Euler's Formula | 欧拉公式

📚 Euler’s Formula | 欧拉公式

Euler’s formula is one of the most beautiful and profound results in mathematics, connecting exponential functions with trigonometric functions through complex numbers. For A-Level AQA Mathematics students, understanding this formula bridges the gap between pure algebra and geometry, providing powerful tools for solving problems in complex number theory, calculus, and even physics.

欧拉公式是数学中最优美且影响深远的成果之一，它通过复数将指数函数与三角函数联系在一起。对于 A-Level AQA 数学学生而言，理解这一公式能够弥合纯代数与几何之间的鸿沟，为求解复数理论、微积分乃至物理学中的问题提供强大工具。

1. Statement of Euler’s Formula | 欧拉公式的表述

Euler’s formula states that for any real number θ (theta), the following relationship holds:

eⁱᶿ = cos θ + i sin θ

Here, e is the base of natural logarithms (approximately 2.71828), i is the imaginary unit defined by i² = −1, and θ is measured in radians. The formula reveals that the exponential function with an imaginary exponent can be expressed as a combination of cosine and sine functions.

欧拉公式表明，对于任意实数 θ（theta），以下关系成立：

eⁱᶿ = cos θ + i sin θ

其中 e 是自然对数的底数（约等于 2.71828），i 是虚数单位，定义为 i² = −1，θ 以弧度为单位。该公式揭示出具有虚数指数的指数函数可以表示为余弦函数和正弦函数的组合。

When θ = π (180°), we obtain Euler’s identity:

eⁱᵖⁱ + 1 = 0

This identity is often called the most beautiful equation in mathematics because it links five fundamental constants: e, i, π, 1, and 0.

当 θ = π（180°）时，我们得到欧拉恒等式：

eⁱᵖⁱ + 1 = 0

这一恒等式常被称为数学中最优美的方程，因为它将五个基本常数联系在了一起：e、i、π、1 和 0。

2. Derivation from Maclaurin Series | 从麦克劳林级数推导

To understand why Euler’s formula is true, we begin with the Maclaurin series expansions of the exponential, sine, and cosine functions. These series are valid for all real values of x, and we extend them to complex values.

为理解欧拉公式为何成立，我们首先考虑指数函数、正弦函数和余弦函数的麦克劳林级数展开。这些级数对所有实数 x 都成立，而我们可以将其推广到复数情形。

The Maclaurin series for eˣ is:

eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + …

For sin x and cos x, we have:

sin x = x − x³/3! + x⁵/5! − x⁷/7! + …

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …

Now substitute x = iθ into the exponential series. Recall that i² = −1, i³ = −i, i⁴ = 1, i⁵ = i, and so on. The powers of i cycle through i, −1, −i, and 1 with period 4.

现在将 x = iθ 代入指数函数的级数。回忆 i² = −1、i³ = −i、i⁴ = 1、i⁵ = i，依此类推。i 的幂以 4 为周期循环于 i、−1、−i 和 1。

Expanding eⁱᶿ:

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

= 1 + iθ − θ²/2! − iθ³/3! + θ⁴/4! + iθ⁵/5! − …

Now separate the real and imaginary parts:

Re(eⁱᶿ) = 1 − θ²/2! + θ⁴/4! − θ⁶/6! + … = cos θ

Im(eⁱᶿ) = θ − θ³/3! + θ⁵/5! − … = sin θ

Therefore, eⁱᶿ = cos θ + i sin θ, which confirms Euler’s formula.

展开 eⁱᶿ：

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

= 1 + iθ − θ²/2! − iθ³/3! + θ⁴/4! + iθ⁵/5! − …

现在将实部和虚部分开：

实部 Re(eⁱᶿ) = 1 − θ²/2! + θ⁴/4! − θ⁶/6! + … = cos θ

虚部 Im(eⁱᶿ) = θ − θ³/3! + θ⁵/5! − … = sin θ

因此，eⁱᶿ = cos θ + i sin θ，这验证了欧拉公式。

3. Geometric Interpretation on the Complex Plane | 复平面上的几何解释

Euler’s formula has a beautiful geometric interpretation. On the complex plane, the number eⁱᶿ = cos θ + i sin θ represents a point on the unit circle centred at the origin. The real part cos θ is the x-coordinate, and the imaginary part sin θ is the y-coordinate.

