Exercise 10B.2: Finding Complex n-th Roots Using De Moivre's Theorem | 练习10B.2：运用棣莫弗定理求复数n次方根

📚 Exercise 10B.2: Finding Complex n-th Roots Using De Moivre's Theorem | 练习10B.2：运用棣莫弗定理求复数n次方根

In IB Mathematics Analysis & Approaches, Exercise 10B.2 challenges students to solve equations of the form zⁿ a positive integer. This topic sits at the heart of complex number theory and requires a solid grasp of polar form, De Moivre' n the Argand plane. This revision article breaks down the method step by step, providing a full worked example, graphical insights, common pitfalls, and practice questions to help you master these exam-style problems.

在IB数学分析与方法课程中，练习10B.2要求学生解形如 zⁿ = w 的方程，其中 w 是复数，n 是正整数。这一主题属于复数理论的核心，要求牢固掌握极坐标形式、棣莫弗定理以及复平面上根的几何意义。本文逐步拆解求解方法，提供完整的例题演示、图形化理解、常见易错点和练习题目，帮助你彻底攻克这类考试题型。

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1. Understanding the Problem: Solving zⁿ = w | 理解问题：解方程 zⁿ = w

Exercise 10B.2 typically presents an equation such as z³ = 8i or z⁴ nown complex number. Unlike real-number equations, a complex equation of degree n always has exactly n solutions, counted with multiplicity. Therefore, we must find all distinct complex roots in Cartesian form a + bi or in polar form re^iθ.

练习10B.2通常会给出类似 z³ = 8i 或 z⁴ = -16 的方程，右边是一个已知复数。与实数方程不同，复数域中n次方程恰好有n个解（计重数）。因此，我们必须求出所有不同的复数根，并写成笛卡尔形式 a + bi 或极坐标形式 re^iθ。

The standard approach is to express w in polar form, then apply De Moivre's Theorem 'in reverse' to extract the nth roots. The solutions are equally spaced around a circle centred at the origin on the complex plane, providing a clear geometric interpretation.

标准解法是先将 w 写成极坐标形式，然后逆向运用棣莫弗定理求出 n 次方根。所有解均匀分布在复平面上以原点为圆心的圆周上，这一几何解释非常直观。

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2. Polar Form of Complex Numbers | 复数的极坐标形式

omplex number w = a + bi can be written in polar form as w = r(cos θ + i sin θ), where r = |w| = √a² + b² modulus and θ = arg(w) is the argument, measured from the positive real axis. Using Euler's formula, this is often abbreviated as w = re^iθ.

在求根之前，我们回顾一下任意复数 w = a + bi 的极坐标表示：w = r(cos θ + i sin θ)，其中 r = |w| = √a² + b² 是模，θ = arg(w) 是辐角（从正实轴开始度量）。根据欧拉公式，这常简写为 w = re^iθ。

For example, if w = 8i, then a = 0, b = 8. The modulus is r = √0² + 8² e the point lies on the positive imaginary axis. Hence, 8i = 8(cos(π/2) + i sin(π/2)) = 8e^iπ/2.

例如，若 w = 8i，则 a = 0，b = 8。模为 r = √0² + 8² = 8，辐角为 θ = π/2（90°），因为该点位于正虚轴上。因此，8i = 8(cos(π/2) + i sin(π/2)) = 8e^iπ/2。

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3. De Moivre's Theorem Statement | 棣莫弗定理的陈述

De Moivre's Theorem states that for any real number θ and integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). In exponential form this is simply (e^iθ)ⁿ = e^inθ. This theorem is the engine behind raising complex numbers to powers and, crucially for our exercise, extracting roots.

棣莫弗定理指出，对任意实数 θ 和整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。用指数形式表示就是 (e^iθ)ⁿ = e^inθ。该定理是计算复数乘方和——对我们练习尤为关键的——开方运算的核心引擎。

When finding roots, we consider the equation zⁿ = w. If we set z = s(cos φ + i sin φ), then by the theorem zⁿ = sⁿ(cos(nφ) + i sin(nφ)). Equating this to w = r(cos θ + i sin θ) gives sⁿ nφ = θ + 2πk, where k is an integer, because sine and cosine are periodic with period 2π.

