Chapter I, Problem 25E

Write the number in polar form with argument between 0 and 2π.

25. −3 + 3i

To determine

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

Answer to Problem 25E

The number $-3+3i$ written in polar form as $-3+3i=3\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$.

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 $-3+3i$.

Obtain the argument of the complex number $-3+3i$.

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

Thus, the argument of argument of the complex number $-3+3i$ is $\theta ={\tan }^{-1}\left(-1\right)=\frac{3\pi }{4}$

Obtain the modulus of the complex number $-3+3i$.

$\begin{array}{c}r=\left|-3+3i\right|=\\ =\sqrt{{\left(-3\right)}^2+3^2}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

Thus, the value of $r=3\sqrt{2}$.

Therefore, the polar form of the complex number $-3+3i$ is $-3+3i=3\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$.