Write $(-\sqrt{3} + i)^{11}$ as $x+iy$, where $x,y \in \mathbb{R}$ Write $(-\sqrt{3} + i)^{11}$ as $x+iy$, where $x,y \in \mathbb{R}$
I was wondering if there was a pattern or method of calculating this more efficiently. I feel like i am not really using complex analysis.
Let $x = -\sqrt{3} + i$
$\Rightarrow x^2 = 2 - 2i\sqrt{3}$
$\Rightarrow x^3 = 6i$
$\Rightarrow x^4 = -8 - 8i\sqrt{3}$
$\Rightarrow x^8 = -128 + 128i\sqrt{3}$
$\Rightarrow x^8 \times x^3 =  (-\sqrt{3} + i)^{11}$ 

Given that $x=-\sqrt{3}+i$ we have that $\vert x\vert=2$ and $\arg(x)=\frac{2\pi}{3}$ so in exponential form
$$ x=2\exp\left(\frac{2\pi}{3}i\right)$$
Thus
\begin{eqnarray}
(-\sqrt{3}+i)^{11}&=&2^{11}\exp\left(\frac{22\pi}{3}i\right)\\
&=&2048\exp\left(\frac{4\pi}{3}i\right)\\
&=&2048\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)\\
&=&-1024\sqrt{3}-1024i
\end{eqnarray}
