Edexcel Further Maths Core Pure 2: De Moivre’s Theorem and Roots of Unity — Edexcel进阶数学 Core Pure 2：棣莫弗定理与单位根

一、棣莫弗定理的陈述：从复数的模-幅角形式出发 | De Moivre’s Theorem: Statement from the Modulus-Argument Form

在 Edexcel 进阶数学 Core Pure 2 中，棣莫弗定理（De Moivre’s Theorem）是连接复数代数与三角函数的桥梁。任何一个非零复数都可以写成模-幅角形式 z = r(cos θ + i sin θ)，其中 r = |z| 是模（modulus），θ = arg z 是幅角（argument）。棣莫弗定理告诉我们，对这个形式取 n 次幂时，规则极其简洁：模取 n 次幂，幅角乘以 n。

In Edexcel Further Maths Core Pure 2, De Moivre’s Theorem is the bridge that connects complex algebra with trigonometry. Any non-zero complex number can be written in modulus-argument form z = r(cos θ + i sin θ), where r = |z| is the modulus and θ = arg z is the argument. De Moivre’s Theorem tells us that when we raise this form to the power n, the rule is remarkably clean: raise the modulus to the power n, and multiply the argument by n.

定理的正式表述是：对于任意整数 n，[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)。当 r = 1 时，它退化为 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。这个结果对正整数 n 可以用数学归纳法严格证明，对负整数 n 和零则需要借助倒数与三角函数的奇偶性来推广。理解证明本身，能帮助你记住”模相乘、幅角相加”这个更深层的乘法本质。

The formal statement is: for any integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ). When r = 1, it reduces to (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. This result can be proved rigorously by mathematical induction for positive integers n, and extended to negative integers and zero using reciprocals and the parity properties of sine and cosine. Understanding the proof itself helps you remember the deeper multiplicative essence: moduli multiply, arguments add.

定理之所以重要，是因为它把”乘方”这种看似复杂的运算，转化成了两个独立的、更简单的操作：先对模做实数乘方，再对幅角做整数乘法。这一思想会贯穿 Core Pure 2 的多个考点，从求高次幂、推导三角恒等式，一直到求复数的 n 次方根。

The theorem matters because it turns the seemingly complicated operation of “raising to a power” into two independent, simpler operations: first take a real power of the modulus, then multiply the argument by an integer. This idea runs through several Core Pure 2 topics, from finding high powers and deriving trigonometric identities, all the way to finding the nth roots of a complex number.

二、用棣莫弗定理求复数的幂：一个三步行法 | Raising Complex Numbers to Powers: A Three-Step Method

考试中一个非常常见的题型是：已知 z = 1 + i√3，求 z⁶ 或 z¹⁰。直接二项式展开会非常痛苦，而棣莫弗定理给出了一套标准的三步行法。第一步：把 z 写成模-幅角形式。对 z = 1 + i√3，模 r = √(1² + (√3)²) = 2，幅角 θ = arctan(√3/1) = π/3，因此 z = 2(cos π/3 + i sin π/3)。

A very common exam question is: given z = 1 + i√3, find z⁶ or z¹⁰. Direct binomial expansion would be extremely painful, but De Moivre’s Theorem gives a standard three-step method. Step one: write z in modulus-argument form. For z = 1 + i√3, the modulus is r = √(1² + (√3)²) = 2 and the argument is θ = arctan(√3/1) = π/3, so z = 2(cos π/3 + i sin π/3).

第二步：对模和幅角分别应用定理。z⁶ = 2⁶(cos(6 × π/3) + i sin(6 × π/3)) = 64(cos 2π + i sin 2π)。第三步：把结果化简回笛卡尔形式。因为 cos 2π = 1 且 sin 2π = 0，所以 z⁶ = 64(1 + 0) = 64。这个答案干净漂亮，整个过程不超过一分钟，而二项式展开 z⁶ 却要展开六项再合并，极易出错。

Step two: apply the theorem to the modulus and argument separately. z⁶ = 2⁶(cos(6 × π/3) + i sin(6 × π/3)) = 64(cos 2π + i sin 2π). Step three: simplify the result back to Cartesian form. Since cos 2π = 1 and sin 2π = 0, we get z⁶ = 64(1 + 0) = 64. The answer is clean and beautiful, and the whole process takes under a minute, whereas binomial-expanding z⁶ requires expanding and combining six terms, which is extremely error-prone.