欧拉公式具有优美的几何意义。在复平面上，数 eⁱᶿ = cos θ + i sin θ 表示以原点为圆心、半径为 1 的单位圆上的一点。实部 cos θ 是 x 坐标，虚部 sin θ 是 y 坐标。

As θ increases from 0 to 2π, the point eⁱᶿ moves counterclockwise around the unit circle. For example:

When θ = 0: e⁰ = 1 (the point (1, 0))
When θ = π/2: eⁱᵖⁱᐟ² = i (the point (0, 1))
When θ = π: eⁱᵖⁱ = −1 (the point (−1, 0))
When θ = 3π/2: e³ⁱᵖⁱᐟ² = −i (the point (0, −1))
When θ = 2π: e²ⁱᵖⁱ = 1 (returning to the starting point)

随着 θ 从 0 增加到 2π，点 eⁱᶿ 沿单位圆逆时针移动。例如：

当 θ = 0 时：e⁰ = 1（点 (1, 0)）
当 θ = π/2 时：eⁱᵖⁱᐟ² = i（点 (0, 1)）
当 θ = π 时：eⁱᵖⁱ = −1（点 (−1, 0)）
当 θ = 3π/2 时：e³ⁱᵖⁱᐟ² = −i（点 (0, −1)）
当 θ = 2π 时：e²ⁱᵖⁱ = 1（回到起点）

This geometric interpretation shows that complex exponentiation corresponds to rotation in the complex plane, which is fundamental to understanding the modulus-argument form of complex numbers.

这种几何解释表明，复数的指数运算对应于复平面上的旋转，这对于理解复数的模-辐角形式至关重要。

4. Complex Numbers in Modulus-Argument Form | 复数的模-辐角形式

Euler’s formula gives us a compact way to express complex numbers. Any non-zero complex number z = a + bi can be written in polar form using its modulus r and argument θ:

z = r(cos θ + i sin θ) = reⁱᶿ

where r = √(a² + b²) is the modulus and θ = arctan(b/a) is the argument, adjusted to the correct quadrant.

欧拉公式为我们提供了表示复数的简洁方式。任何非零复数 z = a + bi 都可以利用其模 r 和辐角 θ 写成极坐标形式：

z = r(cos θ + i sin θ) = reⁱᶿ

其中 r = √(a² + b²) 是模，θ = arctan(b/a) 是辐角，需要根据 z 所在的象限调整。

For example, the complex number z = 1 + i has modulus r = √(1² + 1²) = √2 and argument θ = π/4. Therefore, we can write:

1 + i = √2 · eⁱᵖⁱᐟ⁴

This exponential form is especially useful for multiplication, division, and finding powers of complex numbers, as these operations become simple manipulations of exponents.

例如，复数 z = 1 + i 的模为 r = √(1² + 1²) = √2，辐角为 θ = π/4。因此，我们可以写成：

1 + i = √2 · eⁱᵖⁱᐟ⁴

这种指数形式在复数的乘法、除法和幂运算中特别有用，因为这些运算变成了对指数的简单操作。

5. Multiplication and Division Using Euler’s Formula | 利用欧拉公式进行乘法和除法

One of the most powerful applications of Euler’s formula is simplifying complex number arithmetic. If z₁ = r₁eⁱᶿ¹ and z₂ = r₂eⁱᶿ², then:

z₁z₂ = r₁r₂ · eⁱ⁽ᶿ¹⁺ᶿ²⁾

z₁/z₂ = (r₁/r₂) · eⁱ⁽ᶿ¹⁻ᶿ²⁾

This means: to multiply two complex numbers, we multiply their moduli and add their arguments. To divide, we divide the moduli and subtract the arguments.

欧拉公式最强大的应用之一是简化复数运算。如果 z₁ = r₁eⁱᶿ¹ 且 z₂ = r₂eⁱᶿ²，则：

z₁z₂ = r₁r₂ · eⁱ⁽ᶿ¹⁺ᶿ²⁾

z₁/z₂ = (r₁/r₂) · eⁱ⁽ᶿ¹⁻ᶿ²⁾

这意味着：两个复数相乘时，模相乘、辐角相加；相除时，模相除、辐角相减。

Worked example: Calculate (1 + i)⁸.

