Chapter 9, Problem 45RE

To determine

To write: Complex number ${\left(1-\sqrt{3}i\right)}^6$ in the standard form of $\left(a+ib\right)$ .

Answer to Problem 45RE

The mentioned complex number in the standard form is $64+\left(-0i\right)$ .

Explanation of Solution

Given information:

The mentioned complex number is ${\left(1-\sqrt{3}i\right)}^6$ .

Formula used:

By the use of De Moivre’s theorem we know that if $z=r\left(\cos \theta +i\sin \theta \right)$ then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ where n is positive value $\ge 1$ .

Calculation:

Consider the mentioned complex number is ${\left(1-\sqrt{3}i\right)}^6$ .

Recall the De Moivre’s theorem that if $z=r\left(\cos \theta +i\sin \theta \right)$ then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ where n is positive value $\ge 1$ .

Here we will convert the mentioned complex number into the form $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ .

Therefore ${\left(1-\sqrt{3}i\right)}^6\Rightarrow 2{\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}^6\Rightarrow {\left[2\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)\right]}^6$ .

Therefore the mentioned complex number is ${\left(1-\sqrt{3}i\right)}^6$ which is in the form of $z=r\left(\cos \theta +i\sin \theta \right)$ .

Here $r=2$ , $\theta$ is the angle which is equal to $\frac{\pi }{3}$ and $n=6$ .

Therefore by De Moivre’s theorem we have,

  $\begin{array}{l}z={\left[2\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)\right]}^6\Rightarrow z={\left(2\right)}^6\left(\cos \left(6*\frac{\pi }{3}\right)-i\sin \left(6*\frac{\pi }{3}\right)\right)\Rightarrow z=64\left(\cos \left(2\pi \right)-i\sin \left(2\pi \right)\right)\\ z=64\left(1-0i\right)\Rightarrow z=64-0i\end{array}$

Thus the required Complex number ${\left(1-\sqrt{3}i\right)}^6$ in the standard form of $\left(a+bi\right)$ is $64+\left(-0i\right)$ .