Question

I need this answer asap

Answer 1

Answer:

$a=64,b=0$

Step-by-step explanation:

Given;

$z=-1-\sqrt{3}i$

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

$\phi =\tan^{-1}(\frac{-\sqrt{3} }{-1})=\frac{\pi}{3}$

$\arg(z)=2\pi-\frac{\pi}{3} =\frac{5\pi}{3}$

Apply DeMoivre's Theorem;

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

$\Rightarrow z^6=2^6(\cos(6\times \frac{5\pi}{3}) + i\sin(6\times \frac{5\pi}{3}))$

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

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

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

$\therefore a=64,b=0$