Can some one explain to me what is going on here - power of complex number So here is the question and the work to solve it, but I have no idea how one knows to do the first step or what the first step is... 
$$ \begin{align}
(6-i\sqrt{12})^{12} &= \left[\sqrt{48}\left(\cos\left(\frac{\pi}{6}\right) - i\sin\left(\frac{\pi}{6}\right)\right)\right]^{12}\\
&= (\sqrt{48})^{12} \left[\cos\left(\frac{12\pi}{6}\right) - i\sin\left(\frac{12\pi}{6}\right)\right]\\
&=48^6
\end{align}
$$
 A: Step 1 Write your number in polar coordinates, i.e. $z=r(\cos t+i\sin t)$.
Step 2 Use De Moivre's theorem, that $(\cos t+i\sin t)^n=\cos nt+i\sin nt$.
A: Some of the other posts have given general theory let be just point out how to analyse the question. We have $6-i\sqrt{12}$ we want to write it as $\sin$ and $\cos$ so we can use DeMovire. We need first to find its length,
$$\sqrt{6^2+\sqrt{12}^2} =\sqrt{48}=4\sqrt{3}$$
the normalized number is 
$$\frac{6-i\sqrt{12}}{4\sqrt{3}}=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$
Thus we need $\theta$ such that 
$$\cos \theta =\frac{\sqrt{3}}{2}$$
and 
$$\sin \theta =-\frac{1}{2}$$ from our knowledge of trig and the unit circle we see that 
$\theta=\frac{\pi}{6}$. The rest you know...
A: It’s not clear to me what you want for an explanation, but let me try. By direct computation, you can show this, if $r,r'$ are positive reals and $\theta, \theta'$ are angles (or real numbers representing radian measure, if you like):
$$
[r(\cos\theta+i\sin\theta)]\cdot[r'(\cos\theta'+i\sin\theta')]=rr'\bigr(\cos(\theta+\theta')+i\sin(\theta+\theta')\bigr)\,,
$$
using the addition formulas for cosine and sine, namely $\cos(A+B)=\cos A\cos B-\sin A\sin B$ and $\sin(A+B)=\sin A\cos B+\cos A\sin B$. Geometrically, this means that when you multiply two complex numbers of distance $r$ and $r'$ respectively, the new distance is $rr'$, while if their angles (from the positive real axis) are $\theta$ and $\theta'$ respectively, the product is found in the direction $\theta+\theta'$. Distances multiply, angles add.
All follows from this.
