Chapter 8, Problem 37RE

To determine

De Moiré’s theorem and find the indicated power. ${\left(\sqrt{3}+i\right)}^{-4}$

Answer to Problem 37RE

The indicated power of ${\left(\sqrt{3}+i\right)}^{-4}$ is $\frac{1}{32}\left(-1-i\sqrt{3}\right)$ .

Explanation of Solution

Given:

  ${\left(\sqrt{3}+i\right)}^{-4}$

Concept Used:

De Moiré’s theorem that state for any complex number.

  ${\left(\cos \theta +i\sin \theta \right)}^n=\left(\cos n\theta +i\sin n\theta \right)$

Calculations:

First convert the complex number in the polar form the equation is,

  $a+ib=r\left(\cos \theta +\sin \theta \right)$

Now compare real and imaginary parts,

  $\begin{array}{l}a=r\cos \theta ,\\ b=r\sin \theta \end{array}$

Now squaring and adding real and imaginary parts now find the value of $r$ ,

  $\begin{array}{l}{r}^2=a^2+b^2\\ r=\sqrt{a^2+b^2}\end{array}$

Angle can be found by dividing imaginary by real part of the complex number and quadrant can be found using sign of real $x$ and imaginary $y$ part, so the angle can be found accordingly.

  $\tan \theta =\frac{b}{a};\theta {\tan }^{-1}\left(\frac{b}{a}\right)$

So, the given complex number is ${\left(\sqrt{3}+i\right)}^{-4}$

Now by comparing the $a+ib$ , the simplification is $a=\sqrt{3},b=1$ .

Now calculate the value of $r$ ,

  $r=\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=\sqrt{3+1}=\sqrt{4}=2$

Now finding the value of $\theta$ ,

  $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6};$ as $x:+ve,y:-ve\text{so}1st\text{quaderant}\text{.}$

Now after conversion to polar form,

  ${\left(\sqrt{3}+i\right)}^{-4}={\left(2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\right)}^{-4}$

Now by applying the de moivres formula theorem,

  $\begin{array}{l}{\left(\sqrt{3}+i\right)}^{-4}=2^{-4}{\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)}^{-4}\\ {\left(\sqrt{3}+i\right)}^{-4}=2^{-4}\left(\cos \frac{-4\pi }{6}+i\sin \frac{-4\pi }{6}\right)\end{array}$

Upon simplifying,

  $\begin{array}{l}{\left(\sqrt{3}+i\right)}^{-4}=2^{-4}\left(\cos \frac{4\pi }{6}-i\sin \frac{4\pi }{6}\right)\\ {\left(\sqrt{3}+i\right)}^{-4}=2^{-4}\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=2^{-5}\left(-1-\sqrt{3}\right)\\ {\left(\sqrt{3}+i\right)}^{-4}=\frac{1}{32}\left(-1-i\sqrt{3}\right)\end{array}$

Conclusion:

Hence, the indicated power of ${\left(\sqrt{3}+i\right)}^{-4}$ is $\frac{1}{32}\left(-1-i\sqrt{3}\right)$ .