The Modulus-Argument Form and Euler's Form of Complex Numbers | 复数的模辐角形式与欧拉形式

📚 The Modulus-Argument Form and Euler's Form of Complex Numbers | 复数的模辐角形式与欧拉形式

mplex number in terms of its modulus and argument. This gives rise to the modulus-argument form, and, after introducing Euler's formula, to the elegant exponential form z = r e^iθ.

复数是 IB 高级数学的重要部分。理解和运算复数最有力的方式之一，是借助模与辐角来表示复数，由此得到模辐角形式；引入欧拉公式后，又可以写出简洁的指数形式 z = r e^iθ。

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1. Complex Numbers and the Argand Diagram | 复数与阿尔甘图

i² = −1. We call x the real part, written Re(z), and y the imaginary part, written Im(z).

每个复数都可以写成 z = x + i y，其中 x 和 y 为实数，且 i 满足 i² = −1。我们称 x 为实部，记作 Re(z)，称 y 为虚部，记作 Im(z)。

On an Argand diagram, z is represented by the point (x, y), with the horizontal axis being the real axis and the vertical axis being the imaginary axis. This geometric view is the foundation for defining modulus and argument.

在阿尔甘图中，复数 z 用点 (x, y) 表示，其中水平轴为实轴，竖直轴为虚轴。这种几何视角是定义模和辐角的基础。

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2. The Modulus | 复数的模

modulus of z = x + i y is the distance from the origin to the point (x, y). It is denoted |z| and is given by the Pythagorean formula:

复数 z = x + i y 的模，是指原点到点 (x, y) 的距离，记作 |z|，由勾股公式给出：

|z| = √x² + y²

For example, if z = 3 + 4i, then |z| = √3² + 4² = 5. The modulus is always a non-negative real number, and |z| = 0 if and only if z = 0.

例如，若 z = 3 + 4i，则 |z| = √3² + 4² = 5。模总是非负实数，且 |z| = 0 当且仅当 z = 0。

Important properties of the modulus include |z|² = z z̄, |zw| = |z||w|, and |z/w| = |z|/|w| for w ≠ 0. These properties are frequently used in proofs and simplification.

模的重要性质包括 |z|² = z z̄，|zw| = |z||w|，以及当 w ≠ 0 时 |z/w| = |z|/|w|。这些性质常在证明和化简中使用。

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3. The Argument and Principal Argument | 辐角与主辐角

An argument of a non-zero complex number z is an angle θ measured from the positive real axis to the line segment joining the origin to (x, y). Because angles can be increased or decreased by full rotations, the argument is multi-valued:

非零复数 z 的辐角，是指从正实轴出发，到原点和点 (x, y) 连线所成的角 θ。由于角度可以增加或减少整圈，辐角是多值的：

arg z = θ + 2πk, k ∈ ℤ

atics, this interval is commonly −π < Arg z ≤ π, though some textbooks use 0 ≤ Arg z < 2π.

主辐角记作 Arg z，是在一个标准区间内选取的唯一辐角。在 IB 数学中，这个区间通常取 −π < Arg z ≤ π，不过有些教科书也使用 0 ≤ Arg z < 2π。

To find θ correctly, draw the point on the Argand diagram and adjust the basic arctangent ratio for the quadrant. A simple guide for the interval (−π, π] is:

为了正确求 θ，应在阿尔甘图上画出该点，并根据所在象限对基本的 arctan 比值进行调整。对于区间 (−π, π]，一个简便规则如下：

Quadrant	Principal Argument
First quadrant	θ = arctan(y/x)
Second quadrant	θ = arctan(y/x) + π
Third quadrant	θ = arctan(y/x) − π
Fourth quadrant	θ = arctan(y/x)

e argument of z = 0 is undefined, because no direction can be assigned to the zero vector.

z = 0 的辐角没有定义，因为零向量没有确定的方向。

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4. Modulus-Argument Form | 模辐角形式

Let r = |z| and θ = arg z. Then z can be written in modulus-argument form, also called polar form:

设 r = |z|，θ = arg z。则 z 可以写成模辐角形式，也叫极坐标形式：

z = r(cos θ + i sin θ)

ially useful because the modulus and argument are immediately visible from the expression.

