Power sums of p-th roots of unity

I think I can answer my own question (which is really my colleague's question).

Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{k=1}^{p-1}\sum_{1\leq i,i'\leq j} z_i^k \overline{z_{i'}^k} = \sum_{1\leq i,i'\leq j} \sum_{k=1}^{p-1} (z_i/z_{i'})^k. $$ On the right hand side, the inner sum equals $p-1$ when $i=i'$, and it equals $-1$ when $i\neq i'$. Hence $$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = j(p-1)-j(j-1)=j(p-j),$$ and we infer $$ \max_k\left|z_1^k+\dots+z_j^k\right|\geq\sqrt{\frac{j(p-j)}{p-1}}. $$ Choosing $j:=(p-1)/2$, we get $$ \max_{j,k}\left|z_1^k+\dots+z_j^k\right|\geq\frac{\sqrt{p+1}}{2}. $$ That is, $M_p$ is at least $\sqrt{p+1}/2$, and so it is not bounded.

This is closely related to the problems surrounding Turán's power sum method. For example see the chapter in Montgomery's Ten Lectures book, or this paper of Gonek. Lemma 1 (attributed to Cassels) there shows that if $b_j >0$, and $|z_j|=1$ (for $j=1$, $\ldots$, $N$) then for $K>N$ $$ \max_{k\le K} \Big| \sum_{j=1}^{N} b_j z_j^k \Big| \ge \frac{\sqrt{K-N}}{\sqrt{K}} \Big(\sum_{j=1}^{N} b_j^2 \Big)^{\frac 12}. $$ Now apply this in your situation with $N=(p-1)/2$ (this is $j$ in your problem) and $K=(p-1)$ to get essentially the result you have. Of course your proof is easy enough in this case, but I thought you might appreciate the general context.