Chapter 6, Problem 106CR

To determine

To calculate:

The value of the following complex number in the trigonometric form without using calculator:

  $-3\sqrt{3}+3i$

Answer to Problem 106CR

The trigonometric form of the complex number is $z=2\left(\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right)$ .

Explanation of Solution

Given information:

  $-3\sqrt{3}+3i$

Formula used:

The trigonometric form of the complex number $z=a+bi$ is given by the following expression:

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

The modulus or absolute value of complex number $z=a+bi$ is given by the following expression:

  $r=\sqrt{a^2+b^2}$

The argument of complex number $z=a+bi$ is given by the following expression:

  $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

Calculation:

The given complex number is:

  $-3\sqrt{3}+3i$

Compare it with $z=a+bi$ :

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

The modulus is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(-3\sqrt{3}\right)}^2+{\left(3\right)}^2}\\ r=\sqrt{27+9}\\ r=\sqrt{36}\\ r=6\end{array}$

The argument is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ \theta ={\tan }^{-1}\left(\frac{3}{-3\sqrt{3}}\right)\\ \theta ={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\\ \theta =-\frac{\pi }{6}\end{array}$

The trigonometric form of the complex number is calculated as:

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right)\\ z=2\left(\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\right)\\ z=2\left(\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right)\left[\begin{array}{l}\because \cos \left(-x\right)=\cos x\\ \sin \left(-x\right)=-\sin x\end{array}\right]\end{array}$