If $z = cis(2k\pi/5)$, $z \neq 1$, then what is $(z+1/z)^2+(z^2 + 1/z^2)^2=$?

Question

question 20, part c in the picture:

I substituted the first time as $4 \cos^2(2k \pi/5)$ and the second term as $4 \cos^2(4k \pi/5)$, and then tried writing one term in terms of the other using the identity $\cos 2a = 2 \cos^2 a- 1$. I even tried bringing in $\sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?

Answer 1

The sum is $4+z^2+z^{-2}+z^4+z^{-4}$.
Show that it equals $4+z+z^2+z^3+z^4$.

Answer 2

Hint:

If $5t=2k\pi,5\nmid k,\cos t\ne1$

$\cos3t=\cdots==\cos2t$

The roots of $0=\dfrac{4\cos^3t-2\cos^2t-3\cos t+1}{\cos t-1}=4\cos^2t+2\cos t-1=0$ will be $$t=2k\pi,k\equiv\pm1,\pm2\pmod5$$

Now $z+\dfrac1z=2\cos\dfrac{2k\pi}5, \left(z+\dfrac1z\right)^2=\cdots=2\left(1+\cos\dfrac{4k\pi}5\right)$

$z^2+\dfrac1{z^2}=2\cos\dfrac{4k\pi}5,\left(z^2+\dfrac1{z^2}\right)^2=?$