First, express 1 + i in exponential form: 1 + i = √2 · eⁱᵖⁱᐟ⁴.

Then: (1 + i)⁸ = (√2)⁸ · (eⁱᵖⁱᐟ⁴)⁸ = 16 · e²ⁱᵖⁱ = 16 · 1 = 16.

This is much faster than expanding (1 + i)⁸ directly or using binomial expansion.

例题：计算 (1 + i)⁸。

首先将 1 + i 写为指数形式：1 + i = √2 · eⁱᵖⁱᐟ⁴。

然后：(1 + i)⁸ = (√2)⁸ · (eⁱᵖⁱᐟ⁴)⁸ = 16 · e²ⁱᵖⁱ = 16 · 1 = 16。

这比直接展开 (1 + i)⁸ 或使用二项式定理要快得多。

6. De Moivre’s Theorem Connection | 棣莫弗定理的联系

De Moivre’s theorem states that for any real number θ and any integer n:

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

Euler’s formula provides an elegant derivation of this theorem. Since eⁱᶿ = cos θ + i sin θ, we have:

(eⁱᶿ)ⁿ = eⁱⁿᶿ = cos(nθ) + i sin(nθ)

Thus De Moivre’s theorem follows directly from the laws of exponents applied to complex numbers.

棣莫弗定理断言：对任意实数 θ 和任意整数 n，成立：

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

欧拉公式为这一定理提供了优雅的推导。由于 eⁱᶿ = cos θ + i sin θ，我们有：

(eⁱᶿ)ⁿ = eⁱⁿᶿ = cos(nθ) + i sin(nθ)

因此，棣莫弗定理直接由复数指数运算的指数法则得出。

De Moivre’s theorem is extremely useful for finding powers and roots of complex numbers. For instance, to find the cube root of i, we express i = eⁱᵖⁱᐟ² and then take the cube root: the three roots are eⁱᵖⁱᐟ⁶, eⁱ⁵ᵖⁱᐟ⁶, and eⁱ³ᵖⁱᐟ², equally spaced around the unit circle.

棣莫弗定理在求复数的幂和根时极为有用。例如，要求 i 的立方根，我们先将 i 表示为 i = eⁱᵖⁱᐟ²，然后开三次方：三个根分别为 eⁱᵖⁱᐟ⁶、eⁱ⁵ᵖⁱᐟ⁶ 和 eⁱ³ᵖⁱᐟ²，它们在单位圆上等距分布。

7. Trigonometry Identities from Euler’s Formula | 从欧拉公式推导三角恒等式

Euler’s formula provides a powerful method for deriving trigonometric identities. Since eⁱθ = cos θ + i sin θ and eⁱᐟ⁽⁻θ⁾ = cos θ − i sin θ, we can solve for cos θ and sin θ:

cos θ = (eⁱθ + e⁻ⁱθ)/2

sin θ = (eⁱθ − e⁻ⁱθ)/(2i)

欧拉公式为推导三角恒等式提供了强大的方法。由于 eⁱθ = cos θ + i sin θ 且 eⁱᐟ⁽⁻θ⁾ = cos θ − i sin θ，我们可以解出 cos θ 和 sin θ：

cos θ = (eⁱθ + e⁻ⁱθ)/2

sin θ = (eⁱθ − e⁻ⁱθ)/(2i)

These are called Euler’s formulas for the trigonometric functions. To derive the double-angle formula, we consider:

(eⁱθ)² = e²ⁱθ = cos(2θ) + i sin(2θ)

But also: (eⁱθ)² = (cos θ + i sin θ)² = cos²θ − sin²θ + 2i sin θ cos θ

Equating real and imaginary parts gives:

cos(2θ) = cos²θ − sin²θ and sin(2θ) = 2 sin θ cos θ

这些公式称为三角函数的欧拉公式。为推导倍角公式，我们考虑：

(eⁱθ)² = e²ⁱθ = cos(2θ) + i sin(2θ)

同时也有：(eⁱθ)² = (cos θ + i sin θ)² = cos²θ − sin²θ + 2i sin θ cos θ

令实部和虚部分别相等，得到：

cos(2θ) = cos²θ − sin²θ 以及 sin(2θ) = 2 sin θ cos θ

This technique can be extended to derive higher-order multiple-angle formulas, sum-to-product identities, and even integrals involving trigonometric functions.