这种形式有时也简写为 z = r cis θ。它特别有用，因为从表达式中可以立即看出模和辐角。

To convert from Cartesian form x + i y, use r = √x² + y² θ determined by the quadrant of (x, y). To convert back, use x = r cos θ and y = r sin θ.

要从直角坐标形式 x + i y 转换，使用 r = √x² + y²，并由 (x, y) 所在象限决定 θ。要逆转换，则使用 x = r cos θ 和 y = r sin θ。

For example, z = −1 + i has r = √2 and Arg z = 3π/4, so its modulus-argument form is √2(cos(3π/4) + i sin(3π/4)).

例如，z = −1 + i 的模为 r = √2，主辐角为 Arg z = 3π/4，所以它的模辐角形式为 √2(cos(3π/4) + i sin(3π/4))。

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5. Euler's Formula | 欧拉公式

Euler's formula is one of the most beautiful results in mathematics. It states that for a real angle θ, measured in radians:

欧拉公式是数学中最优美的结论之一。它指出，对于以弧度度量的实数角度 θ：

e^iθ = cos θ + i sin θ

This identity can be derived from the Maclaurin series for e^x os x. Substituting x = iθ and separating real and imaginary terms gives the formula above.

这个恒等式可以由 e^x、sin x 和 cos x 的麦克劳林级数推导。代入 x = iθ 并分离实部和虚部即可得到上式。

A famous special case is θ = π, which gives e^iπ = −1, or equivalently e^iπ + 1 = 0. This equation connects five fundamental constants: 0, 1, e, i, and π.

一个著名的特例是 θ = π，得到 e^iπ = −1，即 e^iπ + 1 = 0。这个等式将五个基本常数 0、1、e、i 和 π 联系在了一起。

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6. Euler's Form of a Complex Number | 复数的欧拉形式

By combining Euler's formula with the modulus-argument form, we obtain Euler's form of a complex number:

将欧拉公式与模辐角形式结合，就得到复数的欧拉形式：

z = r e^iθ

Here r = |z| = arg z. This is not a new number; it is simply a compact notation for r(cos θ + i sin θ).

其中 r = |z|，θ = arg z。这并不是一个新数，而只是 r(cos θ + i sin θ) 的紧凑记法。

For example, the number √2 e^(i(3π/4)) is the Euler form of z = −1 + i. Because e^iθ has modulus 1, multiplying a complex number by e^iθ rotates it by θ without changing its distance from the origin.

例如，√2 e^(i(3π/4)) 就是 z = −1 + i 的欧拉形式。因为 e^iθ 的模为 1，所以复数乘以 e^iθ 相当于把它旋转角度 θ，而不改变它到原点的距离。

The table below summarises the equivalent forms:

下表总结了两种等价形式：

Form	Expression
Modulus-argument form	z = r(cos θ + i sin θ)
Euler form	z = r eiθ

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7. Multiplication and Division in Polar and Euler Form | 极坐标形式与欧拉形式下的乘法与除法

antages of Euler form is that multiplication and division become simple. Suppose z1 = r1 e^iθ1 and z2 = r2 e^iθ2. Then:

欧拉形式的一大优势是乘法和除法变得十分简单。设 z1 = r1 e^iθ1，z2 = r2 e^iθ2，则：

z1 z2 = r1 r2 e^(i(θ1 + θ2))

z1 / z2 = (r1 / r2) e^(i(θ1 − θ2)), z2 ≠ 0

In words, to multiply two complex numbers, multiply their moduli and add their arguments. To divide, divide the moduli and subtract the arguments.

用语言描述：两个复数相乘，就是模相乘、辐角相加；两个复数相除，就是模相除、辐角相减。

Geometrically, multiplication by a complex number is a rotation followed by a scaling. If you multiply by a number with modulus r and argument θ, every point on the Argand diagram is rotated by θ and stretched away from the origin by factor r.

