The following is given:
$z^{142}+\frac{1}{z^{142}} (z\neq 0,z\in \mathbb{C})$
A) Prove that for every complex number z on the unit circle, this expression is real.
B) Is it possible that the expression is real for every z on a circle with radius unequal 1 ?
C) Calculate the expression if z is a root of the equation
$z+\frac{1}{z}=1$
Any assistance will be most appreciated, I am not sure I even know where to start to be honest. Thank you.
when $z$ is on the unit circle, $z^n$ is also on the unit circle (for $n\in\mathbf{N}$), and $z^{-1}=\bar{z}$. Your statement follows from that.
$z$ is on the unit circle $\implies z=\exp(i\theta)\implies z^n = \exp(in \theta)$ is also on the unit circle.
$z$ is on the unit circle $\implies z=\exp(i\theta)=\cos(\theta)+i\sin(\theta)$, and $z^{-1}=\exp(-i\theta)=\cos(\theta)-i\sin(\theta)=\bar{z}$.
Here are some hints.
(A) If $z$ is on the unit circle, then it can be written $z = e^{\theta i}$ for some real $\theta$. Also, $\operatorname{conj}(e^{\theta i}) = e^{-\theta i} = 1/e^{\theta i}$ and $\operatorname{Re}(z) = \frac{z + \operatorname{conj}(z)}{2}$.
(B) Compute $z^{142} + 1/z^{142}$ for $z = re^{\theta i}$ with $r \neq 1$. Use the fact that a complex number is real if and only if $\operatorname{Im}(z) = \frac{z - \operatorname{conj}(z)}{2i} = 0$. We're looking to see if there are any $r$ such that $\operatorname{Im}(z^{142}+1/z^{142}) = 0$ for ALL possible choices of $\theta$.
(C) $z + \frac{1}{z} = 1$ can be rewritten $z^2 - z + 1 = 0$. Solve this and put the solution(s) into polar form ($re^{\theta i}$) and see what you get.
