Upper bound on the magnitude of the roots of a complex polynomial

Problem:

Let $z_0$ be a root of the complex polynomial $z^n + a_{n-1}z^{n-1} + ... + a_0$ $(a_k \in \mathbb{C})$.

Prove that $|z_0| \le \zeta$, where $\zeta$ is the only positive root of $z^n - |a_{n-1}|z^{n-1} - ... - |a_0|$.

(the preceding problem - which I've solved - was to prove that the second polynomial has in fact exactly one positive root; to be precise we'd have to assume that at least one of the $a_i$ are not equal to 0 or allow the root to be zero)

I have no idea how to approach this problem. The statement seems to be that "for given nonnegative real $a_i$ the complex polynomial with the greatest root whose coefficients are of magnitude $a_i$ is $z^n -...-a_0$", and I've tried proving this by "rotating" the coefficients one by one and observing how the roots behave, but I've had no success. (maybe it's just because I have no experience at all with complex polynomials)

I haven't studied complex analysis, so it would be great to find a solution that doesn't use results from that area. Hints would be great as well. :)

Answer

If $z_0$ is a root then $-z_0^n = a_0 + \ldots + a_{n-1}z_0^{n-1}$. Now the triangle inequality says that $|z_0|^n \le |a_0| + \ldots + |a_{n-1}||z_0|^{n-1}$.

Since for positive real $x$, $x^n - |a_0| - \ldots - |a_{n-1}|x^{n-1}$ has a unique root $\zeta$, this quantity is negative for $0 \le x \le \zeta$ and positive for $x \ge \zeta$.

This shows that $|z_0| \le \zeta$.