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.

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 
$$z = 256e^{i\frac{\pi}{3}}$$
