Chapter 9, Problem 44RE

To determine

To write: Complex number ${\left[2\left(\cos \frac{5\pi }{16}+i\sin \frac{5\pi }{16}\right)\right]}^4$ in the standard form of $\left(a+ib\right)$ .

Answer to Problem 44RE

The mentioned complex number in the standard form is $\left(-\frac{16}{\sqrt{2}}\right)+i\left(-\frac{16}{\sqrt{2}}\right)$ .

Explanation of Solution

Given information:

The mentioned complex number is ${\left[2\left(\cos \frac{5\pi }{16}+i\sin \frac{5\pi }{16}\right)\right]}^4$ .

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 $\cos \left(\pi +\theta \right)=-\cos \theta$ and $\sin \left(\pi +\theta \right)=-\sin \theta$ .

Calculation:

Consider the mentioned complex number is ${\left[2\left(\cos \frac{5\pi }{16}+i\sin \frac{5\pi }{16}\right)\right]}^4$ .

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 ${\left[2\left(\cos \frac{5\pi }{16}+i\sin \frac{5\pi }{16}\right)\right]}^4$ which is in the form of $z=r\left(\cos \theta +i\sin \theta \right)$ .

Here $r=2$ , $\theta$ is the angle which is equal to $\frac{5\pi }{16}$ and $n=4$ .

Therefore by De Moivre’s theorem we have,

  $z={\left[2\left(\cos \frac{5\pi }{16}+i\sin \frac{5\pi }{16}\right)\right]}^4\Rightarrow z={\left(2\right)}^4\left(\cos \left(4*\frac{5\pi }{16}\right)+i\sin \left(4*\frac{5\pi }{16}\right)\right)\Rightarrow z=16\left(\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right)$ .

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

  $\begin{array}{l}z=16\left(\cos \left(\pi +\frac{\pi }{4}\right)+i\sin \left(\pi +\frac{\pi }{4}\right)\right)\Rightarrow z=16\left(-\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}\right)\\ \Rightarrow z=16\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)\Rightarrow z=\left(-\frac{16}{\sqrt{2}}-i\frac{16}{\sqrt{2}}\right)\Rightarrow z=\left(-\frac{16}{\sqrt{2}}\right)+i\left(-\frac{16}{\sqrt{2}}\right)\end{array}$

Thus the required Complex number ${\left[2\left(\cos \frac{5\pi }{16}+i\sin \frac{5\pi }{16}\right)\right]}^4$ in the standard form of $\left(a+bi\right)$ is $\left(-\frac{16}{\sqrt{2}}\right)+i\left(-\frac{16}{\sqrt{2}}\right)$ .