Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?

Let $w = \dfrac{1 + i\sqrt{7}}{2}$, then $\dfrac{1 - i\sqrt{7}}{2} = \bar{w}$, its complex conjugate. We are required to find the value of $w^4 + \bar{w}^4$.  
Observer that
$$w \bar{w} = 2 \quad {\rm and} \quad w + \bar{w} = 1.$$
Now
$$(w + \bar{w})^2 = w^2 + 2w\bar{w} + \bar{w}^2 \quad \Rightarrow \quad w^2 + \bar{w}^2 = 1 - 2 w \bar{w} = 1 - 2(2) = -3.$$
Squaring the last expression we have
\begin{align*}
(w^2 + \bar{w}^2)^2 &= (-3)^2\\
w^4 + 2w^2 \bar{w}^2 + \bar{w}^4 &= 9\\
\Rightarrow w^4 + \bar{w}^4 &= 9 - 2(w \bar{w})^2\\
&= 9 - 2(2)^2\\
&= 9 - 8\\
&= 1,
\end{align*}
as required to show. 
