📚 Roots of Complex Numbers | 复数的根
s of marks in IB Mathematics Higher Level and in any pre-university course that covers complex numbers. It builds directly on the polar form of a complex number and on de Moivre' y predictable set of n answers that can be plotted, checked and interpreted geometrically.
求复数的 n 次根是 IB 数学高级课程中一项核心且高频考查的技能。它直接建立在复数的极坐标形式与棣莫弗定理之上，能把形如 zⁿ = w 的方程转化为一组数量确定、可以完全预测的 n 个解，并且这些解可以作图、可以验证、还可以从几何上加以解释。
1. Revisiting the Polar Form of Complex Numbers | 重温复数的极坐标形式
Every non-zero complex number z = x + iy can be written in polar form. The modulus r = √x² + y² ckwise from the positive real axis to the line joining the origin to that point. The principal value of the argument is normally taken in the interval (–π, π], although any angle differing from it by a multiple of 2π describes the same point.
任何非零复数 z = x + iy 都可以写成极坐标形式。模 r = √x² + y² 表示点 (x, y) 到原点的距离，辐角 θ = arg z 表示从正实轴逆时针旋转到该点所对应射线的角。辐角的主值通常取在区间 (–π, π] 内，不过任何与它相差 2π 整数倍的角都表示同一个点。
s useful for computation, but it must never be applied blindly: it only determines θ up to a multiple of π, so the quadrant containing the point has to be checked separately before the argument is written down.
直角坐标形式与极坐标形式之间的换算关系是 x = r cos θ 与 y = r sin θ。关系式 tan θ = y/x 在计算中很有用，但绝不能盲目套用：它只能把 θ 确定到相差 π 的整数倍，因此在写出辐角之前，必须单独判断点所在象限。
z = r cis θ, where cis θ = cos θ + i sin θ
Using the abbreviation cis saves a great deal of writing, and it also makes the algebra of powers and roots almost mechanical. A useful habit is to state the modulus and the argument explicitly at the start of every root question, because almost all later errors come from an incorrectly identified argument.
使用缩写 cis 可以省去大量书写，并且让幂与根的运算几乎变成机械操作。一个很好的习惯是：在每道求根题的开头先明确写出模与辐角，因为后面几乎所有的错误都源于辐角判断有误。
2. De Moivre's Theorem and Powers | 棣莫弗定理与幂运算
De Moivre' nθ). Combined with the polar form, this turns the raising of a complex number to a power into two separate, easy operations: raise the modulus to that power, and multiply the argument by that power.
棣莫弗定理指出，对任意整数 n 都有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。与极坐标形式结合之后，复数求幂就被拆成两个简单步骤：模取该次幂，辐角乘以该次幂。
(r cis θ)ⁿ = rⁿ cis(nθ)
most syllabuses, and it remains true for every real n provided that the expression is interpreted using Euler's formula eiθ ing roots, however, we only need the case where n is the reciprocal of a positive integer, together with the fact that the argument is only determined up to a multiple of 2π.
在多数教学大纲中，该定理是针对整数陈述的；而只要借助欧拉公式 eiθ = cos θ + i sin θ 来解释，它对所有实数 n 都成立。不过就求根而言，我们只需要 n 为正整数倒数这一情形，再加上一个关键事实：辐角只能确定到相差 2π 的整数倍。
Negative powers follow from the same rule, since r⁻ⁿ = 1/rⁿ and cis(– on why the polar form is so much more efficient than expanding brackets in Cartesian form.
负整数幂也遵循同一规则，因为 r⁻ⁿ = 1/rⁿ，而 cis(–nθ) 正是 cis(nθ) 的倒数。这正是极坐标形式比在直角坐标下展开括号高效得多的原因。
3. The Problem of Finding Roots | 求根的问题
w. Because the argument of a complex number is periodic with period 2π, this single equation does not have a single solution but exactly n distinct ones.
给定非零复数 w 与整数 n ≥ 2，求 w 的 n 次根，就是找出所有满足 zⁿ = w 的复数 z。由于复数的辐角以 2π 为周期，这个方程不是只有一个解，而是恰好有 n 个不同的解。
The method is a direct reverse of the power rule. Take the positive real nth root of the modulus R, then divide the argument φ by n. To capture all solutions, add every multiple of 2π to φ before dividing, that is, use the angles φ + 2πk for integer k.
