Chapter 14.4, Problem 21WE

To determine

To find: The given complex number in polar form.

Answer to Problem 21WE

  $2\sqrt{6}\left(\cos {135}^0+i\sin {135}^0\right)$ .

Explanation of Solution

Given information:

A complex number is given as $2\sqrt{3}\left(-1+i\right)$ .

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 $2\sqrt{3}\left(-1+i\right)$ .

Converting into polar form -

  $\begin{array}{l}2\sqrt{3}\left(-1+i\right)\\ =-2\sqrt{3}+2\sqrt{3}i\\ r=\sqrt{x^2+y^2}=\sqrt{{\left(-2\sqrt{3}\right)}^2+{\left(2\sqrt{3}\right)}^2}=2\sqrt{6}\\ \cos \theta =\frac{x}{r}=-\frac{2\sqrt{3}}{2\sqrt{6}}=-\frac{1}{\sqrt{2}}\\ \Rightarrow \theta ={135}^0,{225}^0\\ \sin \theta =\frac{y}{r}=\frac{2\sqrt{3}}{2\sqrt{6}}=\frac{1}{\sqrt{2}}\\ \Rightarrow \theta ={45}^0,{135}^0\\ \text{Taking common value of }\theta \text{.}\end{array}$

Hence, given complex number in desired form is $2\sqrt{6}\left(\cos {135}^0+i\sin {135}^0\right)$ .