Chapter 8, Problem 35RE

To determine

De Moiré’s theorem and find the indicated power. ${\left(1-\sqrt{3}i\right)}^4$

Answer to Problem 35RE

The indicated power of ${\left(1-\sqrt{3}i\right)}^4$ is $2^3\left(-1+i\sqrt{3}\right)$ .

Explanation of Solution

Given:

  ${\left(1-\sqrt{3}i\right)}^4$

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 ${\left(1-\sqrt{3}i\right)}^4$

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

Now calculate the value of $r$ ,

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

Now finding the value of $\theta$ ,

  $\theta ={\tan }^{-1}\left(\frac{-\sqrt{3}}{1}\right)=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$ as $x:+ve,y:-ve\text{so}4th\text{quaderant}\text{.}$

Now after conversion to polar form,

  ${\left(1-\sqrt{3}i\right)}^4={\left(2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\right)}^4$

Now by applying the de moivres formula theorem,

  ${\left(1-\sqrt{3}i\right)}^4=2^4\left(\cos \frac{4\times 5\pi }{3}+i\sin \frac{4\times 5\pi }{3}\right)$

Upon simplifying,

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

Conclusion:

Hence, the indicated power of ${\left(1-\sqrt{3}i\right)}^4$ is $2^3\left(-1+i\sqrt{3}\right)$ .