方法就是把求幂法则反过来用。先取模 R 的正实数 n 次方根，再把辐角 φ 除以 n。为了得到全部解，要在除以 n 之前把 2π 的所有整数倍加到 φ 上，也就是使用角 φ + 2πk，其中 k 为整数。
by exactly 2π and we return to a root already listed, so it is enough to take k = 0, 1, 2, …, n – 1. This is why an nth root question always has exactly n answers, no more and no fewer.
其中只有有限多个会给出不同的答案。当 k 增加到 n 时，角正好增加 2π，于是又回到了已经列出的根，因此只需取 k = 0, 1, 2, …, n – 1。这就是为什么 n 次根问题永远恰好有 n 个答案，不多也不少。
4. The General nth Root Formula | n 次根的通项公式
ument of w. Then the n distinct nth roots of w are obtained by the following formula, in which k runs through n consecutive integers.
设 w = R cis φ，其中 R > 0，φ 是 w 的任意一个辐角。那么 w 的 n 个不同的 n 次根由下面的公式给出，其中 k 取遍连续的 n 个整数。
zk = ⁿ√R · cis( (φ + 2πk) / n ), k = 0, 1, 2, …, n – 1
In degree measure, which some examination papers still use, the same formula reads zk = ⁿ√R · cis( (φ + 360°k) / n ). The choice of representative angle for φ makes no difference to the final set of roots: adding 2πm to φ merely relabels the roots cyclically.
若使用某些试卷仍然采用的度数制，同一公式写作 zk = ⁿ√R · cis( (φ + 360°k) / n )。φ 取哪一个代表角并不影响最终得到的根集：给 φ 加上 2πm 只会把各个根循环地重新编号。
Two facts should be memorised alongside the formula. First, every root has the same modulus ⁿ√R, so all n roots lie on one circle centred at the origin. Second, consecutive roots differ in argument by exactly 2π/n, so they are equally spaced around that circle.
除公式本身之外，还有两个事实必须记住。第一，所有根的模都等于 ⁿ√R，因此 n 个根都落在以原点为圆心的同一个圆上。第二，相邻两个根的辐角之差恰为 2π/n，因此它们在圆周上均匀分布。
5. Worked Example: The Cube Roots of Unity | 例题：三次单位根
Solve z³ = 1. Writing 1 = 1 cis 0 gives R = 1 and φ = 0, so the three roots are zk = cis(2πk/3) for k = 0, 1, 2.
解方程 z³ = 1。把 1 写成 1 cis 0，得 R = 1、φ = 0，因此三个根为 zk = cis(2πk/3)，k = 0, 1, 2。
The values are 1, cis(2π/3) = –½ + (√3/2)i, and cis(4π/3) = –½ – (√3/2)i. These three points form an equilateral triangle inscribed in the unit circle, with one vertex on the positive real axis.
具体结果为 1、cis(2π/3) = –½ + (√3/2)i 以及 cis(4π/3) = –½ – (√3/2)i。这三个点构成单位圆的内接正三角形，其中一个顶点落在正实轴上。
ntity is used constantly in algebra questions about polynomials with integer coefficients.
把非实根 cis(2π/3) 记作 ω，则另一个非实根就是 ω²，并且三个根满足恒等式 1 + ω + ω² = 0。这个恒等式在涉及整系数多项式的代数题中被反复使用。
6. Worked Example: Square Roots of –3 + 4i | 例题：求 –3 + 4i 的平方根
Find the square roots of w = –3 + 4i. The modulus is R = √((–3)² + 4²) = √25 = 5. Since the point lies in the second quadrant, the principal argument is φ = π – arctan(4/3) ≈ 2.2143 radians.
求 w = –3 + 4i 的平方根。模为 R = √((–3)² + 4²) = √25 = 5。由于该点位于第二象限，主辐角为 φ = π – arctan(4/3) ≈ 2.2143 弧度。
Applying the formula with n = 2 gives zk = √5 (φ + 2πk)/2) for k = 0, 1. Using half-angle identities, cos(φ/2) = √((1 + cos φ)/2) = √((1 – 3/5)/2) = 1/√5 and sin(φ/2) = √((1 – cos φ)/2) = √((1 + 3/5)/2) = 2/√5.
取 n = 2 代入公式得 zk = √5 · cis((φ + 2πk)/2)，k = 0, 1。用半角公式可得 cos(φ/2) = √((1 + cos φ)/2) = √((1 – 3/5)/2) = 1/√5，sin(φ/2) = √((1 – cos φ)/2) = √((1 + 3/5
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