Roots of Complex Numbers | 复数的根

📚 Roots of Complex Numbers | 复数的根

Understanding how to find the nth roots of a complex number is a fundamental skill in IB Mathematics Higher Level. It builds on polar form representation and de Moivre’s theorem, allowing us to solve equations like zⁿ = w and interpret solutions geometrically.

理解如何求复数的 n 次根是 IB 数学高级课程中的一项基本技能。它建立在极坐标形式和棣莫弗定理的基础上，使我们能够求解诸如 zⁿ = w 的方程，并从几何角度解释这些解。

1. Revisiting the Polar Form of Complex Numbers | 重温复数的极坐标形式

Any non-zero complex number z = x + iy can be expressed in polar form as z = r(cos θ + i sin θ), where r = √(x² + y²) is the modulus and θ = arg(z) is the argument, typically chosen in (–π, π] for the principal value or any angle differing by multiples of 2π.

任何非零复数 z = x + iy 都可以用极坐标形式表示为 z = r(cos θ + i sin θ)，其中 r = √(x² + y²) 是模，θ = arg(z) 是辐角，主值通常取 (–π, π]，或者任何相差 2π 整数倍的角。

The relationship between the Cartesian and polar forms is given by x = r cos θ, y = r sin θ. The argument satisfies tan θ = y/x, but care must be taken with the quadrant.

直角坐标形式与极坐标形式之间的关系为 x = r cos θ，y = r sin θ。辐角满足 tan θ = y/x，但需注意象限。

z = r cis θ, where cis θ = cos θ + i sin θ

可以用 cis 简写为 z = r cis θ。

2. De Moivre’s Theorem and Powers | 棣莫弗定理与幂运算

De Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). Combined with the polar form, raising a complex number to an integer power becomes straightforward: zⁿ = rⁿ cis(nθ).

棣莫弗定理指出，对于任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。与极坐标形式结合后，复数的整数次幂运算变得简单：zⁿ = rⁿ cis(nθ)。

This theorem is valid for all real n when extended using Euler’s formula, but in the context of finding roots, we only need it for rational n, particularly the reciprocal of an integer.

若使用欧拉公式进行推广，该定理对所有实数 n 均成立，但在求根的背景下，我们只需将其用于有理数 n，尤其是整数的倒数。

(r cis θ)ⁿ = rⁿ cis(nθ)

3. The Problem of Finding Roots | 求根的问题

Given a complex number w ≠ 0, finding the nth roots means determining all complex numbers z such that zⁿ = w. Because the argument of a complex number is periodic with period 2π, the equation yields n distinct solutions.

给定一个非零复数 w，求其 n 次根意味着找出所有满足 zⁿ = w 的复数 z。由于复数的辐角以 2π 为周期，该方程会产生 n 个不同的解。

These roots are derived by taking the nth root of the modulus and dividing the argument (plus all co-terminal angles) by n.

这些根可以通过取模的 n 次方根，并将辐角（以及所有终边相同的角）除以 n 来求得。

4. The General nth Root Formula | n 次根的通项公式

Let w = R cis φ (with R > 0). The n distinct nth roots of w are given by:

设 w = R cis φ（R

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）