## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.5 - Complex Numbers in Polar Form; DeMoivre's Theorem - Exercise Set - Page 767: 43

#### Answer

$$z= 2\left ( \cos \pi +i \sin \pi \right )$$

#### Work Step by Step

We multiply the complex numbers as follows: $$z_1z_2=(1+i)(-1+i)=-1+i-i+i^2=-1-1=-2$$ Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi$. So, for the complex number $z=-2$ we have$$r=\sqrt{(-2)^2+0^2}=\sqrt{4}=2, \\ \cos \theta =\frac{a}{r}=\frac{-2}{2}=-1.$$ So$$\theta =\pi$$Thus, the complex number has the polar form$$z= 2\left ( \cos \pi +i \sin \pi \right ).$$

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