1. Convert the complex number 3cis(-2Pi/3) to standard form & exponential form.
[3cis(-2Pi/3) = 3(cos(-2pi /3)+isin(-2pi /3)). - The Editor]
2. Convert the complex number 3e^{11pi /6} to standard form & polar form.
3. Use De Moivre's Theorem to simplify the following & leave the answer in standard form.
[(-5-5i)⁴ ] x [((sqrt3)+ 3i)² ]

All you need to remember is that

$x+iy=r\mathrm{cis}(\theta )$

when $x=r\cos (\theta )$

and $y=r\sin (\theta )$
That should make the converting to standard form easy. Polar form and exponential form are practically equivalent anyway.

When in doubt, draw an argand diagram, and do it from trig.