Chapter 9.3, Problem 36AYU

To determine

To find: $zw$ and $\frac{z}{w}$.

Answer to Problem 36AYU

Solution:

$zw = 12\left[\cos \left(280°\right) + i\sin \left(280°\right)\right]$ and $\frac{z}{w} = \frac{1}{3}\left[\cos \left(240°\right) + i\sin \left(240°\right)\right]$

Explanation of Solution

Given:

$z = 2\left(\cos 80° + i\sin 80°\right)$; $w = 6\left(\cos 200° + i\sin 200°\right)$

Calculation:

$zw = \left[2\left(\cos 80° + i\sin 80°\right)\right]\left[6\left(\cos 200° + i\sin 200°\right)\right]$

$zw = \left(2 \times 6\right)\left[\cos \left(80° + 200°\right) + i\sin \left(80° + 200°\right)\right]$

By $z_1{z}_2 = r_1{r}_2\left[\cos \left({\theta }_1 + {\theta }_2\right) + \sin \left({\theta }_1 + {\theta }_2\right)\right]$

$zw = 12\left[\cos \left(280°\right) + i\sin \left(280°\right)\right]$

$\frac{z}{w} = \frac{\left[2\left(\cos 80° + i\sin 80°\right)\right]}{\left[6\left(\cos 200° + i\sin 200°\right)\right]}$

$\frac{z}{w} = \frac{2}{6}\left[\cos \left(80° - 200°\right) + i\sin \left(80° - 200°\right)\right]$ By $\frac{z_1}{z_2} = \frac{r_1}{r_2}\left[\cos \left({\theta }_1 - {\theta }_2\right) + i\sin \left({\theta }_1 - {\theta }_2\right)\right]$

$\frac{z}{w} = \frac{1}{3}\left[\cos \left(-120°\right) + i\sin \left(-120°\right)\right]$

$\frac{z}{w} = \frac{1}{3}\left[\cos \left(240°\right) + i\sin \left(240°\right)\right]$