解答 (a) $|z_1| = 4$ , $\arg z_1 = \pi/2$ ; $|z_2| = 9$ , $\arg z_2 = \pi/3$

由 (3.7)

$$\left| \frac{z_1}{z_2} \right| = \frac{4}{9} \quad \text{and} \quad \left| \frac{z_2}{z_1} \right| = \frac{9}{4}$$

由 (3.8)

$$\arg\left(\frac{z_1}{z_2}\right) = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \quad \text{and} \quad \arg\left(\frac{z_2}{z_1}\right) = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$$

$$\text{Thus} \quad \frac{z_1}{z_2} = \frac{4}{9} \left( \cos \frac{\pi}{6} + j \sin \frac{\pi}{6} \right)$$

$$\text{and} \quad \frac{z_2}{z_1} = \frac{9}{4} \left( \cos \frac{\pi}{6} - j \sin \frac{\pi}{6} \right)$$

(b) $|z_1| = 1$ , $\arg z_1 = 3\pi/4$ ; $|z_2| = 2$ , $\arg z_2 = \pi/8$

由 (3.7)

$$\left| \frac{z_1}{z_2} \right| = \frac{1}{2} \quad \text{and} \quad \left| \frac{z_2}{z_1} \right| = 2$$

由 (3.8)

$$\arg\left(\frac{z_1}{z_2}\right) = \frac{3\pi}{4} - \frac{\pi}{8} = \frac{5\pi}{8} \quad \text{and} \quad \arg\left(\frac{z_2}{z_1}\right) = \frac{\pi}{8} - \frac{3\pi}{4} = -\frac{5\pi}{8}$$

$$\text{Thus} \quad \frac{z_1}{z_2} = \frac{1}{2} \left( \cos \frac{5\pi}{8} + j \sin \frac{5\pi}{8} \right)$$

$$\text{and} \quad \frac{z_2}{z_1} = 2 \left( \cos \frac{5\pi}{8} - j \sin \frac{5\pi}{8} \right)$$

例 3.14

求下列复数的模和辐角

$$z = \frac{(1 + j2)^2(4 - j3)^3}{(3 + j4)^4(2 - j)^3}$$

解答

$$\begin{aligned} |z| &= \frac{|1 + j2|^2 |4 - j3|^3}{|3 + j4|^4 |2 - j|^3} \ &= \frac{[\sqrt{(1 + 4)}]^2 [\sqrt{(16 + 9)}]^3}{[\sqrt{(9 + 16)}]^4 [\sqrt{(4 + 1)}]^3} = \frac{1}{25} \sqrt{5} \end{aligned}$$

$$\begin{aligned} \arg z &= 2 \arg(1 + j2) + 3 \arg(4 - j3) - 4 \arg(3 + j4) - 3 \arg(2 - j) \ &= 2(1.107) + 3(-0.643) - 4(0.927) - 3(-0.461) = -2.035 \end{aligned}$$