De Moivre's Theorem | 棣莫弗定理

📚 De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem is a powerful result in complex numbers that connects powers and roots to trigonometric functions. It states that for any real number n and any angle θ, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. This theorem is essential in A-Level mathematics for simplifying complex powers, finding roots, and deriving trigonometric identities.

棣莫弗定理是复数领域中的一项重要成果，它将复数的幂与根和三角函数联系起来。该定理指出：对于任意实数 n 和任意角度 θ，有 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。在 A-Level 数学中，这一定理对于简化复数幂、求根以及推导三角恒等式都至关重要。

1. Statement of the Theorem | 定理表述

For any integer n and any real angle θ, we have:

对于任意整数 n 和任意实数角 θ，有：

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

Here i is the imaginary unit with i² = −1. The expression cos θ + i sin θ is often abbreviated as cis θ, so the theorem becomes (cis θ)ⁿ = cis nθ.

其中 i 是虚数单位，满足 i² = −1。表达式 cos θ + i sin θ 常简写为 cis θ，因此定理可写作 (cis θ)ⁿ = cis nθ。

Although the theorem initially applies to integer powers, it also works for rational powers when we are careful about selecting roots. In A-Level questions, n is usually an integer or a simple fraction.

虽然该定理最初适用于整数次幂，但在小心选择根时，它也对有理数次幂成立。在 A-Level 考试中，n 通常是整数或简单分数。

2. Proof by Induction | 归纳法证明

For positive integers n, De Moivre’s theorem can be proved using mathematical induction. The base case n = 1 is obviously true:

对于正整数 n，可以使用数学归纳法证明棣莫弗定理。基本情形 n = 1 显然成立：

(cos θ + i sin θ)¹ = cos 1θ + i sin 1θ

Assume the result holds for n = k, so (cos θ + i sin θ)ᵏ = cos kθ + i sin kθ. Multiply both sides by (cos θ + i sin θ):

假设当 n = k 时结论成立，即 (cos θ + i sin θ)ᵏ = cos kθ + i sin kθ。两边乘以 (cos θ + i sin θ)：

(cos θ + i sin θ)ᵏ⁺¹ = (cos kθ + i sin kθ)(cos θ + i sin θ)

Expanding using the addition formulae gives cos(kθ + θ) + i sin(kθ + θ) = cos(k+1)θ + i sin(k+1)θ. Thus the statement holds for n = k+1. By induction, it is true for all positive integers n.

利用加法公式展开，得到 cos(kθ + θ) + i sin(kθ + θ) = cos(k+1)θ + i sin(k+1)θ。因此当 n = k+1 时结论也成立。由归纳法可知，该结论对所有正整数 n 成立。

For negative integers, we use the fact that (cis θ)⁻¹ = cos θ − i sin θ = cis(−θ), so the theorem extends naturally.

对于负整数，利用 (cis θ)⁻¹ = cos θ − i sin θ = cis(−θ) 的性质，定理自然推广到负数情形。

3. Connection with Euler’s Formula | 与欧拉公式的联系

De Moivre’s theorem is closely related to Euler’s formula e^{iθ} = cos θ + i sin θ. If we raise both sides to the power n, we get e^{inθ} = (cos θ + i sin θ)ⁿ. But e^{inθ} is also equal to cos nθ + i sin nθ. Comparing the two expressions proves De Moivre’s theorem.

棣莫弗定理与欧拉公式 e^{iθ} = cos θ + i sin θ 密切相关。将欧拉公式两边同时取 n 次幂，得到 e^{inθ} = (cos θ + i sin θ)ⁿ。另一方面，e^{inθ} 也等于 cos nθ + i sin nθ。比较两个表达式即可证明棣莫弗定理。

e^{iθ} = cos θ + i sin θ ⇒ (e^{iθ})ⁿ = e^{inθ} = cos nθ + i sin nθ

This connection is useful because Euler’s formula is often quicker to apply in problems involving exponential forms of complex numbers.

