Chapter 14.4, Problem 7OE

To determine

To find: The given complex number in polar form.

Answer to Problem 7OE

  $2\left(\cos {330}^0+i\sin {330}^0\right)$ .

Explanation of Solution

Given information:

A complex number is given as $\sqrt{3}-i$ .

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 $\sqrt{3}-i$ .

Converting into polar form -

  $\begin{array}{l}\sqrt{3}-i\\ r=\sqrt{x^2+y^2}=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(-1\right)}^2}=2\\ \sin \theta =\frac{y}{r}=\frac{-1}{2}=-\frac{1}{2}\\ \Rightarrow \theta ={150}^0,{330}^0\\ \cos \theta =\frac{x}{r}=\frac{\sqrt{3}}{2}\\ \Rightarrow \theta ={30}^0,{330}^0\\ \text{Taking common values of }\theta ={330}^0\end{array}$

Hence, given complex number in desired form is $2\left(\cos {330}^0+i\sin {330}^0\right)$ .