Chapter 6.6, Problem 37E

Solution

To find: The indicated power of the complex number.

The indicated power of the complex number is $-8$ .

Given information:

The complex number is ${\left(1-\sqrt{3}i\right)}^3$ .

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 $\tan \theta =\frac{b}{a}$ .

By the De Moivre’s Theorem,

If $z=r\left(\cos \theta +i\sin \theta \right)$ and $n$ is a positive integer then,

  $\begin{array}{c}{z}^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n\\ =r^n\left(\cos n\theta +i\sin n\theta \right)\end{array}$

Calculation:

From the given complex number,

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

Substitute $1$ for $a$ and $-\sqrt{3}$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{1^2+{\left(-\sqrt{3}\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

Substitute $1$ for $a$ and $-\sqrt{3}$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

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

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

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

Find the indicated power of the complex number.

  $\begin{array}{c}{\left(1-\sqrt{3}i\right)}^3=2^3\left(\cos \left(-\frac{\pi }{3}\cdot 3\right)+i\sin \left(-\frac{\pi }{3}\cdot 3\right)\right)\\ =8\left(\cos \left(-\pi \right)+i\sin \left(-\pi \right)\right)\\ =8\left(-1+i\cdot 0\right)\\ =-8\end{array}$

Therefore, the indicated power of the complex number is $-8$ .