Answer:
[tex]a=64,b=0[/tex]
Step-by-step explanation:
Given;
[tex]z=-1-\sqrt{3}i[/tex]
[tex]r=\sqrt{(-1)^2+(-\sqrt{3})^2} =2[/tex]
[tex]\phi =\tan^{-1}(\frac{-\sqrt{3} }{-1})=\frac{\pi}{3}[/tex]
[tex]\arg(z)=2\pi-\frac{\pi}{3} =\frac{5\pi}{3}[/tex]
Apply DeMoivre's Theorem;
[tex]z^n=r^n(\cos(n \theta) + i\sin(n \theta))[/tex]
[tex]\Rightarrow z^6=2^6(\cos(6\times \frac{5\pi}{3}) + i\sin(6\times \frac{5\pi}{3}))[/tex]
[tex]\Rightarrow z^6=64(\cos(10\pi) + i\sin(10\pi))[/tex]
[tex]\Rightarrow z^6=64(1+ 0i)[/tex]
[tex]\Rightarrow z^6=64+ 0i[/tex]
[tex]\therefore a=64,b=0[/tex]
