Complex numbers problem I have to solve 
where n is equal to n=80996.
 A: Using $\displaystyle  a^nb^n=(ab)^n\text{ and }\frac {a^n}{b^n}=\left(\frac ab\right)^n$ where $a,b$  are complex numbers and $n\ne0$ in case of division
$$\text{and }\displaystyle\frac{x+iy}{p+iq}=\frac{(x+iy)(p-iq)}{(p+iq)(p-iq)}=\frac{xp+yq+i(py-xq)}{p^2+q^2},$$
$$\frac{(\sqrt3+5i)^n}{(4+2\sqrt3i)^n}=\frac{(\sqrt3+5i)^n}{2^n(2+\sqrt3i)^n}=\frac1{2^n}
\left(\frac{\sqrt3+5i}{2+\sqrt3i}\right)^n$$
$$=\frac1{2^n}\left(\frac{(\sqrt3+5i)(2-\sqrt3i)}{(2+\sqrt3i)(2-\sqrt3i)}\right)^n=\left(\frac72\right)^n\left(\frac{\sqrt3+i}{7}\right)^n=\left(\frac{\sqrt3}2+\frac i2\right)^n$$
$$=\left(\cos\frac\pi6+i\sin\frac\pi6\right)^n$$
Now use de Moivre's formula
A: Wolfram Alpha confirms that
$$\left(\frac{\sqrt3+5i}{4+2\sqrt 3i}\right)^6 = -1$$
You can take it from there.
A: $$\frac{\sqrt3+5i}{4+2\sqrt3i}=\frac{\sqrt3+5i}{4+2\sqrt3i}\cdot\frac{4-2\sqrt3i}{4-2\sqrt3i}=\frac{14\sqrt3+14i}{28}=\frac{\sqrt3+i}2=\cos\frac{\pi}6+i\sin\frac{\pi}6;$$$$\left(\frac{\sqrt3+5i}{4+2\sqrt3i}\right)^n=\left(\cos\frac{\pi}6+i\sin\frac{\pi}6\right)^n.$$
A: $$\left(\frac{\sqrt3+5i}{4+2\sqrt 3i}\right)^n=\left(\frac{14\sqrt3+14i}{28}\right)^n=(\sqrt3/2+(1/2)i)^n=$$
$$=\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}i\right)^n=\cos\frac{n\pi}{6}+\sin\frac{n\pi}{6}i$$