这种联系在涉及复数指数形式的问题中非常有用，因为使用欧拉公式往往更加快捷。

4. Raising Complex Numbers to Powers | 复数的幂运算

Given a complex number z = r(cos θ + i sin θ), De Moivre’s theorem gives:

对于复数 z = r(cos θ + i sin θ)，棣莫弗定理给出：

zⁿ = rⁿ (cos nθ + i sin nθ)

To use this, first write the complex number in modulus–argument form. The modulus r is the distance from the origin, and the argument θ is the angle from the positive real axis.

使用该公式时，首先要把复数写成模–辐角形式。模 r 是到原点的距离，辐角 θ 是与正实轴之间的夹角。

Example: Find (1 + i)⁵. First, 1 + i = √2 (cos 45° + i sin 45°). Then (1 + i)⁵ = (√2)⁵ cis(5 × 45°) = 4√2 cis 225° = 4√2(−√2/2 − i√2/2) = −4 − 4i.
示例：求 (1 + i)⁵。首先，1 + i = √2 (cos 45° + i sin 45°)。于是 (1 + i)⁵ = (√2)⁵ cis(5 × 45°) = 4√2 cis 225° = 4√2(−√2/2 − i√2/2) = −4 − 4i。

Remember to reduce angles by 360° when needed to find the principal argument.

注意在需要时减去 360° 以求得辐角主值。

5. Finding Roots of Complex Numbers | 复数的根

De Moivre’s theorem also gives a method for finding all n-th roots of a complex number. If z = r cis θ, then the n distinct n-th roots are:

棣莫弗定理还提供了一种求复数所有 n 次根的方法。若 z = r cis θ，则 n 个互异的 n 次根为：

z^{1/n} = r^{1/n} cis((θ + 2πk)/n) for k = 0, 1, …, n−1

Here k takes each integer from 0 to n−1, giving exactly n distinct roots. All roots lie on a circle of radius r^{1/n} in the complex plane, equally spaced by angles 2π/n.

这里的 k 从 0 取到 n−1，恰好给出 n 个不同的根。所有根位于复平面中半径为 r^{1/n} 的圆上，并且以 2π/n 的角度间隔均匀分布。

For example, to find the cube roots of 1, set z = 1 = 1 cis 0. The roots are cis(2πk/3) for k = 0, 1, 2, giving 1, cis(2π/3) = −1/2 + i√3/2, and cis(4π/3) = −1/2 − i√3/2.
例如，求 1 的立方根。设 z = 1 = 1 cis 0。根为 cis(2πk/3)，k = 0, 1, 2，得到 1，cis(2π/3) = −1/2 + i√3/2 以及 cis(4π/3) = −1/2 − i√3/2。

6. Solving Polynomial Equations | 解多项式方程

The n-th roots of a complex number are exactly the solutions to the equation zⁿ = w, where w is a given complex number. For example, solving z³ = 8 means finding the cube roots of 8.

复数的 n 次根恰好是方程 zⁿ = w 的解，其中 w 是给定的复数。例如，解 z³ = 8 就是求 8 的立方根。

Write w in modulus–argument form: 8 = 8 cis 0. Then the three solutions are:

将 w 写成模–辐角形式：8 = 8 cis 0。于是三个解为：

z = 2 cis(2πk/3), k = 0, 1, 2

Expanding these gives z = 2, z = −1 + i√3, and z = −1 − i√3. You can verify that each satisfies z³ = 8.

展开得到 z = 2，z = −1 + i√3 以及 z = −1 − i√3。可以验证每个都满足 z³ = 8。

In AQA A-Level questions, you may be asked to solve equations like zⁿ = k or (z + a)ⁿ = b. The general approach is the same: isolate z, convert to polar form, then apply the root formula.

