Chapter 8, Problem 5T

To determine

To evaluate: The product $z_1{z}_2$ and quotient $\frac{z_1}{z_2}$ of the complex numbers $z_1=4\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ and $z_2=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$.

Answer to Problem 5T

The product $z_1{z}_2$ and quotient $\frac{z_1}{z_2}$ of the complex numbers $z_1=4\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ and $z_2=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$ are $\left(\cos \left(\pi \right)+i\sin \left(\pi \right)\right)$ and $2\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)$.

Explanation of Solution

Given:

The two complex numbers are,

$z_1=4\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$

$z_2=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

Formula used: Multiplication rule for two complex numbers in polar form.

$z_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)$ (1)

Calculation:

Substitute 4 for $r_1$, 2 for $r_2$, $\frac{7\pi }{12}$ for ${\theta }_1$ and $\frac{5\pi }{12}$ for ${\theta }_2$ in the equation (1) to find $z_1{z}_2$.

$\begin{array}{c}{z}_1{z}_2=\left(4\right)\left(2\right)\left(\cos \left(\frac{7\pi }{12}+\frac{5\pi }{12}\right)+i\sin \left(\frac{7\pi }{12}+\frac{5\pi }{12}\right)\right)\\ =8\left(\cos \left(\frac{7\pi +5\pi }{12}\right)+i\sin \left(\frac{7\pi +5\pi }{12}\right)\right)\\ =8\left(\cos \left(\pi \right)+i\sin \left(\pi \right)\right)\end{array}$

Formula used: Division rule for two complex numbers in polar form.

$\frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)$ (2)

Calculation:

Substitute 4 for $r_1$, 2 for $r_2$, $\frac{7\pi }{12}$ for ${\theta }_1$ and $\frac{5\pi }{12}$ for ${\theta }_2$ in the equation (2) to find $\frac{z_1}{z_2}$.

$\begin{array}{c}\frac{z_1}{z_2}=\frac{4}{2}\left(\cos \left(\frac{7\pi }{12}-\frac{5\pi }{12}\right)+i\sin \left(\frac{7\pi }{12}-\frac{5\pi }{12}\right)\right)\\ =2\left(\cos \left(\frac{7\pi -5\pi }{12}\right)+i\sin \left(\frac{7\pi -5\pi }{12}\right)\right)\\ =2\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)\end{array}$

The product $z_1{z}_2$ and quotient $\frac{z_1}{z_2}$ of the complex numbers $z_1=4\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ and $z_2=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$ are $\left(\cos \left(\pi \right)+i\sin \left(\pi \right)\right)$ and $2\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)$.