极坐标形式下的除法

现在

$$\begin{aligned}\frac{1}{\cos \theta + j \sin \theta} &= \frac{1}{\cos \theta + j \sin \theta} \frac{\cos \theta - j \sin \theta}{\cos \theta - j \sin \theta} \ &= \frac{\cos \theta - j \sin \theta}{\cos^2 \theta + \sin^2 \theta} \ &= \cos \theta - j \sin \theta, \quad \text{since } \cos^2 \theta + \sin^2 \theta = 1\end{aligned}$$

因此如果

$$z_1 = r_1(\cos \theta_1 + j \sin \theta_1) \quad \text{and} \quad z_2 = r_2(\cos \theta_2 + j \sin \theta_2)$$

则

$$\begin{aligned}\frac{z_1}{z_2} &= \frac{r_1(\cos \theta_1 + j \sin \theta_1)}{r_2(\cos \theta_2 + j \sin \theta_2)} \ &= \frac{r_1}{r_2} (\cos \theta_1 + j \sin \theta_1)(\cos \theta_2 - j \sin \theta_2) \quad (\text{from above}) \ &= \frac{r_1}{r_2} [(\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) + j(\sin \theta_1 \cos \theta_2 - \cos \theta_1 \sin \theta_2)]\end{aligned}$$

或

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} [\cos(\theta_1 - \theta_2) + j \sin(\theta_1 - \theta_2)] \quad (3.6)$$

利用三角恒等式 (2.25b, d)。因此

$$\left| \frac{z_1}{z_2} \right| = \frac{r_1}{r_2} = \frac{|z_1|}{|z_2|} \quad (3.7)$$

且

$$\arg\left(\frac{z_1}{z_2}\right) = \theta_1 - \theta_2 = \arg z_1 - \arg z_2 \quad (3.8)$$

同样可能需要一些调整以确保 $-\pi < \arg(z_1/z_2) \leq \pi$ .

例 3.13

对于以下复数对，求 $z_1/z_2$ 和 $z_2/z_1$ .

• (a) $z_1 = 4(\cos \pi/2 + j \sin \pi/2)$ , $z_2 = 9(\cos \pi/3 + j \sin \pi/3)$
• (b) $z_1 = \cos 3\pi/4 + j \sin 3\pi/4$ , $z_2 = 2(\cos \pi/8 + j \sin \pi/8)$