Chapter 8.3, Problem 74E

To determine

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

Answer to Problem 74E

  ${\left(\sqrt{3}-i\right)}^{-10}=\frac{1}{2048}\left(1-\sqrt{3}\right)i$

Explanation of Solution

Given information: Complex number is ${\left(\sqrt{3}-i\right)}^{-10}$

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

  ${\left(\sqrt{3}-i\right)}^{-10}$

Let us consider complex number:

  $z=\sqrt{3}-i$

Polar form of complex number is

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

Calculating the value of $r$ :

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

Calculating the value of ${\theta }_1$

  $\begin{array}{l}\tan \theta =\frac{-1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\frac{-1}{\sqrt{3}}\\ \theta =2\pi -\frac{\pi }{6}\\ \theta =\frac{11\pi }{6}\end{array}$

Hence, complex number in polar form is

  $z=2\left(\cos \frac{11\pi }{6}+isin\frac{11\pi }{6}\right)$

Now, ${\left(\sqrt{3}-i\right)}^{-10}$ is calculated using De Moivre’s Theorem as follows:

  $\begin{array}{l}{\left(\sqrt{3}-i\right)}^{-10}={\left(2\left(\cos \frac{11\pi }{6}+isin\frac{11\pi }{6}\right)\right)}^{-10}\\ {\left(\sqrt{3}-i\right)}^{-10}={\left(2\right)}^{-10}\left(\cos \left(-10\times \frac{11\pi }{6}\right)+isin\left(-10\times \frac{11\pi }{6}\right)\right)\\ {\left(\sqrt{3}-i\right)}^{-10}=\frac{1}{{\left(2\right)}^{10}}\left(\cos \left(-\frac{110\pi }{6}\right)+isin\left(-\frac{110\pi }{6}\right)\right)\\ {\left(\sqrt{3}-i\right)}^{-10}=\frac{1}{1024}\left(\frac{1}{2}+i\left(-\frac{\sqrt{3}}{2}\right)\right)\\ {\left(\sqrt{3}-i\right)}^{-10}=\frac{1}{2048}-\frac{\sqrt{3}}{2048}i\\ {\left(\sqrt{3}-i\right)}^{-10}=\frac{1}{2048}\left(1-\sqrt{3}\right)i\end{array}$

Conclusion:

Hence, ${\left(\sqrt{3}-i\right)}^{-10}=\frac{1}{2048}\left(1-\sqrt{3}\right)i$