从几何上看，复数的乘法是先旋转后缩放。如果乘以一个模为 r、辐角为 θ 的复数，则阿尔甘图上的每个点都会旋转 θ，并相对原点按比例 r 拉伸。

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8. De Moivre's Theorem | 棣莫弗定理

De Moivre's theorem follows immediately from Euler form, but it is often stated in modulus-argument form:

棣莫弗定理可以直接由欧拉形式得到，但它通常以模辐角形式表达：

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

Equivalently, for any integer n and any complex number z = r e^iθ:

等价地，对于任意整数 n 和复数 z = r e^iθ：

zⁿ = rⁿ e^{i n θ}

d powers of complex numbers quickly. For example, to compute (1 + i)⁶, first write 1 + i = √2 e^iπ/4, then (1 + i)⁶ = (√2)⁶ e^(i(6 × π/4)) = 8 e^(i(3π/2)) = −8i.

该定理用于快速求复数的幂。例如，计算 (1 + i)⁶ 时，先写 1 + i = √2 e^iπ/4，则 (1 + i)⁶ = (√2)⁶ e^(i(6 × π/4)) = 8 e^(i(3π/2)) = −8i。

De Moivre's theorem also generates trigonometric identities by expanding cos nθ + i sin nθ using the binomial theorem.

棣莫弗定理还能通过二项式展开 cos nθ + i sin nθ，从而推导三角恒等式。

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9. Powers and n-th Roots | 幂与 n 次方根

t complex solutions when w ≠ 0. The general formula for the n-th roots of a complex number is:

由于辐角具有周期性，当 w ≠ 0 时方程 zⁿ = w 恰好有 n 个不同的复数解。复数 n 次方根的一般公式为：

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

Here r is the modulus of the original number and θ is one of its arguments. The n roots all lie on a circle of radius r^1/n and are separated by equal angles of 2π/n.

其中 r 是原复数的模，θ 是它的一个辐角。这 n 个根都位于半径为 r^1/n 的圆上，并且相邻两根之间的夹角均为 2π/n。

A classic example is the cube roots of unity. Since 1 = eⁱ⁰, the roots are:

一个经典例子是 1 的三次单位根。因为 1 = eⁱ⁰，所以三个根为：

1, e^2πi/3, e^4πi/3

Writing these in Cartesian form gives 1, −1/2 + (√3/2)i, and −1/2 − (√3/2)i. They are often labelled 1, ω, and ω², with the property 1 + ω + ω² = 0.

写成直角坐标形式为 1、−1/2 + (√3/2)i、−1/2 − (√3/2)i。它们常记为 1、ω、ω²，并满足性质 1 + ω + ω² = 0。

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10. Applications and Common Examination Pitfalls | 应用与常见考试陷阱

Polar and Euler forms are particularly useful in problems involving multiplication, division, powers, and roots. They are less useful for addition and subtraction, where Cartesian form is usually preferred.

极坐标形式和欧拉形式特别适合处理乘法、除法、幂和方根问题。对于加减法，它们不太方便，通常优先使用直角坐标形式。

• Always check the quadrant when finding an argument. A calculator arctan result can be incorrect by π if the point is in the second or third quadrant.

  求辐角时一定要检查象限。如果点在第二或第三象限，计算器给出的 arctan 结果可能会相差 π。

• Remember that e^iθ is periodic with period 2π, so adding 2π to θ does not change the complex number.

  记住 e^iθ 具有 2π 的周期性，所以给 θ 加上 2π 不改变复数的值。

• When finding roots, use θ + 2πk before dividing by n. Forgetting the 2πk term gives only one root instead of n roots.

  求方根时，要先写出 θ + 2πk，再除以 n。忘记 2πk 项只会得到一个根，而不是 n 个根。

• Use radians for Euler's formula and for polar forms. Degrees are not acceptable unless the question explicitly allows them.

  欧拉公式和极坐标形式中要使用弧度。除非题目明确允许，否则不能使用角度制。

• Be careful with the principal argument interval. The answer may depend on whether your course uses (−π, π] or [0, 2π).

  注意主辐角的区间。答案可能取决于你的课程使用 (−π, π] 还是 [0, 2π)。

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11. Summary | 总结

ument provide a second coordinate system. The modulus-argument form z = r(cos θ + i sin θ)

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