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?

We have $\dfrac {1+i \sqrt 7}{2} =  \sqrt 2 \left(\cos \theta + i \sin \theta \right)$ where $ \cos \theta = \dfrac{1}{2 \sqrt 2}$
We need to calculate $2 \Re \left(\dfrac {1+i \sqrt 7}{2} \right)^4 = 2 \times (\sqrt 2)^4 \times \cos 4\theta = 8 \cos 4 \theta$
We have $\cos 2 \theta = 2 \cos^2 \theta -1 = -\dfrac{3}{4}$ and use that to get $\cos 4 \theta = \dfrac{1}{8}$
Hence we see that the sum evaluates to $8 \cos 4 \theta = 1$