在 AQA A-Level 考试中，可能会要求解形如 zⁿ = k 或 (z + a)ⁿ = b 的方程。一般方法相同：先化简出 z，再化为极坐标形式，最后应用求根公式。

7. Expressing cos nθ and sin nθ | 表示 cos nθ 与 sin nθ

By expanding (cos θ + i sin θ)ⁿ using the binomial theorem and equating real and imaginary parts with cos nθ + i sin nθ, we can express cos nθ and sin nθ as polynomials in cos θ and sin θ.

利用二项式定理展开 (cos θ + i sin θ)ⁿ，并将实部、虚部分别与 cos nθ + i sin nθ 比较，就可以将 cos nθ 和 sin nθ 表示为 cos θ 和 sin θ 的多项式。

For n = 2:

当 n = 2 时：

(cos θ + i sin θ)² = cos²θ − sin²θ + i(2 sinθ cosθ)

Comparing with cos 2θ + i sin 2θ gives cos 2θ = cos²θ − sin²θ and sin 2θ = 2 sinθ cosθ.

与 cos 2θ + i sin 2θ 比较，得到 cos 2θ = cos²θ − sin²θ 以及 sin 2θ = 2 sinθ cosθ。

For n = 3, expanding gives cos 3θ = 4 cos³θ − 3 cosθ and sin 3θ = 3 sinθ − 4 sin³θ. These are standard results often required in exam questions.

当 n = 3 时，展开得到 cos 3θ = 4 cos³θ − 3 cosθ 以及 sin 3θ = 3 sinθ − 4 sin³θ。这些是考试中常要求推导的标准结果。

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

De Moivre’s theorem is a powerful tool for deriving identities involving multiple angles. A typical method is to write cos nθ as the real part of (cos θ + i sin θ)ⁿ and then expand using the binomial theorem.

棣莫弗定理是推导多倍角三角恒等式的一种强大工具。典型方法是：将 cos nθ 写成 (cos θ + i sin θ)ⁿ 的实部，然后用二项式定理展开。

Example: Express tan 4θ in terms of tan θ.

示例：用 tan θ 表示 tan 4θ。

Write (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ. Expanding:

写出 (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ。展开：

cos 4θ = cos⁴θ − 6 cos²θ sin²θ + sin⁴θ, sin 4θ = 4 cos³θ sinθ − 4 cosθ sin³θ

Then tan 4θ = sin 4θ / cos 4θ. Dividing numerator and denominator by cos⁴θ gives a rational expression in tan θ.

于是 tan 4θ = sin 4θ / cos 4θ。分子分母同时除以 cos⁴θ，即可得到关于 tan θ 的有理表达式。

This technique can be generalised to any integer n, and often appears in A-Level further algebra questions.

这种技巧可以推广到任意整数 n，在 A-Level 的代数综合题中经常出现。

9. Summing Trigonometric Series | 三角级数求和

De Moivre’s theorem can be combined with geometric series to sum expressions like C + iS = Σ (cis θ)ⁿ. This is useful for finding sums of cos nθ or sin nθ.

棣莫弗定理可以与等比数列结合，用于求和形如 C + iS = Σ (cis θ)ⁿ 的表达式。这对于求 cos nθ 或 sin nθ 的和很有用。

Consider the sum S = 1 + cos θ + cos 2θ + … + cos nθ. Let T = sin θ + sin 2θ + … + sin nθ. Then S + iT is a geometric series with common ratio cis θ:

考虑和式 S = 1 + cos θ + cos 2θ + … + cos nθ。设 T = sin θ + sin 2θ + … + sin nθ。则 S + iT 是公比为 cis θ 的等比数列：

S + iT = 1 + cis θ + cis 2θ + … + cis nθ = (1 − cis(n+1)θ) / (1 − cis θ)

By simplifying the right-hand side using half-angle identities, we can extract the real part S and the imaginary part T.

利用半角公式化简右端，就可以分离出实部 S 和虚部 T。

This method shows the deep connection between complex numbers and trigonometry, and is a favourite for challenging exam problems.

