Chapter 6.6, Problem 84E

To determine

To calculate:

The power of following complex number in the standard form:

  $4{\left(1-\sqrt{3}i\right)}^3$

Answer to Problem 84E

The power of $4{\left(1-\sqrt{3}i\right)}^3$ in the standard form is $32$ .

Explanation of Solution

Given information:

  $4{\left(1-\sqrt{3}i\right)}^3$

Formula used:

The DeMoivre’s theorem for complex numbers is given as:

If $z=r\left(\cos +i\sin \theta \right)$ is complex number it has positive integer $n$ then

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

Calculation:

The modulus value and argument of the given number is calculated as:

  $\begin{array}{l}r=\sqrt{a^2+b^2}=\sqrt{{\left(1\right)}^2+{\left(-\sqrt{3}\right)}^2}=\sqrt{1+4}=2\\ \theta ={\tan }^{-1}\left(\frac{b}{a}\right)={\tan }^{-1}\left(\frac{-\sqrt{3}}{1}\right)={\tan }^{-1}\left(-\sqrt{3}\right)=-\frac{\pi }{3}\\ \theta =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}\end{array}$

So, the given complex number of standard form is written in the trigonometric form as:

  $z=2\left(\cos \left(\frac{5\pi }{3}\right)+i\sin \left(\frac{5\pi }{3}\right)\right)$

By using DeMoivre’s theorem:

  $\begin{array}{l}{z}^n=r^n\left(\cos n\theta +i\sin n\theta \right)\\ 4{\left(1-\sqrt{3}i\right)}^3=4{\left[2\left(\cos \left(\frac{5\pi }{3}\right)+i\sin \left(\frac{5\pi }{3}\right)\right)\right]}^3\\ 4{\left(1-\sqrt{3}i\right)}^3=4{\left(2\right)}^3\left(\cos \left(\frac{15\pi }{3}\right)+i\sin \left(\frac{15\pi }{3}\right)\right)\\ 4{\left(1-\sqrt{3}i\right)}^3=32\left(\cos \left(5\pi \right)+i\sin \left(5\pi \right)\right)\\ 4{\left(1-\sqrt{3}i\right)}^3=32\left[1+0i\right]\\ 4{\left(1-\sqrt{3}i\right)}^3=32\end{array}$