在求根时，我们考虑方程 zⁿ = w。设 z = s(cos φ + i sin φ)，则由定理得 zⁿ = sⁿ(cos(nφ) + i sin(nφ))。将此式与 w = r(cos θ + i sin θ) 对应，得到 sⁿ = r 以及 nφ = θ + 2πk，其中 k 为整数，因为正弦与余弦函数的周期为 2π。

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4. Deriving the General Formula for n-th Roots | 推导 n 次方根的通式

From sⁿ = r we obtain the principal root s = r^1/n .., n - 1; after that the values repeat. Thus the n distinct n-th roots of w are given by:

由 sⁿ = r 可得主根 s = r^1/n（模的正实 n 次方根）。辐角方程化为 φ = (θ + 2πk)/n，k 为整数。当 k = 0, 1, 2, ..., n - 1 时得到不同的根，此后重复。因此，w 的 n 个不同的 n 次方根为：

z_k = r^1/n [ cos((θ + 2πk)/n) + i sin((θ + 2πk)/n) ] , k = 0, 1, ..., n-1

或指数形式：z_k = r^1/n e^{i(θ + 2πk)/n}

This formula is your most reliable tool for Exercise 10B.2. It tells you that all roots lie on a circle of radius r^1/n, and their arguments are equally spaced by 2π/n radians.

这组公式是解答练习10B.2最可靠的工具。它告诉你，所有根都位于半径为 r^1/n 的圆周上，并且辐角均匀间隔 2π/n 弧度。

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5. Step-by-Step Worked Example (Part 1) | 逐步解答示例（第一部分）

Let's apply this to a typical Exercise 10B.2 problem: Find all complex numbers z such that z³ = 8i. ve, its modulus is 8 and its principal argument is π/2. Thus w = 8(cos(π/2) + i sin(π/2)).

我们用一个典型的练习10B.2题目来演示：求所有满足 z³ = 8i 的复数 z。首先将 8i 写成极坐标形式。如前所述，其模为 8，主辐角为 π/2。因此 w = 8(cos(π/2) + i sin(π/2))。

Here n = 3, so we need the cube roots. The modulus of each root is 8^1/3 = 2. The arguments are (π/2 + 2πk)/3 = π/6 + 2πk/3. We generate roots for k = 0, 1, 2.

此处 n = 3，即求立方根。每个根的模为 8^1/3 = 2。辐角为 (π/2 + 2πk)/3 = π/6 + 2πk/3。分别取 k = 0, 1, 2 生成各根。

For k = 0: φ₀ = π/6. So z₀ = 2(cos(π/6) + i sin(π/6)) = 2(√3/2 + i(1/2)) = √3 + i.

对于 k = 0：φ₀ = π/6。故 z₀ = 2(cos(π/6) + i sin(π/6)) = 2(√3/2 + i(1/2)) = √3 + i。

We continue in the next section to find the remaining two roots and verify our work.

我们将在下一节继续求出另外两根并验证结果。

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6. Step-by-Step Worked Example (Part 2): Finding All Roots | 逐步解答示例（第二部分）：求出所有根

For k = 1: φ₁ = π/6 + 2π/3 = π/6 + 4π/6 = 5π/6. Then z₁ = 2(cos(5π/6) + i sin(5π/6)) = 2(-√3/2 + i(1/2)) = -√3 + i.

对于 k = 1：φ₁ = π/6 + 2π/3 = π/6 + 4π/6 = 5π/6。因此 z₁ = 2(cos(5π/6) + i sin(5π/6)) = 2(-√3/2 + i(1/2)) = -√3 + i。

4π/3 = π/6 + 8π/6 = 9π/6 = 3π/2. So z₂ = 2(cos(3π/2) + i sin(3π/2)) = 2(0 + i(-1)) = -2i.

对于 k = 2：φ₂ = π/6 + 4π/3 = π/6 + 8π/6 = 9π/6 = 3π/2。于是 z₂ = 2(cos(3π/2) + i sin(3π/2)) = 2(0 + i(-1)) = -2i。

The three cube roots of 8i are therefore √3 + i, -√3 + i, and -2i. A quick check: (-2i)³ = -8i³ s. Notice the symmetry: the roots are equally spaced by 120° (2π/3) on a circle of radius 2.

因此，8i 的三个立方根为 √3 + i, -√3 + i 和 -2i。快速验证：(-2i)³ = -8i³ = -8(-i) = 8i，结果正确。注意到根的对称性：它们在半径为 2 的圆周上以 120°（2π/3）均匀分布。

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7. Representing Roots on the Complex Plane | 在复平面上表示根

Plotting the roots often appears as a follow-up question in IB exams. The three roots z₀, z₁, z₂ form the vertices of an equilateral triangle inscribed in a circle of radius 2 centred at the origin. The arguments are π/6, 5π/6, and 3π/2, clearly separated by 2π/3.