关键技巧在于幅角要处理”转圈”问题。当 nθ 超过 2π 时，cos(nθ) 和 sin(nθ) 会自动给出正确的值，因为三角函数以 2π 为周期。所以即使 z¹⁰ 的幅角是 10π/3，你也无需担心：cos(10π/3) = cos(4π/3)，因为两者相差 2π。养成先把 nθ 减去若干个 2π、落到主值区间 [0, 2π) 再求值的习惯，能避免符号错误。

The key technique is handling the “winding” of the argument. When nθ exceeds 2π, cos(nθ) and sin(nθ) still give the correct values because the trigonometric functions are periodic with period 2π. So even if the argument of z¹⁰ is 10π/3, you need not worry: cos(10π/3) = cos(4π/3) because the two differ by 2π. Get into the habit of subtracting multiples of 2π from nθ to land in the principal range [0, 2π) before evaluating, and you will avoid sign errors.

三、指数形式与欧拉公式：三种表示法的统一 | Exponential Form and Euler’s Formula: Unifying the Three Forms

Core Pure 2 引入了一个更紧凑的记法：指数形式。欧拉公式 e^(iθ) = cos θ + i sin θ 把指数函数与三角函数联系了起来。借助它，模-幅角形式 z = r(cos θ + i sin θ) 可以写成 z = re^(iθ)。这个形式看起来简洁，但在求 n 次方根时威力巨大，因为指数运算的规则可以直接使用。

Core Pure 2 introduces a more compact notation: the exponential form. Euler’s formula e^(iθ) = cos θ + i sin θ links the exponential function to the trigonometric functions. With it, the modulus-argument form z = r(cos θ + i sin θ) can be written as z = re^(iθ). This form looks elegant, but its real power shows when finding nth roots, because the usual rules of exponents apply directly.

三种形式各有所长：笛卡尔形式 z = a + bi 最适合加减法；模-幅角形式最适合乘方与理解几何意义；指数形式最适合求根与书写简洁。例如，(re^(iθ))ⁿ = rⁿe^(inθ)，这一行就完整表达了棣莫弗定理，读者一眼就能看出”模取 n 次幂、幅角乘 n”的规则。考试中你应该能在这三种形式之间快速、准确地转换。

The three forms each have their strengths: the Cartesian form z = a + bi is best for addition and subtraction; the modulus-argument form is best for powers and for understanding geometric meaning; the exponential form is best for finding roots and for concise writing. For example, (re^(iθ))ⁿ = rⁿe^(inθ) expresses De Moivre’s Theorem in a single line, and the reader can see at a glance the rule “raise the modulus to the n, multiply the argument by n”. In the exam you should be able to convert quickly and accurately among all three forms.

一个常见误区是把 e^(iθ) 当成普通的实数指数来”开方”或”取对数”。要注意，幅角 θ 具有多值性：e^(iθ) = e^(i(θ+2πk)) 对任意整数 k 都成立。这个多值性正是下一节求 n 次方根时会产生 n 个不同根的根本原因，也是学生最容易忽略的细节。

A common misconception is treating e^(iθ) like an ordinary real exponent and trying to “take roots” or “take logarithms” carelessly. Note that the argument θ is multi-valued: e^(iθ) = e^(i(θ+2πk)) for any integer k. This multi-valued nature is the very reason why finding nth roots produces n distinct roots, as we will see in the next section, and it is the detail students most often overlook.

四、单位根：解方程 zⁿ = 1 的几何之美 | Roots of Unity: The Geometry of Solving zⁿ = 1

单位根（roots of unity）是方程 zⁿ = 1 的 n 个解。用指数形式求解非常直接：设 z = re^(iθ)，代入 zⁿ = 1 得 rⁿe^(inθ) = 1 = e^(i·2πk)。比较模得到 rⁿ = 1，故 r = 1（模非负）；比较幅角得到 nθ = 2πk，故 θ = 2πk/n，其中 k = 0, 1, …, n-1。

The roots of unity are the n solutions to the equation zⁿ = 1. Solving in exponential form is very direct: let z = re^(iθ), substitute into zⁿ = 1 to get rⁿe^(inθ) = 1 = e^(i·2πk). Comparing moduli gives rⁿ = 1, hence r = 1 (the modulus is non-negative); comparing arguments gives nθ = 2πk, hence θ = 2πk/n, where k = 0, 1, …, n-1.

