A complex polynomial inequality

Question

Problem

Let $P(z) = (z - z_1)(z - z_2) \cdots (z - z_n)$ with $|z_i| \geq 1$, $i = 1, 2, \dots, n$. Prove that if $0 < r < 1$, then for any $z$, with $|z| = 1$,

$$\left| \frac{P(z)}{P(rz)} \right| \leq \left( \frac{2}{1 + r} \right)^n.$$

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My Approach

To simplify things I thought maybe it would be a good idea to compare just one term from the LHS and RHS, then that would mean we need to prove

$$\left| \frac{z - q}{rz - q} \right| \leq \frac{2}{1 + r} $$

for any $r, z, q$ with $0 < r < 1$, $|z| = 1$, and $|q| \geq 1$.

Now since I had no idea I thought it would be better to first consider the case $z=1$ and $z=-1$ and maybe then try to prove the given inequality

$$\left| \frac{1 - q}{r - q} \right| \leq \frac{2}{r + 1}.$$

Let $z = 1$, then $$\left| \frac{z - q}{rz - q} \right| = \left| \frac{1 - q}{r - q} \right|.$$ By the triangle inequality, we have $|1 - q| \leq 1 + |q|$ and $|r - q| \geq |q| - r$, so

$$\left| \frac{1 - q}{r - q} \right| \leq \frac{1 + |q|}{|q| - r}.$$

But even after trying the triangle inequality many times, well, I wasn't able to get something useful. Therefore I need help with this.

P.S. I don't know complex analysis.

Answer 1

First note that if the claim is true then it must in particular hold for $n=1$. Conversely, if it holds for $n=1$ then multiplying the inequalities for $P_j(z) = z-z_j$ gives the desired inequality for $P$. This means that your approach is correct, the claim is true exactly if

$$\left| \frac{z - q}{rz - q} \right| \leq \frac{2}{1 + r} $$

for any $r, z, q$ with $0 < r < 1$, $|z| = 1$, and $|q| \geq 1$.

We can write $q= z Re^{it}$ with $R \ge 1$ and $0 \le t < 2 \pi$. Then $$ \left| \frac{z - q}{rz - q} \right|^2 = \frac{|1 - Re^{it}|^2}{|r - Re^{it}|^2} = \frac{1-2R \cos(t) + R^2}{r^2-2rR \cos(t) + R^2} = \frac 1r - \frac{(1-r)(R^2-r)}{r(r^2-2rR \cos(t) + R^2)} \, . $$ The right-hand side is decreasing in $\cos(t)$, so that the maximum is attained for $\cos(t) = -1$. It follows that $$ \left| \frac{z - q}{rz - q} \right|^2 \le \left(\frac{1+R}{r+R}\right)^2 \, , $$ and that is decreasing in $R$, so that the maximum is attained for $R=1$, and $$ \left| \frac{z - q}{rz - q} \right|^2 \le \left(\frac{2}{r+1}\right)^2 \, . $$

Answer 2

Multiplying by $\bar z$ each term in the LHS fraction and noting that $|q\bar z|=|q|$ since $|z|=1$, it is enough to prove that for $|q|\ge 1$ $$\left|\frac{1-q}{r-q}\right|\le \frac{2}{1+r}$$

Since $f(q)= \frac{1-q}{r-q}$ is analytic on the disc exterior and goes to $1$ at infinity, the maximum modulus theorem says that we need to prove the inequality only for $|q|=1$

Here we can apply Jack's Lemma and note that the maximum is attained at some $q$ for which $\frac{qf'(q)}{f(q)}$ is real

(this follows writing $f(q)=R(q)e^{i\Psi(q)}$ and noting that if $q=e^{it}$ the maximum modulus is attained at $q$ implies $\frac{\partial R}{\partial t}=0$ there and taking $\frac{\partial}{\partial t}$ in $\log f =\log R+i\Psi$ and using the chain rule leads to the result mentioned)

A simple computation gives $$\frac{qf'(q)}{f(q)}=\frac{(1-r)q}{(1-q)(r-q)}$$ so that has to be real so equal to its conjugate (clearly $q\ne 1$ at a maximum since $f(1)=0$)

Conjugating and using $\bar q=1/q$ and simplifying we get that $-(r-q)=qr-1$ so $q=-1$ and indeed the maximum modulus is $\frac{2}{1+r}$ and we are done!

Answer 3

Hint: with some easy transformations, it boils down to showing that the function $$\bar D \ni z \mapsto \frac{z-1}{z-\frac{1}{r}}$$

achieves its maximum modulus at $z=-1$. (Clearly the minimum modulus is achieved at $z=1$).

The red circle is the level curve of the function $$\frac{d(\cdot, 1)}{d(\cdot, \frac{1}{r})}$$

through $z$. The point $-1$ is always left outside.