在复平面上绘制根经常作为IB考试的后续小问。三个根 z₀, z₁, z₂ 构成一个等边三角形的顶点，内接于以原点为圆心、半径为 2 的圆。辐角分别为 π/6、5π/6 和 3π/2，正好间隔 2π/3。

This geometric insight helps check your work: if the points are not equally spaced or do not lie on the expected circle, you have made an error in the modulus or argument calculation. You can also observe that the sum of all n-th roots is always zero, which is a useful sanity check.

这一几何直觉有助于检验你的答案：如果各点没有均匀分布或不在预期的圆周上，说明模或辐角的计算存在错误。也可观察到所有 n 次方根之和恒为零，这是一个有用的快速检验方法。

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8. Common Pitfalls and How to Avoid Them | 常见错误及避免方法

Pitfall 1: principal argument and find only one root. Always include the full set k = 0, 1, ..., n-1.

常见错误 1：忘记加上 2πk。许多学生只使用了主辐角，结果只得到一个根。切记必须包含 k = 0, 1, ..., n-1 的完整集合。

Pitfall 2: Miscalculating the modulus. When w has a negative real part, make sure to compute the modulus as √a² + b² and not just a. Also remember that the nth root of the modulus is always a positive real number.

常见错误 2：模计算错误。当 w 具有负实部时，务必按 √a² + b² 计算模，而不是简单地取 a。同时记住模的 n 次方根始终为正实数。

Pitfall 3: argument is -π/2 or 3π/2, not π/2. A sketch on the Argand diagram prevents sign mistakes.

常见错误 3：辐角选择错误。根据 w 所在的象限确定正确的主辐角。例如，若 w = -8i，其辐角为 -π/2 或 3π/2，而非 π/2。在复平面草图上标记可以避免符号错误。

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9. Practice Exercise (10B.2-style) | 练习（10B.2类型）

Test your understanding with this question: Solve the equation z⁴ = -16. Follow the same method: express -16 in polar form (modulus 16, argument π), take the fourth root of the modulus ( 16^1/4 ents (π + 2πk)/4 for k = 0, 1, 2, 3. Write your answers in both polar and Cartesian forms.

用下面的问题检验你的理解：解方程 z⁴ = -16。按照同样的方法：将 -16 写成极坐标形式（模 16，辐角 π），求模的四次方根（16^1/4 = 2），再计算辐角 (π + 2πk)/4，k = 0, 1, 2, 3。分别用极坐标形式和笛卡尔形式写出答案。

Try it yourself before looking at the solution below. This problem reinforces the idea that roots are symmetrically arranged, and you should obtain arguments π/4, 3π/4, 5π/4, 7π/4.

在查看下方答案之前请先尝试完成。这道题巩固了根对称分布的思想，你应该得到辐角 π/4、3π/4、5π/4、7π/4。

Solution: The four roots are 2(cos(π/4) + i sin(π/4)) = √2 + i√2, 2(cos(3π/4) + i sin(3π/4)) = -√2 + i√2, 2(cos(5π/4) + i sin(5π/4)) = -√2 - i√2, and 2(cos(7π/4) + i sin(7π/4)) = √2 - i√2. They form a square inscribed in a circle of radius 2.

答案：四个根为 2(cos(π/4) + i sin(π/4)) = √2 + i√2, 2(cos(3π/4) + i sin(3π/4)) = -√2 + i√2, 2(cos(5π/4) + i sin(5π/4)) = -√2 - i√2, 以及 2(cos(7π/4) + i sin(7π/4)) = √2 - i√2。它们构成一个内接于半径为 2 的圆的正方形。

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10. Check Your Understanding | 检查理解

Now challenge yourself with a slightly harder problem: Find all solutions to z⁵ = 1 + i. This involves a non-standard argument. First compute modulus: |1 + i| = √2. Argument: π/4. The modulus of each root is (√2)^1/5 = 2^1/10 ots in exact exponential form and approximate Cartesian form if needed. Be careful with the fractional powers of 2.

现在用一个稍难的问题挑战一下自己：求 z⁵ = 1 + i 的所有解。这涉及非特殊角。首先计算模：|1 + i| = √2。辐角：π/4。每个根的模为 (√2)^1/5 = 2^1/10。辐角为 (π/4 + 2πk)/5 = π/20 + 2πk/5，k = 0,1,2,3,4。请用精确的指数形式表示这些根，必要时给出近似的笛卡尔形式。注意处理好 2 的分数次幂。

er using De Moivre and confirming it returns 1 + i. This self-checking habit will boost your exam confidence.

你可以通过将其中一个根（比如 k = 0）用棣莫弗定理计算 5 次方，验证它是否回到 1 + i，从而检验答案。这种自我检查的习惯将提升你在考试中的信心。

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