Euler's Formula | 欧拉公式

📚 Euler’s Formula | 欧拉公式

Euler’s formula is one of the most elegant and powerful results in mathematics, linking exponential functions, complex numbers, and trigonometry in a single equation. In this article, we will explore its derivation, interpretations, and applications that are essential for A-Level Mathematics.

欧拉公式是数学中最优雅且强大的结论之一，它将指数函数、复数和三角学统一在一个简洁的等式中。本文将探讨其推导、含义以及在 A-Level 数学中的关键应用。

1. The Formula and Its Significance | 公式及其意义

Euler’s formula states that for any real number θ, e^(iθ) = cos θ + i sin θ. Here, i is the imaginary unit defined by i² = −1.

欧拉公式表明，对于任意实数 θ，有 e^(iθ) = cos θ + i sin θ，其中 i 是虚数单位，满足 i² = −1。

This single equation connects the exponential function with trigonometric functions through complex numbers. It provides a bridge between algebra and geometry, turning multiplication and powers of complex numbers into simpler operations.

这一个等式通过复数将指数函数与三角函数联系起来，在代数与几何之间架起桥梁，使复数的乘法和幂运算变得更加简单。

2. Derivation Using Maclaurin Series | 借助麦克劳林级数推导

We can derive Euler’s formula by using the Maclaurin series expansions for e^x, sin x, and cos x.

我们可以利用 e^x、sin x 和 cos x 的麦克劳林级数展开来推导欧拉公式。

Recall the series:

回忆这些级数：

e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + …

sin x = x − x³/3! + x⁵/5! − x⁷/7! + …

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …

Replace x by iθ in the exponential series:

在指数级数中将 x 替换为 iθ：

e^(iθ) = 1 + iθ + (iθ)²/2! + (iθ)³/3! + (iθ)⁴/4! + (iθ)⁵/5! + …

Since i² = −1, i³ = −i, i⁴ = 1, i⁵ = i, the series becomes:

因为 i² = −1，i³ = −i，i⁴ = 1，i⁵ = i，级数变为：

e^(iθ) = (1 − θ²/2! + θ⁴/4! − …) + i(θ − θ³/3! + θ⁵/5! − …)

The real part matches cos θ and the imaginary part matches sin θ, so e^(iθ) = cos θ + i sin θ.

实部恰好是 cos θ，虚部恰好是 sin θ，因此 e^(iθ) = cos θ + i sin θ。

3. The Special Case: e^(iπ) + 1 = 0 | 特例：e^(iπ) + 1 = 0

Setting θ = π in Euler’s formula gives e^(iπ) = cos π + i sin π = −1 + 0i = −1.

在欧拉公式中令 θ = π，得到 e^(iπ) = cos π + i sin π = −1 + 0i = −1。

Adding 1 to both sides yields the famous identity e^(iπ) + 1 = 0, which combines five fundamental constants: 0, 1, e, i, and π.

两边同时加上 1，便得到著名恒等式 e^(iπ) + 1 = 0，它将五个基本常数 0、1、e、i 和 π 联系在一起。

This identity is often regarded as a symbol of mathematical beauty. It is also a quick way to evaluate powers of e^(iθ) at special angles.

这个恒等式常被视为数学之美的象征。它也是快速计算特殊角度下 e^(iθ) 取值的简便工具。

4. Complex Numbers in Polar Form | 复数的极坐标形式

Euler’s formula gives a compact polar representation of complex numbers. A complex number with modulus r and argument θ can be written as:

欧拉公式给出了复数的紧凑极坐标表示。模长为 r、辐角为 θ 的复数可以写为：

z = r e^(iθ) = r (cos θ + i sin θ)

This is often called the exponential form of a complex number. It is especially useful when multiplying or dividing complex numbers.

这常被称为复数的指数形式，在复数乘除运算时特别有用。

If z₁ = r₁ e^(iθ₁) and z₂ = r₂ e^(iθ₂), then:

若 z₁ = r₁ e^(iθ₁)，z₂ = r₂ e^(iθ₂)，则：

z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)), z₁/z₂ = (r₁/r₂) e^(i(θ₁−θ₂))

Multiplying complex numbers therefore corresponds to multiplying their moduli and adding their arguments.

因此，复数相乘对应模长相乘、辐角相加。

5. De Moivre’s Theorem from Euler’s Formula | 从欧拉公式导出棣莫弗定理

Raising both sides of Euler’s formula to the power n gives:

将欧拉公式两边同时取 n 次幂，得到：

(e^(iθ))ⁿ = e^(i nθ) = cos(nθ) + i sin(nθ)

But also, (e^(iθ))ⁿ = (cos θ + i sin θ)ⁿ. Therefore:

而同时 (e^(iθ))ⁿ = (cos θ + i sin θ)ⁿ，因此：

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

This is De Moivre’s theorem, which is extremely useful for deriving multiple-angle identities and finding powers and roots of complex numbers.

这就是棣莫弗定理，它对于推导倍角公式以及求复数的幂和根极为有用。

6. Expressing sin θ and cos θ with Exponentials | 用指数表达 sin θ 和 cos θ

From Euler’s formula we can also write:

由欧拉公式我们还可以写出：

e^(iθ) = cos θ + i sin θ, e^(−iθ) = cos θ − i sin θ

Adding and subtracting these two equations gives:

将两式相加、相减得到：

cos θ = (e^(iθ) + e^(−iθ))/2, sin θ = (e^(iθ) − e^(−iθ))/(2i)

These identities allow integrals and differential equations involving sine and cosine to be converted into exponential problems, which are often easier to handle.

