Chapter 9.8, Problem 1CFU

To determine

To evaluate the given product by traditional multiplication and to compare the results using De Moivre’s theorem.

Answer to Problem 1CFU

  $-4-4i$

Explanation of Solution

Given:

The given product is $(1+i)(1+i)(1+i)(1+i)(1+i)$

Calculation:

  $\begin{array}{l}(1+i)(1+i)(1+i)(1+i)(1+i)\\ ={(}^1\end{array}$ .

De Moiver’ theorem will be quite easy as compared to traditional multiplication and it will be preferred more.

Computing the product as follows-

  $z=1+i$

Finding the polar form

  $\begin{array}{l}r=\sqrt{1^2+1^2}=\sqrt{2}\\ \theta ={45}^{\circ }=\frac{\pi }{4}\end{array}$

Let us write the trigonometric form as follows-

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right)\\ z=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\end{array}$

Now using De Moiver’ theorem as follows-

  $\begin{array}{l}{z}^n=r^n(\cos n\theta +i\sin n\theta )\\ z^5={\left(\sqrt{2}\right)}^5\left(\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right)\\ z^5={\left(\sqrt{2}\right)}^5\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\\ z^5=-\frac{{\left(\sqrt{2}\right)}^6}{2}-\frac{{\left(\sqrt{2}\right)}^6}{2}i\\ z^5=-4-4i\end{array}$