The Polar Form of a Complex Number | 复数的极形式

📚 The Polar Form of a Complex Number | 复数的极形式

A complex number z = x + iy can be located on the Argand diagram by its horizontal real part x and vertical imaginary part y. The polar form replaces these Cartesian coordinates with two new pieces of information: the distance from the origin, called the modulus, and the angle measured from the positive real axis, called the argument. This representation is especially useful for multiplying, dividing and raising complex numbers to powers.

复数 z = x + iy 可以在阿尔冈图上用水平实部 x 和竖直虚部 y 定位。极形式用两个新的信息替换这些笛卡尔坐标：到原点的距离，称为模；以及从正实轴测得的角度，称为辐角。这种表示法在复数的乘法、除法和乘方运算中特别有用。

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1. From Cartesian Coordinates to Modulus-Argument Form | 从笛卡尔坐标到模-辐角形式

On an Argand diagram, every complex number z = x + iy corresponds to a point (x, y). The polar form uses the distance r from the origin O to the point P and the angle θ between the positive real axis and the line OP. We write z = r(cos θ + i sin θ), where r = |z| and θ = arg z. This is called the modulus-argument form.

在阿尔冈图上，每个复数 z = x + iy 都对应一个点 (x, y)。极形式使用从原点 O 到点 P 的距离 r，以及正实轴与线段 OP 之间的夹角 θ。我们写作 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg z。这称为模-辐角形式。

z = x + iy = r(cos θ + i sin θ)

The modulus r is always real and non-negative, while the argument θ is an angle measured in radians. The polar form is particularly concise when dealing with products, quotients and powers.

模 r 始终是实数且非负，而辐角 θ 是以弧度为单位测量的角度。在处理乘积、商和幂时，极形式特别简洁。

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2. The Modulus | 模

The modulus r is the length of the vector from the origin to the point representing z. It is a real, non-negative quantity. From Pythagoras’ theorem, r = |z| = √(x² + y²). When y = 0, the modulus reduces to the usual absolute value of a real number.

模 r 是从原点到表示 z 的点的向量长度。它是一个实数且非负的量。根据勾股定理，r = |z| = √(x² + y²)。当 y = 0 时，模就退化为通常实数的绝对值。

r = |z| = √(x² + y²)

For example, if z = 3 + 4i, then r = √(3² + 4²) = √(9 + 16) = √25 = 5. The modulus tells us how far the point is from the origin, regardless of direction.

例如，如果 z = 3 + 4i，那么 r = √(3² + 4²) = √(9 + 16) = √25 = 5。模告诉我们该点距离原点有多远，与方向无关。

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

The argument θ is the angle measured from the positive real axis to the line segment representing z, usually in radians. A given non-zero complex number has infinitely many arguments because adding or subtracting 2π gives the same direction. To make the argument unique, AQA uses the principal argument, often written as Arg z, which lies in the interval -π < θ ≤ π.

辐角 θ 是从正实轴到表示 z 的线段所测得的角度，通常以弧度为单位。一个非零复数有无穷多个辐角，因为加上或减去 2π 会得到相同的方向。为了使辐角唯一，AQA 使用主辐角，通常写作 Arg z，其取值范围为 -π < θ ≤ π。

-π < Arg z ≤ π

For any non-zero z = x + iy, the tangent of the argument satisfies tan θ = y/x. However, since tan has period π, you must determine the correct quadrant from the signs of x and y. The argument of z = 0 is undefined.

对于任何非零的 z = x + iy，辐角的正切满足 tan θ = y/x。然而，由于 tan 的周期为 π，你必须根据 x 和 y 的符号确定正确的象限。z = 0 的辐角没有定义。

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4. Finding the Argument by Quadrant | 按象限求辐角

Because the inverse tangent function only returns angles between -π/2 and π/2, it cannot on its own distinguish between opposite quadrants. A reliable method is to find the acute reference angle α = arctan(|y/x|) for x ≠ 0, then use the signs of x and y to assign the principal argument.

由于反正切函数只返回 -π/2 到 π/2 之间的角度，它本身无法区分相反的象限。一个可靠的方法是先求出锐参考角 α = arctan(|y/x|)（x ≠ 0），然后利用 x 和 y 的符号来确定主辐角。

Quadrant | 象限	Signs of x and y | x、y 的符号	Principal argument θ | 主辐角 θ
First | 第一	x > 0, y > 0	θ = α
Second | 第二	x < 0, y > 0	θ = π – α
Third | 第三	x < 0, y < 0	θ = -π + α
Fourth | 第四	x > 0, y < 0	θ = -α

If x = 0, the complex number lies on the imaginary axis. In that case, θ = π/2 when y > 0 and θ = -π/2 when y < 0.

如果 x = 0，复数位于虚轴上。此时，当 y > 0 时 θ = π/2，当 y < 0 时 θ = -π/2。

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5. Converting from Polar Form to Cartesian Form | 从极形式转换为笛卡尔形式

Given the modulus r and argument θ, the real and imaginary parts are found by projecting onto the axes. The conversion formulas are x = r cos θ and y = r sin θ.

给定模 r 和辐角 θ，实部和虚部可以通过投影到坐标轴上求得。转换公式为 x = r cos θ，y = r sin θ。

x = r cos θ,    y = r sin θ

For example, if z = 2(cos π/3 + i sin π/3), then x = 2 cos π/3 = 2 × ½ = 1 and y = 2 sin π/3 = 2 × √3/2 = √3. Therefore z = 1 + i√3.

例如，如果 z = 2(cos π/3 + i sin π/3)，那么 x = 2 cos π/3 = 2 × ½ = 1，y = 2 sin π/3 = 2 × √3/2 = √3。因此 z = 1 + i√3。

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6. Converting from Cartesian Form to Polar Form | 从笛卡尔形式转换为极形式

To convert z = x + iy into polar form, first calculate the modulus using r = √(x² + y²). Then find the reference angle α = arctan(|y/x|) and use the quadrant table to determine θ. Finally write z = r(cos θ + i sin θ).

要将 z = x + iy 转换为极形式，首先使用 r = √(x² + y²) 计算模。然后求出参考角 α = arctan(|y/x|)，并利用象限表确定 θ。最后写出 z = r(cos θ + i sin θ)。

For example, consider z = -1 + i√3. The modulus is r = √((-1)² + (√3)²) = √(1 + 3) = 2. The reference angle is α = arctan(√3/1) = π/3. Since x < 0 and y > 0, the point is in the second quadrant, so θ = π – π/3 = 2π/3. Hence z = 2(cos 2π/3 + i sin 2π/3).

例如，考虑 z = -1 + i√3。模为 r = √((-1)² + (√3)²) = √(1 + 3) = 2。参考角为 α = arctan(√3/1) = π/3。由于 x < 0 且 y > 0，该点

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