于是 n 个单位根是 z_k = e^(2πik/n) = cos(2πk/n) + i sin(2πk/n)，k = 0, 1, …, n-1。几何上，它们均匀分布在单位圆上，相邻两根之间的幅角差恒为 2π/n，构成了正 n 边形的顶点。例如 z⁴ = 1 的四个根是 1, i, -1, -i，恰好是单位圆上正方形的四个顶点。这种”旋转对称”的几何图像，是理解单位根求和等于零的关键：n 个对称分布的向量相加，结果自然为零。

The n roots of unity are therefore z_k = e^(2πik/n) = cos(2πk/n) + i sin(2πk/n), for k = 0, 1, …, n-1. Geometrically, they are evenly spaced around the unit circle, with a constant argument difference of 2π/n between consecutive roots, forming the vertices of a regular n-gon. For example, the four roots of z⁴ = 1 are 1, i, -1, -i, which are exactly the four vertices of a square on the unit circle. This “rotational symmetry” picture is the key to understanding why the sum of the roots of unity is zero: n symmetrically distributed vectors add up to zero.

单位根还有两个常考性质。第一，所有 n 个单位根之和为 0，即 1 + ω + ω² + … + ωⁿ⁻¹ = 0（ω 为任意 n 次本原单位根）。第二，单位根成对共轭：cos(2πk/n) + i sin(2πk/n) 与 cos(2πk/n) – i sin(2πk/n) 互为共轭，因此它们的乘积为 1、实部相同、虚部相反。这些性质常与复系数多项式、根的对称性等题目结合考查。

Roots of unity also have two frequently tested properties. First, the sum of all n roots of unity is 0, that is 1 + ω + ω² + … + ωⁿ⁻¹ = 0 (where ω is any primitive nth root of unity). Second, roots of unity come in conjugate pairs: cos(2πk/n) + i sin(2πk/n) and cos(2πk/n) – i sin(2πk/n) are conjugates, so their product is 1, their real parts are equal, and their imaginary parts are opposite. These properties are often combined with questions on complex-coefficient polynomials and symmetry of roots.

五、一般复数的 n 次方根：模开 n 次方、幅角加 2πk 后平分 | nth Roots of a General Complex Number: Root the Modulus, Divide the Argument

把单位根的方法推广到一般复数 w = r(cos θ + i sin θ) 的 n 次方根，是 Core Pure 2 的核心计算题。设根为 z = s(cos φ + i sin φ)，由 zⁿ = w 比较模得 sⁿ = r，故 s = r^(1/n)（取正的 n 次方根）；比较幅角得 nφ = θ + 2πk，故 φ = (θ + 2πk)/n，其中 k = 0, 1, …, n-1。

Generalising the roots-of-unity method to the nth roots of a general complex number w = r(cos θ + i sin θ) is a core calculation in Core Pure 2. Let the root be z = s(cos φ + i sin φ); from zⁿ = w, comparing moduli gives sⁿ = r, so s = r^(1/n) (the positive nth root); comparing arguments gives nφ = θ + 2πk, so φ = (θ + 2πk)/n, where k = 0, 1, …, n-1.

因此 w 的 n 个 n 次方根是 z_k = r^(1/n)[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]，k = 0, 1, …, n-1。记忆口诀是”模开 n 次方，幅角加 2πk 再除以 n”。关键陷阱在于那个 +2πk：许多学生只写出 k = 0 的那一个根，漏掉了其余 n-1 个根。务必记住，方程 zⁿ = w 在复数域内恰好有 n 个根（重根按重数计）。

The n nth roots of w are therefore z_k = r^(1/n)[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], for k = 0, 1, …, n-1. A useful mnemonic is “root the modulus, add 2πk to the argument and divide by n”. The key trap is that +2πk term: many students write only the k = 0 root and miss the other n-1 roots. Always remember that the equation zⁿ = w has exactly n roots over the complex numbers (counting multiplicity).

几何上，这 n 个根都落在以原点为圆心、半径为 r^(1/n) 的圆上，且等间距分布，相邻根之间的幅角差为 2π/n。换句话说，它们把以原点为圆心、半径 r^(1/n) 的圆”均匀分割”成 n 段圆弧。这个几何图像可以用来快速检查答案：如果你算出的几个根没有等距分布在同一个圆上，那一定是哪里算错了。

Geometrically, these n roots all lie on the circle centred at the origin with radius r^(1/n), and are equally spaced, with an argument difference of 2π/n between consecutive roots. In other words, they evenly divide the circle of radius r^(1/n) into n equal arcs. This geometric picture can be used to check your answer quickly: if the roots you calculated are not equally spaced on a single circle, then something has gone wrong.

