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 $x = -\sqrt{3} + i$
$\Rightarrow x^2 = 2 - 2i\sqrt{3}$
$\Rightarrow x^3 = 6i$
$\Rightarrow x^4 = -8 - 8i\sqrt{3}$
$\Rightarrow x^8 = -128 + 128i\sqrt{3}$
$\Rightarrow x^8 \times x^3 =  (-\sqrt{3} + i)^{11}$ 
 A: 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}
A: Yes, there is a pattern.
Recall that complex numbers can be written as $re^{2\pi i\theta}$.
When $\theta \in\mathbb Q$, there exists a pattern we can exploit.
This is that if $x = re^{2\pi ip/q}$, then $x^q = r^q e^{2\pi ip}$, and $e^{2\pi ip} =1$.
So, we have that $x^q = r^q$.
If you've seen any abstract algebra, this is exactly the order of the element of a multiplicative group.
Now, you have that $x^3 = 6i$.
We have that $i = e^{\frac{\pi}{2}i} = e^{2\pi i/4}$, so just from this I know that $(x^3)^4 = (6i)^4 = 6^4 = 24$.
So, $x^{11} = 24(-\sqrt{3}+i)^{-1}$, which is much easier to compute than just iterated multiplication.
You can also use the binomial theorem for things like this, but I find using the polar form (either by directly converting, or by repeatedly multiplying until you find something like $6i$ that has a very simple polar form, and then converting) to be much tidier.
A: If you arenvt comfortable with polar coorindates:
The trick is to realize that if $z = \frac {-\sqrt {3}+1}2$.  Then $z^2=\overline{z}$ and $z^3=i$.  
From there all values are cyclic. $z^{12}=1$ and $z^{12k+3j+m}=i^jz^m $.
So $z^{11}=i^3z^2=-\overline z $.
Now $w=-\sqrt {3}+1=2z $ so $w^{11}=2048\overline z=-1024\sqrt 3-1024i $.
But it's better to learn polar coordinates and take the guesswork out.
