Solution: Without loss of generality, let us assume that $p(z)$ is a monic polynomial. Let $z_1, z_2, \dots, z_n$ be zeros of $p$ with $\vert z_i \vert = 1$ and $z_i$'s not necessarily distinct. So we can write \[ p(z) = \prod_{i=1}^{n} \left( z - z_1 \right) . \] Let us denote $2z p^\prime (z) - n p(z)$ by $q(z)$. Note that \begin{align*} q(z) & = 2z \sum_{j=1}^n \left( \prod_{\substack{i = 1 \\ i\neq j }}^{n} \left( z - z_i \right) \right) - \underbrace{p(z) - p(z) - p(z) - \dots - p(z)}_{n\text{-times}} \\ & = p(z) \left[ \sum_{i = n}^n \left( \frac{2z}{z - z_i} - 1 \right) \right] \\ & = \sum_{i= 1}^n p(z) \left( \frac{z + z_i}{z - z_i} .\right) \end{align*} Thus, \[ \frac{q(z)}{p(z)} = \sum_{i = 1} ^n \frac{z + z_i}{z - z_i}. \]

Let us suppose that $\vert z \vert \neq 1$. We will show that $\frac{q(z)}{p(z)} \neq 0$. Note that \begin{align*} \text{Re} \left( \frac{q(z)}{p(z)} \right) & = \text{Re} \sum_{i=1}^{n} \frac{z + z_i}{z - z_i} \\ & = \sum_{i=1}^{n} \text{Re} \left( \frac{z + z_i}{z - z_i} \right) \\ & = \sum_{i=1}^{n} \frac{\vert z \vert ^2 - \vert z_i \vert ^2}{\vert z - z_i \vert ^2} \\ & = \sum_{i=1}^{n} \frac{\vert z \vert - 1}{\vert z - z_i \vert ^2} \neq 0. \end{align*} Since, $\text{Re} \left( \frac{q(z)}{p(z)} \right) \neq 0$ for $\vert z \vert =1$, so the zeros of $\frac{q(z)}{p(z)}$ does not lie outside the unit circle and hence all zeros of $q(z)$ lie on the unit circle.