Chapter 6.6, Problem 8E

Solution

To find: The polar form of the complex number.

The polar form of the complex number is $z=3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$ .

Given data:

The complex number is $3-3i$ .

Formula used:

The polar form of a complex number $z=a+bi$ is given by,

  $z=r\left(\cos \theta +i\sin \theta \right)$

Here, $r=\sqrt{a^2+b^2}$ and $\tan \theta =\frac{b}{a}$ .

Calculation:

From the given complex number,

  $\begin{array}{c}a=3\\ b=-3\end{array}$

Substitute $3$ for $a$ and $-3$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(3\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{9+9}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

Substitute $3$ for $a$ and $-3$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{-3}{3}\\ \theta ={\tan }^{-1}\left(-1\right)\end{array}$

The reference angle is,

  $\begin{array}{c}{\theta }^{\prime }={\tan }^{-1}\left(1\right)\\ =\frac{\pi }{4}\end{array}$

Since the complex number is in QIV, then,

  $\begin{array}{c}\theta =2\pi -{\theta }^{\prime }\\ =2\pi -\frac{\pi }{4}\\ =\frac{7\pi }{4}\end{array}$

Substitute $3\sqrt{2}$ for $r$ and $\frac{7\pi }{4}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

Therefore, the polar form of the complex number is $z=3\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$ .