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 $x = -\sqrt{3} + i$
$\Rightarrow x^2 = 2 - 2i\sqrt{3}$
$\Rightarrow x^3 = 6i$
$\Rightarrow x^4 = -8 - 8i\sqrt{3}$
$\Rightarrow x^8 = -128 + 128i\sqrt{3}$
$\Rightarrow x^8 \times x^3 =  (-\sqrt{3} + i)^{11}$ 

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.
