Chapter 9, Problem 10CT

To determine

The product $z\cdot w$, where $z=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)$ and $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$, and express the answer in polar form and exponential form.

Answer to Problem 10CT

Solution:

The product $z\cdot w$ in the polar form is $6\left(\cos \frac{107\pi }{180}+i\sin \frac{107\pi }{180}\right)$ and the exponential form is $6e^{i\cdot \frac{107\pi }{180}}$.

Explanation of Solution

Given information:

$z=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)$ and $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$

Explanation:

Let $z=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)$ and $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$.

$z\cdot w=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)\cdot 3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$

First express the complex numbers in exponential form.

$z\cdot w=2e^{i\cdot \frac{17\pi }{36}}\cdot 3e^{i\cdot \frac{11\pi }{90}}$

The product of $z_1=r_1{e}^{i{\theta }_1}$ and $z_2=r_2{e}^{i{\theta }_2}$ is find as $z_1\cdot z_2=r_1{r}_2{e}^{i\left({\theta }_1+{\theta }_2\right)}$

$\Rightarrow z\cdot w=2\cdot 3e^{i\cdot \left(\frac{17\pi }{36}+\frac{11\pi }{90}\right)}$

Thus $z\cdot w=6e^{i\cdot \frac{107\pi }{180}}$

In the polar form, $z\cdot w=6\left(\cos \frac{107\pi }{180}+i\sin \frac{107\pi }{180}\right)$.

Therefore, the product $z\cdot w$ in polar formis $6\left(\cos \frac{107\pi }{180}+i\sin \frac{107\pi }{180}\right)$, and the exponential form is $6e^{i\cdot \frac{107\pi }{180}}$.