# A difficult inequality involving complex numbers

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.

• Note that the very strong condition is a premise -- being strong makes it easier for you to prove something from it, since you can assume that it holds without needing to argue for it. – Henning Makholm Nov 8 '14 at 13:23
• Doesn't this follow from the Turan inequality for the sum of powers? – Jack D'Aurizio Nov 8 '14 at 13:53

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 $0$,

$$f(z) = \sum_{i=1}^n \sum_{r=0}^\infty z_i^r\cdot z^r = \sum_{r=0}^\infty \left(\sum_{i=1}^n z_i^r\right)z^r,$$

$$R = \min_{1\leqslant i\leqslant n} \left\lvert \frac{1}{z_i}\right\rvert = \frac{1}{\max\limits_{1\leqslant i \leqslant n} \lvert z_i\rvert}.$$

But the assumption is that the coefficients of the Taylor series are bounded, hence the radius of convergence is at least $1$.

• Excellent answer, very clean, very elegant – Alen Nov 8 '14 at 14:08

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.

• What general case? I don't see what's more general than what you've already done, if ${z_1}$ doesn't have the maximum absolute value, simply reorder the indices – Alen Nov 8 '14 at 13:45
• Well, maybe both $|z_1|$ and $|z_2|$ are maximal. In this case my proof does not work. – Crostul Nov 8 '14 at 13:53
• Ah, you're right – Alen Nov 8 '14 at 14:02
• sorry, how did you get the first inequality? – FunctionOfX Nov 8 '14 at 20:24
• nvm, i get it now – FunctionOfX Nov 8 '14 at 20:47