Show $2^{n-1} (z^n+1)-(z+1)^n=0$ for $n \geq 2$ implies $|z|=1$. I've been trying to prove the following question:
For $n \geq 2$, if $z \in \mathbb{C}$ solves $2^{n-1} (z^n+1)-(z+1)^n=0$, then $|z|=1$.
I rewrite the equation as $\frac{z^n+1}{2}=\left(\frac{z+1}{2}\right)^n$. I don't know how to proceed from here. Any help would be appreciated! Thanks!
 A: I can't think of an elementary proof, but we can use the general criteria of Cohn, discovered in 1922. This characterizes the class of polynomials whose roots lie only on the unit circle.
$\mathrm{ Cohn \ Theorem}$ : Suppose that $p(z) = \sum a_kz^k$ is a complex polynomial of degree $n$. Then all the roots of $p$ lie on the unit circle, if and only if :

*

*There is a $\mu \in \mathbb C$, $|\mu |=1$ so that $a_{n-k} = \mu\overline{a_{k}}$ for all $0 \leq k \leq n$. Such a polynomial is called self-inversive.


*Any root $z_0$ of $p'(z)$ (where $p'$ is the derivative of $p$) satisfies $|z_0| \leq 1$.
Actually Cohn's theorem is a little more general than this, but I'll leave the generalization for future discussion.
Proof via Cohn's theorem : Let $p(z) = 2^{n-1}(1+z^n) - (1+z)^n$.One can write the condition for self-inversiveness as : there exists a $|\mu|=1$ so that $p(z) = \mu z^n  \overline{p}(\frac 1z)$, where $\bar{p}$ is the polynomial $p$ with each coefficient conjugated. However, our $p$ has real coefficients.
You can compute $z^n \bar{p}(\frac 1z)$ and check that $\mu = 1$ works. When $\mu=1$ works, the polynomial is referred to as a reciprocal polynomial, because its coefficients read the same back-to-front, which allows for some nice parametrization.
But now we get to $p'(z) = n2^{n-1}z^{n-1} - n(z+1)^{n-1}$. We are finding roots of $p'$. At a root, we must have $2^{n-1}z^{n-1} = n(z+1)^{n-1}$, and (ignoring $z 
=0$ which doesn't work) rearranging and taking the absolute value, we get :$$
2^{n-1} = \left|1+\frac 1z \right|^{n-1} \implies \left|1+ \frac 1z\right| = 2
$$
and therefore if $|z| > 1$ we can't have $\left|1+ \frac 1z\right| = 2$ by the triangle inequality. The result for $p(z)$ follows by Cohn's theorem.
