De Moivre's Theorem | 棣莫弗定理

📚 De Moivre’s Theorem | 棣莫弗定理

De Moivre’s Theorem is one of the most elegant and powerful results in A-Level Further Mathematics. It connects complex numbers, trigonometry, and exponential growth through a single compact formula, and it is essential for finding powers and roots of complex numbers, deriving trigonometric identities, and solving equations.

棣莫弗定理是 A-Level 进阶数学中最优雅且最强大的结论之一。它通过一个简洁的公式将复数、三角学与指数增长联系起来，是求复数幂与根、推导三角恒等式以及解方程的核心工具。

1. Polar Form of a Complex Number | 复数的极坐标形式

Before stating the theorem, we must express a complex number in polar form. A complex number \( z = a + bi \) can be written as \( z = r(\cos \theta + i\sin \theta) \), where \( r = |z| = \sqrt{a^2 + b^2} \) is the modulus and \( \theta = \arg(z) \) is the argument satisfying \( \tan \theta = \frac{b}{a} \).

在陈述定理之前，我们需要将复数表示为极坐标形式。复数 \( z = a + bi \) 可写成 \( z = r(\cos \theta + i\sin \theta) \)，其中 \( r = |z| = \sqrt{a^2 + b^2} \) 是模长，\( \theta = \arg(z) \) 是辐角，满足 \( \tan \theta = \frac{b}{a} \)。

For example, the complex number \( 1 + i\sqrt{3} \) has modulus 2 and argument \( \frac{\pi}{3} \), so its polar form is \( 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) \).

例如，复数 \( 1 + i\sqrt{3} \) 的模为 2，辐角为 \( \frac{\pi}{3} \)，因此其极坐标形式为 \( 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) \)。

2. Statement of De Moivre’s Theorem | 定理陈述

De Moivre’s Theorem states that for any real number \( n \) and any complex number in polar form \( r(\cos \theta + i\sin \theta) \):

棣莫弗定理指出：对于任意实数 \( n \) 和极坐标形式 \( r(\cos \theta + i\sin \theta) \) 的复数：

\([r(\cos \theta + i\sin \theta)]^n = r^n(\cos n\theta + i\sin n\theta)\)

When \( r = 1 \), the simpler form is \( (\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta \). This holds for integer values of \( n \), and with care, for rational values when finding roots.

当 \( r = 1 \) 时，简化形式为 \( (\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta \)。该等式对整数 \( n \) 成立；在求根时，对有理数 \( n \) 也需谨慎使用。

3. Proof by Induction for Positive Integers | 正整数情形的数学归纳法证明

For positive integers \( n \), the theorem can be proved by induction. The base case \( n = 1 \) is trivial: \( (\cos \theta + i\sin \theta)^1 = \cos \theta + i\sin \theta \).

对于正整数 \( n \)，可用数学归纳法证明。基础情形 \( n = 1 \) 显然成立：\( (\cos \theta + i\sin \theta)^1 = \cos \theta + i\sin \theta \)。

Assume the statement holds for \( n = k \). Then multiply both sides by \( \cos \theta + i\sin \theta \):

假设命题对 \( n = k \) 成立。两边同乘 \( \cos \theta + i\sin \theta \)：

\((\cos \theta + i\sin \theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos \theta + i\sin \theta)\)

Expanding using standard trigonometric addition formulas gives \( \cos(k+1)\theta + i\sin(k+1)\theta \), so the theorem holds for \( n = k+1 \). By induction, it holds for all positive integers.

利用标准三角和角公式展开，可得 \( \cos(k+1)\theta + i\sin(k+1)\theta \)，因此命题对 \( n = k+1 \) 成立。由归纳法，命题对所有正整数成立。

4. Extending to Negative and Rational Powers | 推广到负指数与有理指数

For negative integers, let \( n = -m \). Using \( (\cos \theta + i\sin \theta)^{-1} = \cos \theta – i\sin \theta \), we obtain \( (\cos \theta + i\sin \theta)^{-m} = \cos m\theta – i\sin m\theta = \cos(-m\theta) + i\sin(-m\theta) \).

对于负整数，令 \( n = -m \)。利用 \( (\cos \theta + i\sin \theta)^{-1} = \cos \theta – i\sin \theta \)，可得 \( (\cos \theta + i\sin \theta)^{-m} = \cos m\theta – i\sin m\theta = \cos(-m\theta) + i\sin(-m\theta) \)。

For rational powers such as \( n = \frac{1}{q} \), the theorem generates \( q \) distinct roots. This is because the argument is only defined modulo \( 2\pi \), so we must include \( + 2k\pi \) inside the cosine and sine.

