Question

I need this answer asap

Answer 1

`a=64,b=0`

Explanation:

Given;

`z=-1-√(3)i`

`r=\sqrt{(-1)^2+(-√(3))^2} =2`

`\phi =\tan^(-1)((-√(3) )/(-1))=(\pi)/(3)`

`\arg(z)=2\pi-(\pi)/(3) =(5\pi)/(3)`

Apply DeMoivre's Theorem;

`z^n=r^n(\cos(n \theta) + i\sin(n \theta))`

`\Rightarrow z^6=2^6(\cos(6* (5\pi)/(3)) + i\sin(6* (5\pi)/(3)))`

`\Rightarrow z^6=64(\cos(10\pi) + i\sin(10\pi))`

`\Rightarrow z^6=64(1+ 0i)`

`\Rightarrow z^6=64+ 0i`

`\therefore a=64,b=0`