solving complex numbers with powers algebraically Find algebraically the value of :$\left(2^{0.5} + 6^{0.5} - \left( 2^{0.5} - 6^{0.5} \right)i  \right)^4$ 
Below are my works
I try to simplify inside. but i found that i can't add $2^{0.5}$ and $6^{0.5}$ together.
 A: Simplify by manipulating the inner expression:
$$\begin{align}\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{6})i &= (\sqrt{2} + \sqrt6i) + (\sqrt{6} - \sqrt2i)\\
&= (\sqrt{2} + \sqrt6i) - (\sqrt{2} + \sqrt6i)i
\\&= (\sqrt{2} + \sqrt6i)(1 - i)\end{align}$$
Now, let $$\begin{align}z &= (\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{6})i)^4 \\
&= (\sqrt{2} + \sqrt6i)^4(1 - i)^4\end{align}$$
Then,
$$\begin{align}|z| &= |\sqrt{2} + \sqrt6i|^4\cdot|1 - i|^4 \\
&= (\sqrt{2 + 6})^4 \cdot (\sqrt{1 +1})^4 \\
&= 256\end{align}$$
On the other hand, $$\begin{align}\arg{z} &= 4\arg(\sqrt{2} + \sqrt6i) + 4 \arg(1 - i) \\
&= 4 \tan^{-1}\frac{\sqrt{6}}{\sqrt{2}} -4 \tan^{-1}1 \\
&= \frac{\pi}{3}\end{align}$$
Hence 
$$z = 256e^{i\frac{\pi}{3}}$$
A: Set $\displaystyle\sqrt2+\sqrt6=R\cos\phi, \sqrt6-\sqrt2=R\sin\phi$ where real $R\ge0$
Squaring and adding we get $\displaystyle R^2=2(2+6)=16\implies R=4$
On division, $\displaystyle\tan\phi=\frac{R\sin\phi}{R\cos\phi}=\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}=\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3$ which is $\tan15^\circ$ (Find here)
From the dentition of atan2, $\displaystyle\phi=15^\circ$
So, we have $\sqrt2+\sqrt6+(\sqrt6-\sqrt2)i=4(\cos15^\circ+i\sin15^\circ)$
Now apply de Moivre's Identity
A: Hints:
$$\sqrt2\pm\sqrt6=\sqrt2(1\pm\sqrt3)\implies \left(\sqrt2+\sqrt 6-(\sqrt2-\sqrt6)i\right)^4=$$
$$=4\left(1+\sqrt3-(1-\sqrt3)i\right)^4=4\left[(1+\sqrt3)^2-(1-\sqrt3)^2-2(1+\sqrt3)(1-\sqrt3)i\right]^2=$$
$$=4\left[4\sqrt3+4i\right]^2=64(\sqrt3+i)^2=64(2+2\sqrt3i)=\ldots$$
