Chapter 9, Problem 12CT

To determine

The value of $w^5$ , where $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$ , and express the answer in polar form and exponential form.

Answer to Problem 12CT

Solution:

The value of $w^5$ in polar form of is $243\left(\cos \frac{11\pi }{18}+i\sin \frac{11\pi }{18}\right)$ , and the exponential form is $243e^{i\cdot \frac{11\pi }{18}}$ .

Explanation of Solution

Given information:

  $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$ .\

  $w^5={\left[3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)\right]}^5$

First express the complex numbers in exponential form.

  $w^5={\left(3e^{i\cdot \frac{11\pi }{90}}\right)}^5$

Use De Moivre’s theorem.

If $z=r{e}^{i\theta }$ is a complex number, then $z^n=r^n{e}^{i\left(n\theta \right)}$ .

  $\Rightarrow w^5=3^5{e}^{i\cdot \left(5\cdot \frac{11\pi }{90}\right)}=243e^{i\cdot \left(\frac{11\pi }{18}\right)}$

Thus $w^5=243e^{i\cdot \left(\frac{11\pi }{18}\right)}$

In the polar form, $w^5=243\left(\cos \frac{11\pi }{18}+i\sin \frac{11\pi }{18}\right)$ .

Therefore, the value of $w^5$ in polar form is $243\left(\cos \frac{11\pi }{18}+i\sin \frac{11\pi }{18}\right)$ , and the exponential form is $243e^{i\cdot \frac{11\pi }{18}}$ .