Chapter 14, Problem 14CLR

To determine

To express: The expression ${\left(2-2i\sqrt{3}\right)}^5$ in $x+yi$ form by using DeMoivre’s theorem.

Answer to Problem 14CLR

The required result in standard for is ${\left(2-2i\sqrt{3}\right)}^5=512\left(1+\sqrt{3}i\right)$ .

Explanation of Solution

Given:

The expression is ${\left(2-2i\sqrt{3}\right)}^5$ .

Calculation:

Compare the expression ${\left(2-2i\sqrt{3}\right)}^5$ with $z=x+yi$ .

  $\begin{array}{c}x=2\\ y=-2\sqrt{3}\end{array}$

Substitute 2 for $x$ and $-2\sqrt{3}$ for $y$ in the equation $r=\sqrt{x^2+y^2}$ to find $r$ .

  $\begin{array}{c}r=\sqrt{{\left(2\right)}^2+{\left(-2\sqrt{3}\right)}^2}\\ =\sqrt{4+12}\\ =\sqrt{16}\\ =4\end{array}$

Find the angle by using the formula $\theta =\arctan \left(\frac{y}{x}\right)$ .

  $\begin{array}{c}\theta =\arctan \left(\frac{-2\sqrt{3}}{2}\right)\\ =300°\end{array}$

So, the polar form of $\left(2-2i\sqrt{3}\right)$ is $4\left(\cos 300°+i\sin 300°\right)$ .

The DeMoivre’s theorem states that if $z^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n$ and $n$ is a positive integer then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ . So,

  $\begin{array}{c}{\left(2-2i\sqrt{3}\right)}^5={\left[4\left(\cos 300°+i\sin 300°\right)\right]}^5\\ ={\left(4\right)}^5\left(\cos 1500°+i\sin 1500°\right)\\ =1024\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\ =512\left(1+\sqrt{3}i\right)\end{array}$

Therefore, the required result in standard for is ${\left(2-2i\sqrt{3}\right)}^5=512\left(1+\sqrt{3}i\right)$ .