Chapter 9, Problem 43RE

To determine

To write: Complex number ${\left[\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\right]}^4$ in the standard form of $\left(a+ib\right)$ .

Answer to Problem 43RE

The mentioned complex number in the standard form is $0+4i$ .

Explanation of Solution

Given information:

The mentioned complex number is ${\left[\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\right]}^4$ .

Formula used:

By the use of De Moivre’s theorem we know that if $z=r\left(\cos \theta +i\sin \theta \right)$ then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ where n is positive value $\ge 1$ .

Also the trigonometric formula $\cos \left(2\pi +\theta \right)=\cos \theta$ and $\sin \left(2\pi +\theta \right)=\sin \theta$ .

Calculation:

Consider the mentioned complex number is ${\left[\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\right]}^4$ .

Recall the De Moivre’s theorem that if $z=r\left(\cos \theta +i\sin \theta \right)$ then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ where n is positive value $\ge 1$ .

Therefore the mentioned complex number is ${\left[\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\right]}^4$ which is in the form of $z=r\left(\cos \theta +i\sin \theta \right)$ .

Here $r=\sqrt{2}$ , $\theta$ is the angle which is equal to $\frac{5\pi }{8}$ and $n=4$ .

Therefore by De Moivre’s theorem we have,

  $z={\left[\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\right]}^4\Rightarrow z={\left(\sqrt{2}\right)}^4\left(\cos \left(4*\frac{5\pi }{8}\right)+i\sin \left(4*\frac{5\pi }{8}\right)\right)\Rightarrow z=4\left(\cos \left(\frac{5\pi }{2}\right)+i\sin \left(\frac{5\pi }{2}\right)\right)$ .

Recall the trigonometric formula $\cos \left(2\pi +\theta \right)=\cos \theta$ and $\sin \left(2\pi +\theta \right)=\sin \theta$ .

  $\begin{array}{l}z=4\left(\cos \left(2\pi +\frac{\pi }{2}\right)+i\sin \left(2\pi +\frac{\pi }{2}\right)\right)\Rightarrow z=4\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ \Rightarrow z=4\left(0+i*1\right)\Rightarrow z=\left(0+4i\right)\end{array}$

Thus the required Complex number ${\left[\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\right]}^4$ in the standard form of $\left(a+bi\right)$ is $0+4i$ .