Chapter 9, Problem 31RE

To determine

To write: Complex number $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ in the standard form of $\left(a+bi\right)$ .

Answer to Problem 31RE

The mentioned complex number in the standard form is $-\frac{3}{2}+i\frac{3\sqrt{3}}{2}$ .

Explanation of Solution

Given information:

The mentioned complex number is $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ .

Formula used:

By the use of De Moivre’stheorem 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 any integral value.

Also trigonometric formula $\cos \left(\pi -\theta \right)=-\cos \theta$ and $\sin \left(\pi -\theta \right)=\sin \theta$ .

Calculation:

Consider the mentioned complex number is $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ .

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 any integral value.

Therefore the mentioned complex number is $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ which is in the form of $z=r\left(\cos \theta +i\sin \theta \right)$ .

Here $r=3$ , $\theta$ is the angle which is equal to $\frac{2\pi }{3}$ and $n=1$ .

Also recall the trigonometric formula $\cos \left(\pi -\theta \right)=-\cos \theta$ and $\sin \left(\pi -\theta \right)=\sin \theta$ .

Therefore by De Moivre’s theorem we have,

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

Thus the required Complex number $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ in the standard form of $\left(a+bi\right)$ is $-\frac{3}{2}+i\frac{3\sqrt{3}}{2}$ .