# Show that $\sqrt[k]{|z-a_1|\cdots |z-a_k|}$ has a max greater than $R$, and a min less than $R$

This is a homework problem.

For $|z| \le R$ and $|a_j| < R$ for $j=1,\ldots, k$, not all zero, show that $\sqrt[k]{|z-a_1|\cdots |z-a_k|}$ has a max greater than $R$, and a min less than $R$.

In the first part of the problem, I show that when $|z| = R$,

$$\left|\frac{R(z-a)}{R^2-\bar{a}z}\right| = 1.$$

This of course implies that $|R^2-\bar{a_j}z| = R|z-a_k|$. Then, since $\prod_{j=1}^k R^2-\bar{a_j}z$ is analytic, the maximum and minimum modulus theorems apply, and so any max/min must appear when $|z| = R$.

Then, it is evident that $$R|z-a_j| = R\sqrt{R^2-\bar{a_j}z-a_j\bar{z}+|a_j|^2},$$

so the max and min of $\sqrt[k]{|z-a_1|\cdots |z-a_j|}$ is the max and min of $$\sqrt[k]{\left(R^2-2\Re(\bar{a_j}z)+|a_j|^2\right)^{1/2}},$$

respectively.

I essentially need to show that there exists some $z$ such that the above is greater than $R$, and that there is some $z$ such that the above is less than $R$. But I am now stuck as far as how to do that.

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@Brian Thanks for the tag edit; I sat there thinking "I'm forgetting one..." :) –  Arkamis Feb 14 '13 at 1:23

Let $f(z) = (z-a_1)\cdots (z-a_k)$ and let $g(z) = \frac{1}{R^k}\prod_{j=1}^k R^2-\bar{a}_j z$. These two different analytic functions are equal in magnitude when $|z| = R$.
Choose some $z$ in the convex hull formed by $\{a_n\}$ such that $|z-a_k| < R$ for all $a_k$. We can do this, e.g. a point in the neighborhood of the barycenter of the hull. Then, $|z-a_k| \le c_kR$ for $0 < c_k < 1$ and hence $|f(z)| \le CR^n$, $0 < C <1$. So on the boundary of $|z| \le R$ there must be a point smaller than this, and so $|f(z)| < R^n \implies \sqrt[n]{|f(z)|} < R$ when $|z| = R$.
Similarly consider $g(0) = R^n$. By the maximum principle, there is a point on the boundary larger than this, so $|g(z)| > R^n \implies \sqrt[n]{|g(z)|} > R$ when $|z|=R$. Finally, since $|g(z)| = |f(z)|$ when $|z| = R$, $\sqrt[n]{|g(z)|} = \sqrt[n]{|f(z)|}$.