Answer:
third option
Step-by-step explanation:
Using De Moivre's theorem
([tex]\sqrt{3}[/tex] + i)³ , then
| [tex]\sqrt{3}[/tex] + i |
= [tex]\sqrt{(\sqrt{3})^2+1^2 }[/tex]
= [tex]\sqrt{3+1}[/tex] = [tex]\sqrt{4}[/tex] = 2
arg([tex]\sqrt{3}[/tex] + i) = [tex]tan^{-1}[/tex]( [tex]\frac{1}{\sqrt{3} }[/tex] ) = [tex]\frac{\pi }{6}[/tex]
Thus
([tex]\sqrt{3}[/tex] + i)³
= 2³ [ cos(3 ×[tex]\frac{\pi }{6}[/tex] ) + isin( 3 × [tex]\frac{\pi }{6}[/tex] ) ]
= 8 [ cos([tex]\frac{\pi }{2}[/tex]) + isin([tex]\frac{\pi }{2}[/tex] ) ]
