Question

Convert the complex number z equals to a minus one upon cos pi by 3 plus i sin pi by 3 in the polar form

Answer 1

Answer:

$\cos(\frac{2\pi}{3})+i\sin\frac{2\pi}{3}$

Step-by-step explanation:

The given complex number is

$z=\frac{-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}$

Multiplying the numerator and denominator by $\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}$

$z=\frac{-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}\times \frac{\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}}}$

$\implies z=\frac{-(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3})}{(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3})}$

$\implies z=\frac{-(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3})}{\cos^2\frac{\pi}{3}+\sin^2\frac{\pi}{3}}$

$\implies z=\frac{-\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}{1}$

$\implies z=\cos(-\frac{\pi}{3})+i\sin\frac{\pi}{3}$

$\implies z=\cos(\pi-\frac{\pi}{3})+i\sin(\pi-\frac{\pi}{3})$

$\implies z=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})$

This is the required complex number in polar form.

Hope this helps.