对于有理数幂如 \( n = \frac{1}{q} \)，该定理会产生 \( q \) 个不同的根。因为辐角只在模 \( 2\pi \) 意义下确定，所以必须在余弦和正弦内部添加 \( + 2k\pi \)。

5. Finding Powers of Complex Numbers | 求复数的幂

One of the simplest applications is raising a complex number to an integer power. Convert to polar form, apply De Moivre’s Theorem, then simplify.

最简单的应用之一是将复数提升到整数次幂。先转换为极坐标形式，再应用棣莫弗定理，最后化简。

Example: Compute \( (1 + i)^6 \).

示例：计算 \( (1 + i)^6 \)。

First, \( 1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \). Then:

首先，\( 1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \)。于是：

\((1+i)^6 = (\sqrt{2})^6 \left(\cos\frac{6\pi}{4} + i\sin\frac{6\pi}{4}\right) = 8\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = -8i\)

Notice that \( \cos\frac{3\pi}{2} = 0 \) and \( \sin\frac{3\pi}{2} = -1 \), giving the clean result \( -8i \).

注意 \( \cos\frac{3\pi}{2} = 0 \) 且 \( \sin\frac{3\pi}{2} = -1 \)，得到简洁结果 \( -8i \)。

6. Finding nth Roots of Complex Numbers | 求复数的 n 次方根

To find all \( n \)th roots of a complex number \( z = r(\cos \theta + i\sin \theta) \), we use:

要求复数 \( z = r(\cos \theta + i\sin \theta) \) 的所有 \( n \) 次方根，我们使用：

\(w_k = r^{1/n}\left(\cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, 2, \ldots, n-1\)

Each value of \( k \) gives a different root. These \( n \) roots are evenly spaced around a circle in the complex plane, separated by angles \( \frac{2\pi}{n} \).

每个 \( k \) 值给出一个不同的根。这 \( n \) 个根在复平面上均匀分布在一个圆周上，相邻根之间的夹角为 \( \frac{2\pi}{n} \)。

Example: Find the cube roots of \( -8 \).

示例：求 \( -8 \) 的立方根。

Write \( -8 = 8(\cos \pi + i\sin \pi) \). Then \( r^{1/3} = 2 \), and for \( k = 0, 1, 2 \):

将 \( -8 \) 写成 \( 8(\cos \pi + i\sin \pi) \)。则 \( r^{1/3} = 2 \)，对 \( k = 0, 1, 2 \)：

k	Root
0	\(2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 1 + i\sqrt{3}\)
1	\(2(\cos \pi + i\sin \pi) = -2\)
2	\(2(\cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3}) = 1 – i\sqrt{3}\)

Always check that the arguments of the roots differ by \( \frac{2\pi}{3} \).

始终检查各根的辐角相差 \( \frac{2\pi}{3} \)。

7. Deriving Trigonometric Identities | 推导三角恒等式

De Moivre’s Theorem provides a systematic way to derive multiple-angle identities. By expanding \( (\cos \theta + i\sin \theta)^n \) using the binomial theorem and equating real and imaginary parts, we obtain formulas for \( \cos n\theta \) and \( \sin n\theta \).

棣莫弗定理为推导倍角恒等式提供了一种系统方法。利用二项式定理展开 \( (\cos \theta + i\sin \theta)^n \)，并比较实部和虚部，即可得到 \( \cos n\theta \) 和 \( \sin n\theta \) 的公式。

For \( n = 3 \):

当 \( n = 3 \) 时：

\((\cos \theta + i\sin \theta)^3 = \cos 3\theta + i\sin 3\theta\)

Expanding the left side gives \( \cos^3\theta + 3i\cos^2\theta\sin\theta – 3\cos\theta\sin^2\theta – i\sin^3\theta \). Equating real and imaginary parts:

展开左边得 \( \cos^3\theta + 3i\cos^2\theta\sin\theta – 3\cos\theta\sin^2\theta – i\sin^3\theta \)。比较实部和虚部：

Real part: \( \cos 3\theta = \cos^3\theta – 3\cos\theta\sin^2\theta = 4\cos^3\theta – 3\cos\theta \)
Imaginary part: \( \sin 3\theta = 3\cos^2\theta\sin\theta – \sin^3\theta = 3\sin\theta – 4\sin^3\theta \)

These identities are standard and often appear in exam questions.

这些恒等式是标准结果，常出现在考试题目中。

8. Expressing \(\cos^n\theta\) and \(\sin^n\theta\) | 用倍角形式表示 \(\cos^n\theta\) 和 \(\sin^n\theta\)

Conversely, De Moivre’s Theorem helps express powers of sine and cosine as sums of multiple-angle terms. Let \( z = \cos \theta + i\sin \theta \). Then \( z^{-1} = \cos \theta – i\sin \theta \), so:

反过来，棣莫弗定理帮助将正弦和余弦的幂表示为倍角项之和。令 \( z = \cos \theta + i\sin \theta \)，则 \( z^{-1} = \cos \theta – i\sin \theta \)，于是：

\(\cos \theta = \frac{z + z^{-1}}{2}, \quad \sin \theta = \frac{z – z^{-1}}{2i}\)

Raising these expressions to a power and using the binomial theorem yields identities such as \( \cos^3\theta = \frac{1}{4}(\cos 3\theta + 3\cos\theta) \).