这种技巧可以推广到推导更高阶的倍角公式、和差化积公式，甚至涉及三角函数的积分计算。

8. Solving Differential Equations with Complex Exponentials | 用复指数求解微分方程

Euler’s formula is invaluable in solving linear differential equations. Consider the second-order differential equation:

y” + y = 0

The characteristic equation is r² + 1 = 0, giving r = ±i. The general solution can be expressed using complex exponentials:

y = Aeⁱˣ + Be⁻ⁱˣ

Using Euler’s formula, this can be rewritten as:

y = C cos x + D sin x

where C and D are constants determined by initial conditions. This shows how complex exponentials naturally lead to real sinusoidal solutions.

欧拉公式在求解线性微分方程中具有不可替代的价值。考虑二阶微分方程：

y” + y = 0

其特征方程为 r² + 1 = 0，解得 r = ±i。通解可以用复指数形式表示：

y = Aeⁱˣ + Be⁻ⁱˣ

利用欧拉公式，可以改写为：

y = C cos x + D sin x

其中 C 和 D 是由初始条件确定的常数。这表明复指数如何自然地导出实的正弦解。

9. Applications to Physics and Engineering | 在物理学和工程学中的应用

Euler’s formula appears throughout physics and engineering. In alternating current (AC) circuit analysis, the voltage and current in a circuit are represented as complex exponentials. The impedance of a capacitor or inductor involves the imaginary unit i, and Euler’s formula allows engineers to analyse AC circuits using complex algebra.

欧拉公式贯穿于物理和工程的各个领域。在交流电路分析中，电路中的电压和电流用复指数表示。电容器或电感器的阻抗涉及虚数单位 i，欧拉公式使得工程师能够用复数代数来分析交流电路。

In quantum mechanics, the wave function of a free particle is often expressed as:

ψ(x, t) = A·eⁱ⁽ᵏˣ⁻ᵛᵗ⁾

This complex exponential contains both spatial and temporal oscillations, and Euler’s formula helps physicists interpret the real and imaginary parts of the wave function in terms of physical observables.

在量子力学中，自由粒子的波函数通常表示为：

ψ(x, t) = A·eⁱ⁽ᵏˣ⁻ᵛᵗ⁾

这个复指数同时包含空间和时间的振荡，欧拉公式帮助物理学家根据可观测的物理量来解释波函数的实部和虚部。

Signal processing relies heavily on the Fourier transform, which decomposes signals into complex exponentials. Euler’s formula connects the frequency domain with the time domain, making it possible to analyse and filter signals efficiently.

信号处理高度依赖傅里叶变换，它将信号分解为复指数的组合。欧拉公式将频域与时域联系起来，使得高效分析和过滤信号成为可能。

10. Exam-Style Questions and Solutions | 考试风格题目与解答

Let us examine the type of questions you might encounter in your AQA A-Level examination, along with model solutions.

让我们看一下 AQA A-Level 考试中可能出现的题目类型以及规范解答。

Question 1: Express z = −2 + 2i in the form reⁱᶿ, where r > 0 and −π < θ ≤ π.

Solution: r = √((−2)² + 2²) = √8 = 2√2.

The point (−2, 2) lies in the second quadrant, so θ = π − π/4 = 3π/4.

Therefore: z = 2√2 · e³ⁱᵖⁱᐟ⁴.

问题 1：将 z = −2 + 2i 表示为 reⁱᶿ 的形式，其中 r > 0 且 −π < θ ≤ π。

解答：r = √((−2)² + 2²) = √8 = 2√2。

点 (−2, 2) 位于第二象限，所以 θ = π − π/4 = 3π/4。

因此：z = 2√2 · e³ⁱᵖⁱᐟ⁴。

Question 2: Given that z = 4eⁱᵖⁱᐟ³, find z³.

Solution: z³ = (4eⁱᵖⁱᐟ³)³ = 64eⁱᵖⁱ = 64(cos π + i sin π) = 64(−1 + 0i) = −64.