六、复平面中的轨迹：垂直平分线、圆与半直线 | Loci in the Complex Plane: Perpendicular Bisectors, Circles and Half-Lines

轨迹（loci）问题是 Core Pure 2 的另一个高频考点，考查的是复数的几何意义。最常见的三类轨迹如下。第一类，|z – a| = r 表示以点 a 为圆心、半径为 r 的圆，因为 |z – a| 恰好是 z 到 a 的距离。第二类，|z – a| = |z – b| 表示线段 ab 的垂直平分线，因为它描述的是”到两点距离相等”的点集。

Locus problems are another high-frequency topic in Core Pure 2, testing the geometric meaning of complex numbers. The three most common types of locus are as follows. Type one, |z – a| = r, represents the circle centred at a with radius r, because |z – a| is exactly the distance from z to a. Type two, |z – a| = |z – b|, represents the perpendicular bisector of the segment ab, because it describes the set of points equidistant from two given points.

第三类，arg(z – a) = θ 表示从点 a 出发、与正实轴成角 θ 的一条半直线（不含起点 a 本身）。此外还有区间形式，如 arg(z) 介于两个角之间表示一个扇形区域，|z – a| < r 表示圆内部的区域（不含边界）。理解这些轨迹的关键，是把 |z - a| 读作"距离"、把 arg(z - a) 读作"方向角"。

Type three, arg(z – a) = θ, represents a half-line starting from the point a and making an angle θ with the positive real axis (excluding the starting point a itself). There are also interval forms, such as arg(z) lying between two angles representing a sector region, and |z – a| < r representing the interior of the circle (excluding the boundary). The key to understanding these loci is to read |z - a| as "distance" and arg(z - a) as "direction angle".

典型综合题会要求你先求某条件对应的轨迹，再找出轨迹上的最值点或交点。例如”求满足 |z – 3| = 2 的 z 中，模最大的那个 z”：轨迹是圆心 3、半径 2 的圆，到原点距离最大的点就是圆上离原点最远的点，即 z = 5。把代数条件翻译成几何图像，往往比直接做代数运算更快、更直观。

A typical composite question asks you to first find the locus corresponding to a condition, then find the extremum point or intersection point on that locus. For example, “find the z satisfying |z – 3| = 2 that has the largest modulus”: the locus is the circle centred at 3 with radius 2, and the point farthest from the origin is the point on the circle furthest from the origin, namely z = 5. Translating an algebraic condition into a geometric picture is often faster and more intuitive than doing the algebra directly.

七、考试技巧：Core Pure 2 棣莫弗定理的常见失分点 | Exam Technique: Common Pitfalls with De Moivre’s Theorem in Core Pure 2

第一，幅角主值的选择。arg z 通常取主值区间 (-π, π]，但求 n 次方根时必须回到”一般幅角” θ + 2πk，否则会漏根。第二，忘记模的 n 次方根要用正的实数根 r^(1/n)，而不是带符号的根。第三，三角函数的特殊值记错，例如 cos π/3 = 1/2、sin π/6 = 1/2，这些基本功错误在压轴题里尤其致命。

First, the choice of principal argument. The argument arg z is usually taken in the principal range (-π, π], but when finding nth roots you must return to the “general argument” θ + 2πk, otherwise you will miss roots. Second, forgetting that the nth root of the modulus should be the positive real root r^(1/n), not a signed root. Third, misremembering special trigonometric values, such as cos π/3 = 1/2 and sin π/6 = 1/2; these basic errors are especially fatal in the harder final questions.

