<s>[INST] Show that sum of squares of all the values of $(\sqrt{3}+i)^{3/7}$ is $0$. Show that sum of squares of all the values of $(\sqrt{3}+i)^{3/7}$ is $0$.
Attempt:
Let $z=(\sqrt{3}+i)$ then simplifying I get
$z^{3/7}=2^{3/7}\Big\{\cos{\frac{13k+1}{2\times 7}\pi}+i\sin{{\frac{13k+1}{2\times 7}\pi}}\Big\}$, $k=0,1,2\cdots 6$ 
Now sum of squares of all values of $z^{3/7}$ is 
$S=2^{3/7}\Big[\{\cos{\frac{\pi}{7}}+\cos{\frac{13\pi}{7}}+\cos{\frac{25\pi}{7}}+\cos{\frac{37\pi}{7}}+\cos{\frac{49\pi}{7}}+\cos{\frac{61\pi}{7}}+\cos{\frac{73\pi}{7}}\}+i\{\sin{\frac{\pi}{7}}+\sin{\frac{13\pi}{7}}+\sin{\frac{25\pi}{7}}+\sin{\frac{37\pi}{7}}+\sin{\frac{49\pi}{7}}+\sin{\frac{61\pi}{7}}+\sin{\frac{73\pi}{7}}\}\Big]$
How to show that this is zero?
 [/INST] Hint:  Values of $(\sqrt3+i)^{3/7}$ are roots of $x^7-(\sqrt3+i)^3=0$.  Now think Vieta and use $\sum x_i^2 = (\sum x_i)^2-2\sum x_i x_j$.
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