Chapter I, Problem 28E

To determine

To write: The number $8i$ in polar form with argument between 0 and $2\pi$.

Answer to Problem 28E

The number $8i$ written in polar form as $8i=8\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$.

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 $8i$.

Obtain the argument of the complex number $8i$.

$\begin{array}{c}\tan \theta =\frac{b}{a}\\ =\frac{8}{0}\\ =\infty \end{array}$

Thus, the argument of argument of the complex number $8i$ is $\theta ={\tan }^{-1}\left(\infty \right)=\frac{\pi }{2}$

Obtain the modulus of the complex number $8i$.

$\begin{array}{c}r=\left|8i\right|=\\ =\sqrt{{\left(0\right)}^2+8^2}\\ =\sqrt{64}\\ =8\end{array}$

Thus, the value of $r=8$.

Therefore, the polar form of the complex number $8i$ is $8i=8\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$.