Chapter 6, Problem 45RE

Solution

a)

To find: the polar form of ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ De Moivre’s Theorem.

The polar form of ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ is $243\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ .

Given information:

Given complex number: $3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

Formula used:

De Moivre’s Theorem:

Let $z=r\left(\cos \theta +i\sin \theta \right)$ and let $n$ be positive integer. Then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ .

Calculation:

Substitute 3 for $r$ , $\frac{\pi }{4}$ for $\theta$ and $5$ for $n$ in $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ :

  $\begin{array}{c}{z}^n=r^n\left(\cos n\theta +i\sin n\theta \right)\\ z^5={\left(3\right)}^5\left(\cos 5\left(\frac{\pi }{4}\right)+i\sin 5\left(\frac{\pi }{4}\right)\right)\\ =243\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\end{array}$

Thus, $z^5$ is $243\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ .

b)

To find: the standard form expression of ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ .

The standard form of the complex number ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ is $-\frac{243}{\sqrt{2}}-\frac{243}{\sqrt{2}}i$ .

Given information:

Given complex number: $3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

From part (a), the polar form of ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ is $243\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ .

Formula used:

Polar form to Standard form of complex number:

If polar form a complex number $z$ is $r\left(\cos \theta +i\sin \theta \right)$ , then the standard form of $z$ is $a+bi$ , where $a=r\cos \theta ,$

  $b=r\sin \theta ,$ and $r=\sqrt{a^2+b^2}$ .

Calculation:

Compare given complex number $243\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ with polar form $r\left(\cos \theta +i\sin \theta \right)$. Conclude that $r=243,$ and $\theta =\frac{5\pi }{4}$ .

Standard form of complex number:

The standard form of ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ is $a+bi$ . The values of $a$ and $b$

computed as follows:

Substitute the values $r=243,$ and $\theta =\frac{5\pi }{4}$ in the equations $a=r\cos \theta$ and $b=r\sin \theta$:

  $\begin{array}{c}a=r\cos \theta \\ =243\left(\cos \frac{5\pi }{4}\right)\\ =243\left(\cos \left(\pi +\frac{\pi }{4}\right)\right)\\ =243\left(-\cos \left(\frac{\pi }{4}\right)\right)\\ =-243\left(\frac{1}{\sqrt{2}}\right)\end{array}$

  $\begin{array}{c}b=r\sin \theta \\ =243\left(\sin \frac{5\pi }{4}\right)\\ =243\left(\sin \left(\pi +\frac{\pi }{4}\right)\right)\\ =243\left(-\sin \left(\frac{\pi }{4}\right)\right)\\ =-243\left(\frac{1}{\sqrt{2}}\right)\\ =-\frac{243}{\sqrt{2}}\end{array}$

Thus, the standard form of the complex number ${\left[ 3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^5$ is $-\frac{243}{\sqrt{2}}-\frac{243}{\sqrt{2}}i$ .