Moivre's theorem says
[tex]\lbrack r(\cos \theta+i\sin \theta)\rbrack^n=r^n(\cos n\theta+i\sin n\theta)[/tex]in our case
[tex](\cos (\frac{3\pi}{5})+i\sin (\frac{3\pi}{5}))^3[/tex]r=1
[tex](\cos (\frac{3\cdot3\pi}{5})+i\sin (\frac{3\cdot3\pi}{5}))[/tex][tex](\cos (\frac{9\pi}{5})+i\sin (\frac{9\pi}{5}))[/tex]ANSWER
[tex]C.(\cos (\frac{9\pi}{5})+i\sin (\frac{9\pi}{5}))[/tex]