -1 – i√3. The three roots lie evenly spaced on the circle centred at the origin with radius 2, separated by 120°.

n 次单位根（即 zⁿ = 1 的全部解）具有优美的对称性：它们构成正 n 边形的顶点，且全部根之和为 0。这一性质常用于求含单位根的级数和、证明整除性，以及构造多项式因式分解。

The nth roots of unity (all solutions of zⁿ = 1) possess beautiful symmetry: they form the vertices of a regular n-gon, and the sum of all roots is zero. This property is used to evaluate sums involving roots of unity, prove divisibility results, and construct polynomial factorisations.

注意两个易错点：一是不要漏掉 k = 0 的解，二是根的个数总是 n 个，即使 w 的幅角恰好为 0。另外，不同 k 给出的 θ 相差 2π/n，检查答案时确认相邻根的夹角相等即可。

Two pitfalls to note: first, do not omit the solution for k = 0; second, the number of roots is always n, even when the argument of w happens to be 0. Also, successive values of θ differ by 2π/n, so when checking your answers confirm the angles between neighbouring roots are equal.

九、用复数证明三角恒等式：倍角公式的优雅推导 | Deriving Trigonometric Identities with Complex Numbers

AA HL 的一个经典考法是用棣莫弗定理推导倍角公式。展开 (cos θ + i sin θ)ⁿ 有两种方式：左边用棣莫弗定理写成 cos(nθ) + i sin(nθ)，右边用二项式定理展开，然后比较实部和虚部，就得到一组三角恒等式。

A classic AA HL technique is deriving multiple-angle formulae with De Moivre’s Theorem. Expand (cos θ + i sin θ)ⁿ in two ways: on the left use De Moivre’s Theorem to write cos(nθ) + i sin(nθ), on the right expand with the binomial theorem, then compare real and imaginary parts to obtain a family of trigonometric identities.

以 n = 3 为例：(cos θ + i sin θ)³ = cos 3θ + i sin 3θ。展开左边得 cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ。比较实部：cos 3θ = cos³θ – 3 cos θ sin²θ = 4 cos³θ – 3 cos θ；比较虚部：sin 3θ = 3 cos²θ sin θ – sin³θ = 3 sin θ – 4 sin³θ。这就是三倍角公式。

Take n = 3: (cos θ + i sin θ)³ = cos 3θ + i sin 3θ. Expanding the left side gives cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ. Comparing real parts: cos 3θ = cos³θ – 3 cos θ sin²θ = 4 cos³θ – 3 cos θ; comparing imaginary parts: sin 3θ = 3 cos²θ sin θ – sin³θ = 3 sin θ – 4 sin³θ. These are the triple-angle formulae.

这种方法可以推广到任意 n 倍角公式。考试题目通常会给出部分提示，例如”用二项式定理展开并比较实部”，此时只需机械地完成展开和比较即可。注意展开时 i 的幂次规律：i⁰ = 1, i¹ = i, i² = -1, i³ = -i，每四次循环一次。

This method generalises to any n-fold angle formula. Exam questions usually give partial hints, such as “expand using the binomial theorem and compare real parts”, in which case the task is mechanical. Watch the pattern of powers of i during expansion: i⁰ = 1, i¹ = i, i² = -1, i³ = -i, repeating every four powers.

另一个相关技巧是用二项式展开 (cos θ + i sin θ)ⁿ 后把含 i 的项与不含 i 的项分组，直接读出 cos(nθ) 的表达式只含 cos 的偶次幂，sin(nθ) 的表达式只含 sin 的奇次幂，这可以作为验算依据。

A related trick: after expanding (cos θ + i sin θ)ⁿ binomially, group the terms containing i and those without; the expression for cos(nθ) contains only even powers of cos, while sin(nθ) contains only odd powers of sin, which serves as a verification check.

十、IB AA HL 考试题型与解题策略：真题思路拆解 | Exam Question Patterns and Strategies in IB AA HL

AA HL 的复数题目通常以多部分（multi-part）形式出现，由易到难递进。第一问往往是基础计算：把复数写成极坐标形式、求模长和幅角、或化简表达式。这一问分值不大但必须全对，因为它决定后续问题的起点。

Complex number questions in AA HL usually appear in multi-part form, progressing from easy to hard. Part (a) is typically basic computation: writing a complex number in polar form, finding modulus and argument, or simplifying an expression. This part carries few marks but must be fully correct, as it determines the starting point for later parts.

中间问常见类型包括：用棣莫弗定理计算高次幂、求解 zⁿ = w 的全部根、用复数方法证明三角恒等式。解题时先判断题目考察的知识点，再选择对应的工具：求幂和求根用极坐标形式，证明恒等式用二项式展开加比较实虚部。

Middle parts commonly include: computing high powers with De Moivre’s Theorem, finding all roots of zⁿ = w, and proving trigonometric identities using complex numbers. Identify which skill is being tested and choose the matching tool: polar form for powers and roots, binomial expansion with comparison of real and imaginary parts for identities.

