Chapter I, Problem 26E

To determine

To write: The number $1-\sqrt{3}i$ in polar form with argument between 0 and $2\pi$.

Answer to Problem 26E

The number $1-\sqrt{3}i$ written in polar form as $1-\sqrt{3}i=2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$.

Explanation of Solution

The polar form of the complex number $z=a+bi$ is $z=r\left(\cos \theta +i\sin \theta \right)$ where $r=\left|z\right|=\sqrt{a^2+b^2}$  and $\tan \theta =\frac{b}{a}$.

Consider the complex number $1-\sqrt{3}i$.

Obtain the argument of the complex number $1-\sqrt{3}i$.

$\begin{array}{c}\tan \theta =\frac{b}{a}\\ =\frac{-\sqrt{3}}{1}\\ =-\sqrt{3}\end{array}$

Thus, the argument of argument of the complex number $1-\sqrt{3}i$ is $\theta ={\tan }^{-1}\left(-\sqrt{3}\right)=\frac{5\pi }{3}$

Obtain the modulus of the complex number $1-\sqrt{3}i$.

$\begin{array}{c}r=\left|1-\sqrt{3}i\right|=\\ =\sqrt{1^2+{\left(-\sqrt{3}\right)}^2}\\ =\sqrt{1+3}\\ =2\end{array}$

Thus, the value of $r=2$.

Therefore, the polar form of the complex number $1-\sqrt{3}i$ is $1-\sqrt{3}i=2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$.