第四，用棣莫弗定理推导三角恒等式时的方向选择。典型题型是”用棣莫弗定理把 cos 5θ 表示成 cos θ 的多项式”，方法是展开 (cos θ + i sin θ)⁵ 并取实部；反过来”把 cos⁵θ 表示成 cos θ 的倍角之和”则要用 z + 1/z = 2cos θ 这个代换。两个方向都要熟练。第五，最后答案要按要求的形式呈现，评分标准常要求精确值或根式形式，而不是保留一堆小数。

Fourth, the direction choice when using De Moivre’s Theorem to derive trigonometric identities. A typical question is “use De Moivre’s Theorem to express cos 5θ as a polynomial in cos θ”, done by expanding (cos θ + i sin θ)⁵ and taking the real part; conversely, “express cos⁵θ as a sum of multiple-angle terms in cos θ” uses the substitution z + 1/z = 2cos θ. You should be fluent in both directions. Fifth, present the final answer in the required form; the mark scheme often asks for exact values or surd form rather than a string of decimals.

最后，把 n 次方根写完整。标准写法要明确写出 k = 0, 1, …, n-1 的全体根，并用一句话说明这些根等距分布在半径为 r^(1/n) 的圆上。完整、清晰的表达不仅避免漏解扣分，也能在检查时帮你快速发现计算错误。平时练习时建议逐题画出根在复平面上的位置，养成几何直觉。

Finally, write out the nth roots completely. The standard presentation should explicitly list all the roots for k = 0, 1, …, n-1, and include a sentence noting that these roots are equally spaced on the circle of radius r^(1/n). Complete, clear presentation not only avoids losing marks for missing solutions, but also helps you spot calculation errors quickly when checking. In daily practice, it is recommended to sketch the position of the roots on the complex plane for each problem, to build geometric intuition.

八、用棣莫弗定理推导倍角公式：实部虚部分离法 | Deriving Multiple-Angle Formulas with De Moivre’s Theorem: Separating Real and Imaginary Parts

棣莫弗定理的另一个经典用途是推导三角恒等式。以 cos 3θ 和 sin 3θ 为例，由定理可知 (cos θ + i sin θ)³ = cos 3θ + i sin 3θ。把左边按二项式展开：(cos θ + i sin θ)³ = cos³θ + 3cos²θ(i sin θ) + 3cos θ(i sin θ)² + (i sin θ)³。

Another classic use of De Moivre’s Theorem is deriving trigonometric identities. Take cos 3θ and sin 3θ as an example; the theorem gives (cos θ + i sin θ)³ = cos 3θ + i sin 3θ. Expand the left side using the binomial theorem: (cos θ + i sin θ)³ = cos³θ + 3cos²θ(i sin θ) + 3cos θ(i sin θ)² + (i sin θ)³.

利用 i² = -1 化简，得到 cos³θ + 3i cos²θ sin θ – 3cos θ sin²θ – i sin³θ。现在分别比较实部与虚部：实部给出 cos 3θ = cos³θ – 3cos θ sin²θ；虚部给出 sin 3θ = 3cos²θ sin θ – sin³θ。再用 sin²θ = 1 – cos²θ 代换，就能得到教材中的标准形式 cos 3θ = 4cos³θ – 3cos θ。

Using i² = -1 to simplify, we get cos³θ + 3i cos²θ sin θ – 3cos θ sin²θ – i sin³θ. Now compare the real and imaginary parts separately: the real part gives cos 3θ = cos³θ – 3cos θ sin²θ, and the imaginary part gives sin 3θ = 3cos²θ sin θ – sin³θ. Then substituting sin²θ = 1 – cos²θ yields the standard textbook form cos 3θ = 4cos³θ – 3cos θ.

这个方法的核心思路是”实部虚部分离”：先用棣莫弗定理把一个复数的 n 次方等于另一个复数，再把两边都化成 a + bi 的形式，最后让实部对实部、虚部对虚部。对于更高次的情形，如 cos 5θ，二项式展开会变长，但方法完全相同。掌握这一套路后，任何倍角公式都能自行推导，无需死记硬背。

The core idea of this method is “separating real and imaginary parts”: first use De Moivre’s Theorem to equate a complex number raised to the n with another complex number, then rewrite both sides in a + bi form, and finally match real parts to real parts and imaginary parts to imaginary parts. For higher powers, such as cos 5θ, the binomial expansion grows longer but the method is identical. Once you master this routine, you can derive any multiple-angle formula yourself, with no need to memorise them.

九、完整例题：求 z³ = -8 的全部根 | Worked Example: Finding All Roots of z³ = -8

下面通过一道完整例题巩固整个流程。求方程 z³ = -8 的全部根。第一步，把右边写成模-幅角形式：-8 = 8(cos π + i sin π)，因此 r = 8、θ = π。第二步，套用求根公式 z_k = r^(1/n)[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]，其中 n = 3。

Let us consolidate the whole process with a complete worked example. Find all roots of the equation z³ = -8. Step one: write the right side in modulus-argument form, -8 = 8(cos π + i sin π), so r = 8 and θ = π. Step two: apply the root formula z_k = r^(1/n)[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], with n = 3.