这一方法体现了复数与三角函数之间的深刻联系，也是考试中常见的高难度题型。

10. Geometric Interpretation | 几何意义

Multiplication by cis θ corresponds to a rotation by θ in the complex plane. Therefore, raising a complex number to the power n rotates its argument by nθ and scales its modulus by rⁿ.

在复平面中，乘以 cis θ 相当于旋转角度 θ。因此，将复数取 n 次幂会使辐角旋转 nθ，并使模长放大为 rⁿ。

The roots of a complex number form a regular n-gon on a circle centred at the origin. For example, the fifth roots of unity are the vertices of a regular pentagon.

复数的 n 次根在圆心为原点的圆周上构成正 n 边形。例如，1 的五次根是正五边形的五个顶点。

Understanding this geometry helps you check your algebraic answers: the roots must be equally spaced and lie on the correct circle.

理解这种几何关系有助于检验代数答案：根必须等间距分布且位于正确的圆上。

11. Common Exam Pitfalls | 常见考试陷阱

Students often make mistakes with angles when applying De Moivre’s theorem. Always convert the argument to the principal range, usually −180° < θ ≤ 180° or −π < θ ≤ π, after taking powers or roots.

学生在应用棣莫弗定理时常在角度上出错。无论是取幂还是求根之后，都要将辐角化为辐角主值范围，通常是 −180° < θ ≤ 180° 或 −π < θ ≤ π。

Forgetting to include all n roots: the formula requires k = 0, 1, …, n−1, not just k = 0.
忘记包含全部 n 个根：公式要求 k = 0, 1, …, n−1，而不仅仅是 k = 0。
Using degrees in one part and radians in another: stick to the unit specified in the question.
有的地方用度，有的地方用弧度：必须全程使用题目要求的单位。
Confusing modulus and argument when converting between cartesian and polar forms.
在直角坐标与极坐标之间转换时混淆模与辐角。
When using binomial expansion, forgetting the imaginary unit i in terms like i³ = −i.
使用二项式展开时，忘记虚数单位 i，例如 i³ = −i。

Be especially careful when n is fractional. The theorem gives the principal root; other roots are obtained by adding multiples of 2π/n.

特别要注意当 n 为分数时，定理给出的是主根；其余根需要加上 2π/n 的整数倍来获得。

12. Practice Questions | 练习

Try these questions to test your understanding:

尝试以下问题来检验你的理解：

1. Simplify (cos 30° + i sin 30°)¹².
1. 化简 (cos 30° + i sin 30°)¹²。
2. Find all solutions of z⁴ = −16.
2. 求方程 z⁴ = −16 的所有解。
3. Express cos 5θ as a polynomial in cos θ.
3. 将 cos 5θ 表示为 cos θ 的多项式。
4. Show that sin 3θ = 3 sinθ − 4 sin³θ using De Moivre’s theorem.
4. 利用棣莫弗定理证明 sin 3θ = 3 sinθ − 4 sin³θ。
5. Solve z³ + 8 = 0 and plot the roots on an Argand diagram.
5. 解方程 z³ + 8 = 0，并在阿尔冈图上标出各根。

Answers: 1) cos 360° + i sin 360° = 1. 2) z = √2 cis(π/4 + kπ/2), k = 0,1,2,3. 3) cos 5θ = 16 cos⁵θ − 20 cos³θ + 5 cosθ. 4) Expand (cos θ + i sin θ)³ and take the imaginary part. 5) z = −2, 1 ± i√3.

答案：1) cos 360° + i sin 360° = 1。2) z = √2 cis(π/4 + kπ/2)，k = 0,1,2,3。3) cos 5θ = 16 cos⁵θ − 20 cos³θ + 5 cosθ。4) 展开 (cos θ + i sin θ)³ 并取虚部。5) z = −2，1 ± i√3。

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De Moivre’s Theorem | 棣莫弗定理