Answer:
B. [tex]\dfrac{9}{2}-i\dfrac{9\sqrt{3}}{2}[/tex]
Step-by-step explanation:
DeMoivre's theorem:
If the complex number [tex]z=r(\cos \alpha+i\sin\alpha),[/tex] then for natural number n
[tex]z^n=r^n(\cos n\alpha+i\sin n\alpha).[/tex]
In your case,
[tex]z=\sqrt{3}\left(\cos\dfrac{5\pi}{12}+i\sin\dfrac{5\pi}{12}\right)[/tex]
and you have to evaluate [tex]z^4.[/tex]
By DeMoivre's theorem,
[tex]z^4=(\sqrt{3})^4\left(\cos 4\cdot \dfrac{5\pi}{12}+i\sin 4\cdot \dfrac{5\pi}{12}\right)=9\left(\cos \dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right)=9\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)=\dfrac{9}{2}-i\dfrac{9\sqrt{3}}{2}.[/tex]
