Chapter 14.4, Problem 8OE

To determine

To find: The given complex number in polar form.

Answer to Problem 8OE

  $\sqrt{2}\left(\cos {225}^0+i\sin {225}^0\right)$ .

Explanation of Solution

Given information:

A complex number is given as $-1-i$ .

Concepts used: The polar form of a complex number $z=x+yi$ is given by −

  $z=r\left(\cos \theta +i\sin \theta \right)$

where $r=\sqrt{x^2+y^2},\cos \theta =\frac{x}{r},\sin \theta =\frac{y}{r}$ .

Calculation:

Given complex number in polar form is $-1-i$ .

Converting into polar form -

  $\begin{array}{l}-1-i\\ r=\sqrt{x^2+y^2}=\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2}=\sqrt{2}\\ \sin \theta =\frac{y}{r}=\frac{-1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\\ \Rightarrow \theta ={225}^0,{315}^0\\ \cos \theta =\frac{-1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\\ \Rightarrow \theta ={135}^0,{225}^0\\ \text{Taking common values of }\theta ={225}^0\end{array}$

Hence, given complex number in desired form is $\sqrt{2}\left(\cos {225}^0+i\sin {225}^0\right)$ .