先算模的立方根：8^(1/3) = 2。于是 z_k = 2[cos((π + 2πk)/3) + i sin((π + 2πk)/3)]。分别取 k = 0, 1, 2：当 k = 0 时，z₀ = 2(cos π/3 + i sin π/3) = 2(1/2 + i√3/2) = 1 + i√3；当 k = 1 时，z₁ = 2(cos π + i sin π) = 2(-1 + 0) = -2；当 k = 2 时，z₂ = 2(cos 5π/3 + i sin 5π/3) = 2(1/2 – i√3/2) = 1 – i√3。

First compute the cube root of the modulus: 8^(1/3) = 2. Hence z_k = 2[cos((π + 2πk)/3) + i sin((π + 2πk)/3)]. Taking k = 0, 1, 2 in turn: for k = 0, z₀ = 2(cos π/3 + i sin π/3) = 2(1/2 + i√3/2) = 1 + i√3; for k = 1, z₁ = 2(cos π + i sin π) = 2(-1 + 0) = -2; for k = 2, z₂ = 2(cos 5π/3 + i sin 5π/3) = 2(1/2 – i√3/2) = 1 – i√3.

因此 z³ = -8 的三个根是 1 + i√3、-2、1 – i√3。验证一下：这三个根都落在以原点为圆心、半径为 2 的圆上，且相邻两根的幅角差都是 2π/3，均匀分布，符合”n 次方程有 n 个等距分布的根”的几何规律。在答题纸上完整写出这三个根并配上一句几何说明，就是满分作答的标准。

The three roots of z³ = -8 are therefore 1 + i√3, -2, and 1 – i√3. As a check, all three roots lie on the circle centred at the origin with radius 2, and consecutive roots differ in argument by 2π/3, so they are evenly spaced, consistent with the geometric rule that an nth-degree equation has n equally spaced roots. Writing out these three roots in full, together with a one-sentence geometric remark, is the standard for a full-mark answer.

Summary | 总结

本文围绕 Edexcel 进阶数学 Core Pure 2 的棣莫弗定理与单位根主题，系统梳理了核心方法与考点。棣莫弗定理 [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) 把复数的乘方拆成”模取 n 次幂、幅角乘 n”两个独立操作，是求高次幂和推导三角恒等式的利器。

This article has systematically organised the core methods and exam points around De Moivre’s Theorem and roots of unity in Edexcel Further Maths Core Pure 2. De Moivre’s Theorem, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ), splits raising a complex number to a power into two independent operations: raising the modulus to the nth power and multiplying the argument by n. It is a powerful tool for finding high powers and deriving trigonometric identities.

欧拉公式 e^(iθ) = cos θ + i sin θ 引出了指数形式 z = re^(iθ)，它与笛卡尔形式、模-幅角形式共同构成三种等价表示。求 n 次方根时，牢记”模开 n 次方、幅角加 2πk 后除以 n”的规则，就能完整写出 k = 0, 1, …, n-1 的 n 个根，它们等距分布在半径为 r^(1/n) 的圆上。

Euler’s formula e^(iθ) = cos θ + i sin θ leads to the exponential form z = re^(iθ), which together with the Cartesian form and the modulus-argument form makes three equivalent representations. When finding nth roots, remember the rule “root the modulus, add 2πk to the argument and divide by n”, and you can write out all n roots for k = 0, 1, …, n-1, equally spaced on the circle of radius r^(1/n).

轨迹问题则把代数条件翻译为几何图像：|z – a| = r 是圆，|z – a| = |z – b| 是垂直平分线，arg(z – a) = θ 是半直线。掌握这些对应关系，配合特殊角的三角函数值，就能在考试中又快又稳地完成 Core Pure 2 的复数压轴题。

Locus problems translate algebraic conditions into geometric pictures: |z – a| = r is a circle, |z – a| = |z – b| is a perpendicular bisector, and arg(z – a) = θ is a half-line. With these correspondences in hand, along with the special-angle trigonometric values, you can tackle the complex-number final questions of Core Pure 2 quickly and reliably in the exam.

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