Chapter 8.3, Problem 79E

To determine

To calculate: To find the indicated power of complex number using De Moivre’s Theorem

Answer to Problem 79E

  ${\left(2\sqrt{3}+2i\right)}^{-5}=\frac{1}{2048}\left(-\sqrt{3}-i\right)$

Explanation of Solution

Given information: Complex number is ${\left(2\sqrt{3}+2i\right)}^{-5}$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

Polar form of the complex number is given as

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

According to De Moivre’s Theorem,

  $z^n=r^n\left(\cos n\theta +isinn\theta \right)$

Calculation:

Complex number is given as

  ${\left(2\sqrt{3}+2i\right)}^{-5}$

Let us consider complex number:

  $z=2\sqrt{3}+2i$

Polar form of complex number is

  $z=r\left(\cos \theta +isin\theta \right)--(1)$

Calculating the value of $r$ :

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

Calculating the value of $\theta :$

  $\begin{array}{l}\tan \theta =\frac{2}{2\sqrt{3}}\\ \tan \theta =\frac{1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\frac{1}{\sqrt{3}}\\ \theta =\frac{\pi }{6}\end{array}$

Hence, complex number in polar form is

  $z=4\left(\cos \left(\frac{\pi }{6}\right)+isin\left(\frac{\pi }{6}\right)\right)$

Now, ${\left(2\sqrt{3}+2i\right)}^{-5}$ is calculated using De Moivre’s Theorem as follows:

  $\begin{array}{l}{\left(2\sqrt{3}+2i\right)}^{-5}={\left(4\left(\cos \left(-5\times \frac{\pi }{6}\right)+isin\left(-5\times \frac{\pi }{6}\right)\right)\right)}^{-5}\\ {\left(2\sqrt{3}+2i\right)}^{-5}=\frac{1}{{\left(4\right)}^5}\left(\cos \left(-\frac{5\pi }{6}\right)+isin\left(-\frac{5\pi }{6}\right)\right)\\ {\left(2\sqrt{3}+2i\right)}^{-5}=\frac{1}{1024}\left(\left(-\frac{\sqrt{3}}{2}\right)+i\left(-\frac{1}{2}\right)\right)\\ {\left(2\sqrt{3}+2i\right)}^{-5}=\frac{1}{1024}\left(\frac{-\sqrt{3}-i}{2}\right)\\ {\left(2\sqrt{3}+2i\right)}^{-5}=\frac{1}{2048}\left(-\sqrt{3}-i\right)\end{array}$

Conclusion:

Hence, ${\left(2\sqrt{3}+2i\right)}^{-5}=\frac{1}{2048}\left(-\sqrt{3}-i\right)$