压轴问往往把复数与几何结合：例如给定 |z – 3| = |z – i|（这是垂直平分线），或给定 |z| ≤ 2 且 0 ≤ Arg(z) ≤ π/4（这是扇形区域），要求画出区域或求最值。这类题的核心是先把模长等式翻译成几何图形。

Final parts often combine complex numbers with geometry: for example |z – 3| = |z – i|, which is a perpendicular bisector, or |z| ≤ 2 with 0 ≤ Arg(z) ≤ π/4, which is a sector, asking you to sketch the region or find extreme values. The key is translating modulus equations into geometric figures first.

应试策略建议：第一，所有角度统一用弧度；第二，把中间结果保留在极坐标形式直到最后一步，避免过早展开产生计算错误；第三，写完每个小题立即用几何直觉检验答案（例如幂的模长应是 rⁿ，方向应旋转 nθ）。

Exam strategy advice: first, keep all angles in radians; second, retain intermediate results in polar form until the final step to avoid premature expansion errors; third, immediately sanity-check each answer with geometric intuition, for example a power should have modulus rⁿ and direction rotated by nθ.

十一、易错点清单：模长、幅角主值与共轭的陷阱 | Common Pitfalls: Modulus, Principal Argument and Conjugate Traps

第一个高频错误是幅角象限判断失误。计算 arg(z) 时只按 tan θ = b/a 求角度而不看象限，会导致符号错误。稳妥做法是：先画出点 (a, b) 的位置，再决定最终角度是 θ、π – θ、θ + π 还是 -θ。

The first high-frequency error is misjudging the quadrant of the argument. Computing arg(z) from tan θ = b/a alone without considering the quadrant causes sign errors. The safe approach: plot the point (a, b) first, then decide whether the final angle is θ, π – θ, θ + π or -θ.

第二个错误是忘记复数乘以共轭后仍是实数这一性质，在除法中直接约分或猜测答案。第三个错误是把 |z|² 与 z² 混淆：z² 是复数，|z|² 是实数，两者一般不相等（例如 z = 1 + i 时 z² = 2i 而 |z|² = 2）。

The second error is forgetting that a complex number times its conjugate is real, leading to careless cancellation or guessing in division. The third is confusing |z|² with z²: z² is complex while |z|² is real, and they are generally unequal (for z = 1 + i, z² = 2i but |z|² = 2).

第四个错误出现在棣莫弗定理的应用中：只对 (cos θ + i sin θ) 部分取幂，忘记把模长 r 也取 n 次方，或者把 nθ 计算错误。建议把 (r cis θ)ⁿ = rⁿ cis(nθ) 作为完整公式记忆，缺一不可。

The fourth error occurs when applying De Moivre’s Theorem: raising only the (cos θ + i sin θ) part to the power while forgetting to raise the modulus r to the nth power, or miscalculating nθ. Memorise (r cis θ)ⁿ = rⁿ cis(nθ) as one complete formula; neither part can be omitted.

第五个错误涉及方程 zⁿ = w 的根：漏解、角度范围写错、或者把幅角主值误当成唯一解。记住根共有 n 个，均匀分布在圆周上；最后用”相邻根夹角 = 2π/n”来核对。

The fifth error concerns roots of zⁿ = w: missing solutions, writing incorrect angle ranges, or mistaking the principal argument for the only solution. Remember there are n roots evenly spaced on the circle; check using “angle between neighbouring roots = 2π/n”.

Summary | 总结

复数单元是 IB Mathematics Analysis and Approaches HL 的核心内容之一，它将代数、几何与三角学融为一体。本单元的关键线索是”同一种对象，两种表示”：代数形式适合加减和方程求解，极坐标形式（含欧拉公式的指数形式）适合乘除、幂与根。

The complex number unit is one of the core topics of IB Mathematics Analysis and Approaches HL, blending algebra, geometry and trigonometry. The central thread is “one object, two representations”: the Cartesian form suits addition, subtraction and equation solving, while the polar form (including the exponential form via Euler’s formula) suits multiplication, division, powers and roots.

必须掌握的五项核心技能是：一，代数形式的四则运算与共轭技巧；二，模长与幅角（特别是幅角主值）的准确计算；三，棣莫弗定理及其在幂运算中的应用；四，zⁿ = w 全部根的求解；五，用二项式展开比较实虚部证明三角恒等式。

The five core skills you must master are: first, arithmetic in Cartesian form with the conjugate technique; second, accurate computation of modulus and argument, especially the principal argument; third, De Moivre’s Theorem and its use in computing powers; fourth, finding all roots of zⁿ = w; fifth, proving trigonometric identities by binomial expansion and comparing real and imaginary parts.

备考时建议把每种题型各练熟三到五道真题，并坚持用几何直觉检验每个代数结果。掌握了这套”代数形式与极坐标形式自由切换”的能力，复数部分的考试题目就会变得清晰而可控。

When preparing, practise three to five past-paper questions for each question type and always check algebraic results with geometric intuition. Once you can switch fluently between Cartesian and polar forms, the complex number questions in the exam become clear and manageable.

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