Need help to simplify the expression involving powers $$\left(1-\frac{\sqrt{3}-i}{2}\right)^{24}$$
somehow this should be equal to :$$\left(2-\sqrt{3}\right)^{12}$$
but I can't see how...
 A: Hint:
$$
\left( 1 - \dfrac{\sqrt{3} - i}{2} \right)^{24} = \left[\left( 1 - \dfrac{\sqrt{3} - i}{2} \right)^2\right]^{12}
$$
A: $$
a=\frac{\sqrt{3}-i}{2} = \cos 30^\circ-i\sin30^\circ.
$$
Look at the triangle whose vertices are $0$, $a$, and $1$.  Since the distance from $0$ to $1$ and the distance from $0$ to $a$ are both equal to the radius of the unit circle, the triangle is isosceles.  The angle at the center of the circle is $30^\circ$ and the other two angles must be equal to each other.  Since they have to add up to $180^\circ$, they must each be half of the remaining $150^\circ$, hence each $75^\circ$.
The short side of the triangle is just $1-a$.  Hece $1-a=|1-a|(\cos75^\circ+i\sin75^\circ)$.  Now
$$
|1-a|=\left|1-\frac{\sqrt{3}-i}{2}\right| = \left|\frac{2-\sqrt{3}-i}{2}\right| = \frac{2\sqrt{2-\sqrt{3}}}{2} = \sqrt{2-\sqrt{3}}.
$$
Hence
$$
(1-a)^{24} = \left(\sqrt{2-\sqrt{3}}\right)^{24} (\cos(24 \cdot 75^\circ) + i\sin(24 \cdot 75^\circ)) = \left(2-\sqrt{3}\right)^{12}\cdot(1).
$$
($24\cdot75^\circ=1800^\circ = 5\text{ full circles}$, so the cosine is $1$ and the sine is $0$.)
A: Hint: start by squaring $(1-\frac{\sqrt{3}-i}{2})$. Remember that $(a-b)^2 = a^2-2ab+b^2$, and that $i^2=-1$.
A: The easiest way to deal with imaginary numbers is with exponentials. This is definitely the case here. First, put everything over a common denominator, then take the exponential. Remember that $(e^{a})^{b} = e^{a*b}$ and $e^{a+b} = e^{a}*e^{b}.$ When you're done simplifying, take the ln of both sides. In the last step, think about what $(-i)^{12}$ equals.
