Suppose that $z_1,\ldots,z_n$ are complex numbers with the property that there is some constant $C$ such that $$\big|z_1^r+\cdots+z_n^r\big|\leqslant C$$ for all integers $r\geqslant0$. Show that for all $i$ we have $\left|z_i\right|\leqslant1$.
Pretty sure need to perfume some sort of analytic arguments, this 'for all r in non-negative integers' statement is very strong, otherwise this inequality will hardly hold.
On
This is a partial answer. Suppose without loss of generality that $z_1$ has maximum absolute value among $z_1, \dots, z_n$. If we assume that $|z_1| > |z_i|$ for all $i=2, \dots, n$ it's easy to see that $|z_1| \leq 1$.
Otherwise, suppose by contradiction $|z_1| > 1$.
Now, for all $r \geq 1$ the following inequality holds $$C \geq |z_1^r + \dots + z_n^r| \geq |z_1|^r - |z_2|^r - \dots - |z_n|^r$$
dividing by $|z_1|^r$ we get
$$\frac{C}{|z_1|^r} \geq 1 -\frac{|z_2|^r}{|z_1|^r} - \dots \frac{|z_n|^r}{|z_1|^r}$$
sending $r \to + \infty$ we get the contradiction $0 \geq 1 - 0- \dots -0=1$.
Maybe the general case can be treated by using this particular case.
The easiest (in the sense of "least work") way to prove it uses a tiny bit of complex analysis.
Consider the function
$$f(z) = \sum_{i=1}^n \frac{1}{1 - z_i\cdot z}.$$
This is a meromorphic function having simple poles at $\frac{1}{z_i},\, 1 \leqslant i \leqslant n$, and nowhere else. The Taylor series of $f$ with centre $0$,
$$f(z) = \sum_{i=1}^n \sum_{r=0}^\infty z_i^r\cdot z^r = \sum_{r=0}^\infty \left(\sum_{i=1}^n z_i^r\right)z^r,$$
therefore has radius of convergence
$$R = \min_{1\leqslant i\leqslant n} \left\lvert \frac{1}{z_i}\right\rvert = \frac{1}{\max\limits_{1\leqslant i \leqslant n} \lvert z_i\rvert}.$$
But the assumption is that the coefficients of the Taylor series are bounded, hence the radius of convergence is at least $1$.