这些恒等式允许我们将在包含正弦和余弦的积分或微分方程转换成指数形式的问题，通常更容易处理。

7. Deriving Trigonometric Addition Formulae | 推导三角函数的和角公式

Using Euler’s formula, we can quickly derive the addition formulae. Since:

利用欧拉公式，我们可以快速推导和角公式。因为：

e^(i(α+β)) = e^(iα) e^(iβ)

Expanding both sides using cos + i sin gives:

用 cos + i sin 展开两边，得到：

cos(α+β) + i sin(α+β) = (cos α + i sin α)(cos β + i sin β)

Multiplying the right side and equating real and imaginary parts yields:

展开右边并比较实部与虚部，得到：

cos(α+β) = cos α cos β − sin α sin β

sin(α+β) = sin α cos β + cos α sin β

This method is cleaner than geometric proofs and extends naturally to any number of angles.

这种方法比几何证明更简洁，并且自然推广到多个角度。

8. Powers of i and Exponential Periodicity | i 的幂与指数周期性

Euler’s formula helps understand powers of i. Since i = e^(iπ/2), we have:

欧拉公式有助于理解 i 的幂。由于 i = e^(iπ/2)，我们有：

i² = e^(iπ) = −1, i³ = e^(i3π/2) = −i, i⁴ = e^(i2π) = 1

The exponential function e^(iθ) is periodic with period 2π, because e^(i(θ+2π)) = e^(iθ).

指数函数 e^(iθ) 具有以 2π 为周期的周期性，因为 e^(i(θ+2π)) = e^(iθ)。

This periodic nature shows that complex exponentials repeat every full rotation around the Argand diagram.

这种周期性表明复数指数在阿甘图（复平面）上每旋转一整圈后就会重复。

9. Roots of Unity | 单位根

The equation zⁿ = 1 has n distinct complex solutions called the n-th roots of unity. Using Euler’s formula:

方程 zⁿ = 1 有 n 个不同的复数解，称为 n 次单位根。利用欧拉公式：

z = e^(i 2πk/n), for k = 0, 1, 2, …, n−1

For example, the cube roots of unity are 1, e^(i2π/3) and e^(i4π/3), which form an equilateral triangle in the complex plane.

例如，三次单位根为 1、e^(i2π/3) 和 e^(i4π/3)，它们在复平面上构成一个等边三角形。

These roots have applications in solving polynomial equations, Fourier transforms, and many areas of engineering.

这些根在解多项式方程、傅里叶变换以及许多工程领域都有应用。

10. Euler’s Formula in Differential Equations | 欧拉公式在微分方程中的应用

For linear differential equations with constant coefficients, the characteristic equation often has complex roots. Euler’s formula helps express solutions in real form.

对于常系数线性微分方程，特征方程常有复数根。欧拉公式帮助我们将解表达为实形式。

If the roots are a ± bi, the general solution can be written as:

若根为 a ± bi，通解可以写为：

y = e^(ax)(C₁ cos(bx) + C₂ sin(bx))

This comes from writing the complex exponential solutions e^((a+bi)x) and e^((a−bi)x) in terms of sine and cosine using Euler’s formula.

这来源于用欧拉公式将复指数解 e^((a+bi)x) 和 e^((a−bi)x) 写成正弦和余弦的形式。

11. Common Exam Techniques | 常见解题技巧

In A-Level exams, Euler’s formula is often tested through:

在 A-Level 考试中，欧拉公式通常通过以下方式考查：

Converting complex numbers from Cartesian to exponential form and vice versa.

在直角坐标形式与指数形式之间互化复数。
Using e^(iθ) to evaluate trigonometric identities quickly.

利用 e^(iθ) 快速推导三角恒等式。
Applying De Moivre’s theorem to find powers and roots.

应用棣莫弗定理求幂与根。
Solving differential equations with complex characteristic roots.

求解具有复特征根的微分方程。

Always remember that e^(iθ) is a point on the unit circle in the complex plane. This geometric intuition can save time in MCQs.

始终记住 e^(iθ) 是复平面上单位圆上的一个点。这种几何直觉能在选择题中节省时间。

12. Worked Example | 例题精讲

Problem: Express z = −1 + √3 i in exponential form, then compute z⁶.

题目：将 z = −1 + √3 i 表示为指数形式，并计算 z⁶。

Solution: First find the modulus and argument. Modulus r = √((−1)² + (√3)²) = √(1+3) = 2. The point lies in the second quadrant, so arg(z) = π − π/3 = 2π/3. Therefore:

解：先求模与辐角。模长 r = √((−1)² + (√3)²) = √(1+3) = 2。该点位于第二象限，因此辐角为 arg(z) = π − π/3 = 2π/3。所以：

z = 2 e^(i 2π/3)

Then z⁶ = (2 e^(i 2π/3))⁶ = 64 e^(i 4π) = 64 (cos 4π + i sin 4π) = 64(1 + 0i) = 64.

于是 z⁶ = (2 e^(i 2π/3))⁶ = 64 e^(i 4π) = 64 (cos 4π + i sin 4π) = 64(1 + 0i) = 64。

This example shows how the exponential form makes powers of complex numbers almost trivial to compute.

这个例子展示了指数形式如何使复数的幂运算变得极其简单。

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Euler’s Formula | 欧拉公式