Exponential Form of a Complex Number | 复数的指数形式

📚 Exponential Form of a Complex Number | 复数的指数形式

The exponential form of a complex number is a compact way to write complex numbers using the modulus and argument. It connects algebra, trigonometry and exponential functions through Euler's formula, and makes multiplication, division, powers and roots much easier. In AQA A-Level Mathematics, you need to convert between Cartesian, modulus-argument and exponential forms, and use these forms confidently in calculations and geometric problems.

复数的指数形式是一种利用模与辐角来简洁表示复数的方法。它通过欧拉公式把代数、三角和指数函数联系起来，使乘法、除法、乘方与开方运算变得更加容易。在 AQA A-Level 数学中，你需要能够在笛卡儿形式、模-辐角形式和指数形式之间转换，并在计算和几何问题中熟练运用这些形式。

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1. Modulus-Argument Form Review | 模-辐角形式回顾

umber z = x + iy can be written in modulus-argument form z = r(cos θ + i sin θ), where r = |z| = √x² + y² he modulus and θ = arg z is the argument, usually taken in the principal range -π < θ ≤ π.

任何非零复数 z = x + iy 都可以写成模-辐角形式 z = r(cos θ + i sin θ)，其中 r = |z| = √x² + y² 是模，θ = arg z 是辐角，通常取主值范围 -π < θ ≤ π。

z = r(cos θ + i sin θ),   r = |z|,   θ = arg z

The modulus gives the distance from the origin in the Argand diagram, and the argument gives the angle measured anticlockwise from the positive real axis.

模表示复平面上从原点到该点的距离，辐角表示从正实轴逆时针测量的角度。

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2. Euler's Formula and Its Meaning | 欧拉公式及其意义

Euler's formula states that for any real angle θ, e^iθ ity allows us to replace the bracketed trigonometric expression by a single exponential term.

欧拉公式指出，对任意实数角 θ，e^iθ = cos θ + i sin θ。这个优美的恒等式使我们能够用一个指数项代替括号中的三角表达式。

e^iθ = cos θ + i sin θ

The formula can be justified using Maclaurin series for e^x, cos θ and sin θ, although AQA does not require the proof. It is important to remember that θ must be in radians when using exponential form.

该公式可以用 e^x、cos θ 和 sin θ 的麦克劳林级数来证明，但 AQA 不要求掌握证明。重要的是记住，使用指数形式时 θ 必须以弧度为单位。

A special case is e^iπ + 1 = 0, which combines five of the most important constants in mathematics.

一个特例是 e^iπ + 1 = 0，它将数学中五个最重要的常数结合在了一起。

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3. Definition of Exponential Form | 指数形式的定义

Combining the modulus-argument form with Euler's formula gives the exponential form of a complex number: z = r e^iθ. Here r ≥ 0 is the modulus and θ is the argument in radians.

将模-辐角形式与欧拉公式结合，就得到复数的指数形式：z = r e^iθ。其中 r ≥ 0 是模，θ 是以弧度表示的辐角。

z = r e^iθ

For example, z = 1 + i has modulus √2 and argument π/4, so z = √2 e^iπ/4. Similarly, z = -3 has modulus 3 and argument π, so z = 3 e^iπ.

例如，z = 1 + i 的模为 √2，辐角为 π/4，因此 z = √2 e^iπ/4。类似地，z = -3 的模为 3，辐角为 π，所以 z = 3 e^iπ。

• 1 + i = √2 e^iπ/4
• 2i = 2 e^iπ/2
• -1 = e^iπ
• 1 - i = √2 e^-iπ/4

You should be able to move quickly between the three equivalent forms: Cartesian, modulus-argument and exponential.

你应该能够在笛卡儿形式、模-辐角形式和指数形式这三种等价形式之间快速转换。

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4. Converting Between Forms | 不同形式之间的转换

To convert from Cartesian form x + iy to exponential form, first find r = √x² + y², then find θ = arg(x + iy) using arctan(y/x), adjusted for the quadrant.

从笛卡儿形式 x + iy 转换为指数形式时，先求 r = √x² + y²，再用 arctan(y/x) 并根据象限调整求得 θ = arg(x + iy)。

Quadrant	Sign of x, y	Principal argument θ
First	x > 0, y ≥ 0	θ = arctan(y/x)
Second	x < 0, y ≥ 0	θ = arctan(y/x) + π
Third	x < 0, y < 0	θ = arctan(y/x) - π
Fourth	x > 0, y < 0	θ = arctan(y/x)

If z = r e^iθ z = r cos θ + i r sin θ, so the real and imaginary parts are x = r cos θ and y = r sin θ.

