Writing $(\sqrt{3}+3i)^{18}$ and $(i-1)^{-11}$ in the form $a+bi$ I have two current math problems I just can't solve. I'm to express the following in the form $a+bi$:

$(\sqrt{3}+3i)^{18} \qquad\text{and}\qquad (i-1)^{-11}$

The answer for the first one is $12^9$; and for the second, it's $\frac{1}{64}(1-i)$.
The results I get keep ending up into something that isn't even worth mentioning.
 A: Hint:
Use the exponential form of the complex numbers:


*

*The module of  $\sqrt 3+3i$ is $\sqrt{12}=2\sqrt3$, hence
$$\sqrt 3+3i=2\sqrt3\biggl(\frac12+\frac{\sqrt3}2i\Biggr)=2\sqrt3\,\mathrm e^{\tfrac{i\pi}3}$$
and $\;(\sqrt 3+3i)^{18}=(2\sqrt3)^{18}\,\Bigl(\mathrm e^{\tfrac{6i\pi}3}\Bigr)^3=12^9\,1^3=12^9$.

*Proceed in the same way for  $(i-1)^{11}$.

A: This is a solution to the problem with as few prerequisites as possible. We have 
$$(\sqrt{3}+3\cdot i)^{18}= \big[(\sqrt{3}+3\cdot i)^{3}\big]^{6}.
$$
Note that 
\begin{align}
(\sqrt{3}+3\cdot i)^{2}=&(\sqrt{3}+3\cdot i)\cdot (\sqrt{3}+3\cdot i)\\
                       =& 3 +6\sqrt{3}\cdot i-9\\
                       =& -6+6\sqrt{3}\cdot i
\end{align}
and
\begin{align}
(\sqrt{3}+3\cdot i)^{3}=&(\sqrt{3}+3\cdot i)\cdot (-6+6\sqrt{3}\cdot i)\\
                       =& \sqrt{3}\cdot (-6+6\sqrt{3}\cdot i)+3\cdot i\cdot(-6+6\sqrt{3}\cdot i)\\
                       =& -6\sqrt{3}+18\cdot i-18\cdot i-18\sqrt{3}\\
                       =& -24\cdot\sqrt{3}\\
\end{align}
Now it's up to you. In the same way, do the other exercise.