问题 2：已知 z = 4eⁱᵖⁱᐟ³，求 z³。

解答：z³ = (4eⁱᵖⁱᐟ³)³ = 64eⁱᵖⁱ = 64(cos π + i sin π) = 64(−1 + 0i) = −64。

Question 3: Use Euler’s formula to prove the identity sin(3θ) = 3 sin θ − 4 sin³θ.

Solution: Consider (eⁱθ)³ = e³ⁱθ = cos(3θ) + i sin(3θ).

Also: (eⁱθ)³ = (cos θ + i sin θ)³

= cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ

Equating imaginary parts: sin(3θ) = 3 cos²θ sin θ − sin³θ

= 3(1 − sin²θ)sin θ − sin³θ = 3 sin θ − 3 sin³θ − sin³θ = 3 sin θ − 4 sin³θ ✓

问题 3：利用欧拉公式证明恒等式 sin(3θ) = 3 sin θ − 4 sin³θ。

解答：考虑 (eⁱθ)³ = e³ⁱθ = cos(3θ) + i sin(3θ)。

同时：(eⁱθ)³ = (cos θ + i sin θ)³

= cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ

令虚部相等：sin(3θ) = 3 cos²θ sin θ − sin³θ

= 3(1 − sin²θ)sin θ − sin³θ = 3 sin θ − 3 sin³θ − sin³θ = 3 sin θ − 4 sin³θ ✓

11. Common Pitfalls and Tips | 常见错误与考试技巧

Students frequently make errors when working with Euler’s formula. Here are the most common pitfalls and how to avoid them.

学生在使用欧拉公式时经常犯错。以下是最常见的陷阱以及避免方法。

Pitfall \| 陷阱	Explanation \| 说明	Correct approach \| 正确做法
Using degrees instead of radians	Euler’s formula requires θ in radians	Always set your calculator to radians mode
Forgetting to adjust the argument for the quadrant	arctan gives angles in the first or fourth quadrants only	Check the signs of a and b in a + bi and add π if needed
Incorrectly computing powers of i	The cycle i, −1, −i, 1 repeats every 4 powers	Divide the exponent by 4 and use the remainder

记住这些考试技巧：

Always write complex numbers in exponential form before performing multiplication, division, or raising to powers. When finding the argument, draw a quick sketch to confirm the correct quadrant. Remember that adding 2π to an argument does not change the complex number, so you can always adjust the angle to lie in the required interval.

回顾这些考试技巧：

在进行乘法、除法或幂运算之前，始终先将复数写成指数形式。求辐角时，快速画一张草图来确认正确的象限。记住，在辐角上加 2π 不会改变复数本身，因此你总可以调整角度使其落在要求的区间内。

12. Summary and Further Study | 总结与进一步学习

Euler’s formula eⁱᶿ = cos θ + i sin θ is a cornerstone of A-Level mathematics. It provides a bridge between algebra, geometry, and analysis, allowing complex numbers to be expressed in a compact and powerful exponential form. The formula enables us to multiply and divide complex numbers easily, derive trigonometric identities, solve differential equations, and understand the behaviour of oscillations in physics and engineering.

欧拉公式 eⁱᶿ = cos θ + i sin θ 是 A-Level 数学的基石。它在代数、几何和数学分析之间架起了桥梁，使得复数能以简洁而强大的指数形式表达。该公式使我们能够轻松地进行复数的乘除运算、推导三角恒等式、求解微分方程，并理解物理和工程中振荡现象的本质。

To master this topic, practice converting between Cartesian form (a + bi), polar form (r(cos θ + i sin θ)), and exponential form (reⁱᶿ) until the process becomes automatic. Work through past paper questions, particularly those involving roots of unity and trigonometric identities. With consistent practice, you will find Euler’s formula to be one of the most versatile tools in your mathematical toolkit.

为掌握这一主题，请反复练习在笛卡尔形式 (a + bi)、极坐标形式 (r(cos θ + i sin θ)) 和指数形式 (reⁱᶿ) 之间进行转换，直到这一过程变得自动化。认真完成历年真题，特别是涉及单位根和三角恒等式的问题。通过持续练习，你将会发现欧拉公式是你数学工具箱中用途最广泛的工具之一。

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Euler’s Formula | 欧拉公式