若 z = r e^iθ，则 z = r cos θ + i r sin θ，因此实部和虚部分别为 x = r cos θ 和 y = r sin θ。

x = r cos θ,   y = r sin θ

Always draw a quick Argand sketch when finding arguments; this prevents quadrant errors and helps you choose the correct principal value.

求辐角时务必快速画出复平面草图；这能避免象限错误，并帮助你选择正确的主值。

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5. Multiplication and Division Using Exponential Form | 指数形式下的乘法与除法

Exponential form makes multiplication and division extremely tidy. If z₁ = r₁ e^iθ₁ and z₂ = r₂ e^iθ₂, then z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)).

指数形式使乘法和除法变得非常简洁。如果 z₁ = r₁ e^iθ₁ 且 z₂ = r₂ e^iθ₂，那么 z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))。

z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))

For division, z₁/z₂ = (r₁/r₂) e^(i(θ₁-θ₂)), provided r₂ ≠ 0.

对于除法，z₁/z₂ = (r₁/r₂) e^(i(θ₁-θ₂))，其中 r₂ ≠ 0。

z₁/z₂ = (r₁/r₂) e^(i(θ₁-θ₂))

Geometrically, multiplying by a complex number scales by the modulus and rotates by the argument. Dividing scales by the reciprocal of the modulus and rotates by the negative of the argument.

从几何上看，乘以一个复数会按模进行缩放并按辐角进行旋转；除以一个复数则按模的倒数缩放并按相反辐角旋转。

This is far quicker than expanding Cartesian brackets or using trigonometric identities. For example, multiplying 2 e^iπ/3 by 3 e^iπ/4 gives 6 e^i7π/12 immediately.

这比展开笛卡儿括号或使用三角恒等式要快得多。例如，2 e^iπ/3 乘以 3 e^iπ/4 立刻得到 6 e^i7π/12。

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6. Powers and De Moivre's Theorem | 乘方与棣莫弗定理

For an integer n, (r e^iθ)ⁿ = rⁿ e^inθ. This is the exponential form of De Moivre's theorem: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ).

对于整数 n，(r e^iθ)ⁿ = rⁿ e^inθ。这就是棣莫弗定理的指数形式：(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。

zⁿ = rⁿ e^inθ

This is particularly useful for finding high powers of complex numbers. Write the number in exponential form first, then raise the modulus to n and multiply the argument by n. After that, convert back to Cartesian form if required.

这对求复数的高次幂特别有用。先将复数写成指数形式，再把模取 n 次方，并将辐角乘以 n。如有需要，再转换回笛卡儿形式。

For example, (1 + i)¹⁰ can be found by writing 1 + i = √2 e^iπ/4, so (1 + i)¹⁰ = (√2)¹⁰ e^i10π/4 = 32 e^i5π/2 = 32 e^iπ/2 = 32i.

例如，求 (1 + i)¹⁰ 时可先写成 1 + i = √2 e^iπ/4，所以 (1 + i)¹⁰ = (√2)¹⁰ e^i10π/4 = 32 e^i5π/2 = 32 e^iπ/2 = 32i。

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7. Roots of Complex Numbers | 复数根

To find the n-th roots of a complex number z = r e^iθ, write w = s e^iφ and solve wⁿ = z. This gives s = r^1/n and φ = (θ + 2kπ)/n for k = 0, 1, 2, ..., n-1.

求复数 z = r e^iθ 的 n 次根时，设 w = s e^iφ 并解 wⁿ = z。可得 s = r^1/n，φ = (θ + 2kπ)/n，其中 k = 0, 1, 2, ..., n-1。

wₖ = r^1/n e^(i(θ + 2kπ)/n),   k = 0, 1, ..., n-1

Adding 2kπ accounts for the fact that the argument is periodic with period 2π. The n roots are equally spaced around a circle of radius r^1/n, each separated by angle 2π/n.

加上 2kπ 是因为辐角以 2π 为周期。n 个根均匀分布在半径为 r^1/n 的圆上，相邻根之间的夹角为 2π/n。

Remember that a non-zero complex number has exactly n distinct n-th roots. Always give all values of k from 0 to n-1 unless the question specifies a subset.

请记住，一个非零复数恰好有 n 个不同的 n 次根。除非题目指定某个子集，否则要给出 k

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