将这些表达式取幂并应用二项式定理，可得到如 \( \cos^3\theta = \frac{1}{4}(\cos 3\theta + 3\cos\theta) \) 的恒等式。

This technique is particularly useful for integrating powers of trigonometric functions and for solving recurrence relations.

这一技巧在求解三角函数幂的积分以及递推关系时特别有用。

9. Solving Polynomial Equations | 解多项式方程

De Moivre’s Theorem is central to solving equations like \( z^n = w \). The roots found using the formula in Section 6 are precisely the solutions. This is why every non-zero complex number has exactly \( n \) distinct nth roots.

棣莫弗定理是求解 \( z^n = w \) 这类方程的核心。第6节公式中求得的根正好是这些方程的解。这就是为什么每个非零复数恰好有 \( n \) 个不同的 \( n \) 次方根。

Example: Solve \( z^4 = 16 \). Write \( 16 = 16(\cos 0 + i\sin 0) \). The four roots are:

示例：解方程 \( z^4 = 16 \)。将 \( 16 \) 写成 \( 16(\cos 0 + i\sin 0) \)。四个根为：

\(z_k = 2\left(\cos\frac{2k\pi}{4} + i\sin\frac{2k\pi}{4}\right), \quad k = 0, 1, 2, 3\)

These evaluate to \( 2, 2i, -2, -2i \). The solutions form a square on the complex plane.

这些值为 \( 2, 2i, -2, -2i \)。解在复平面上形成一个正方形。

10. Summing Trigonometric Series | 三角级数求和

De Moivre’s Theorem can also be used to sum series of the form \( \sum_{r=0}^{n} \cos r\theta \) or \( \sum_{r=0}^{n} \sin r\theta \). Consider the geometric series:

棣莫弗定理还可用于求形如 \( \sum_{r=0}^{n} \cos r\theta \) 或 \( \sum_{r=0}^{n} \sin r\theta \) 的级数和。考虑几何级数：

\(\sum_{r=0}^{n} (\cos \theta + i\sin \theta)^r = \sum_{r=0}^{n} e^{ir\theta} = \frac{1 – e^{i(n+1)\theta}}{1 – e^{i\theta}}\)

Taking real and imaginary parts after rationalising the denominator gives closed forms for the cosine and sine sums.

在有理化分母后取实部和虚部，即可得到余弦和正弦级数和的封闭形式。

11. Common Pitfalls | 常见易错点

Students frequently make the following errors:

学生常犯以下错误：

Forgetting to include \( 2k\pi \) when finding roots. This leads to missing solutions.
Ignoring the modulus. The factor \( r^n \) must be applied separately.
Using degrees inconsistently. Always work in radians unless told otherwise.
Assuming the theorem works for all real \( n \) without considering principal arguments. When \( n \) is not an integer, the result may be multi-valued.

求根时忘记加 \( 2k\pi \)，导致漏解。

忽略模长，未单独处理 \( r^n \) 因子。

角度单位混用，未统一使用弧度制。

未考虑主辐角，就把定理直接用于非整数指数，导致多值问题。

12. Exam Tips and Conclusion | 考试技巧与总结

In AQA A-Level Further Mathematics, De Moivre’s Theorem appears in pure mathematics papers. You should be comfortable converting between cartesian and polar form, applying the theorem to integer powers, and using the root formula carefully.

在 AQA A-Level 进阶数学中，棣莫弗定理出现在纯数试卷中。你需要熟练地在笛卡尔形式与极坐标形式之间转换，将定理应用于整数次幂，并谨慎使用求根公式。

Key revision points:

关键复习要点：

Skill	Technique
Powers	Convert to polar form, use \( r^n(\cos n\theta + i\sin n\theta) \)
Roots	Use \( r^{1/n} \) and \( \frac{\theta + 2k\pi}{n} \)
Identities	Expand \( (\cos \theta + i\sin \theta)^n \) and equate parts
Equations	Find all \( n \) roots and check symmetry

Mastering De Moivre’s Theorem opens the door to deeper topics such as complex exponentials and phasors. Practice drawing roots on an Argand diagram and always verify your results by substitution.

掌握棣莫弗定理为深入学习复指数和相量等更高阶主题打开了大门。练习在阿甘图上画出各个根，并始终通过代入验证结果。

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De Moivre’s Theorem | 棣莫弗定理