Maximum modulus of a holomorphic function on a disc within a certain sector

Question

Given the polynomial $$f(z) = az^n + b \qquad (n \geq 2)$$ and a modulus $0 < \rho < 1$, can one find a modulus $0 < r < \rho$ such that there is a point $$w \in \{ |z| \leq r \} \cap \{ |z - 1| < 1 \}, \qquad|f(w)| > \frac{|a|r^n}{2} + |b|$$

Answer 1

If $a = 0$, then we have $\lvert f(w)\rvert = \lvert b\rvert$ for all $w$, so the strict inequality cannot be achieved.

If $a \neq 0$, we can divide by $a$ and assume $a = 1$. If $n \geqslant 3$, then

$$\{ w^n : \lvert w\rvert = r, \lvert w-1\rvert < 1\} = \{ z : \lvert z\rvert = r^n\},$$

since for every $r < 1$ the angle of the arc $\{ w : \lvert w\rvert = r, \lvert w-1\rvert < 1\}$ is greater than $\frac{2\pi}{3}$. In that case, we can always find a $w$ with $\lvert f(w)\rvert = r^n + \lvert b\rvert > \frac{1}{2}r^n + \lvert b\rvert$.

For $n = 2$, $\{ w^2 : \lvert w\rvert = r, \lvert w-1\rvert < 1\}$ is not the full circle of radius $r^2$, a small piece near the negative real half-axis is missing. But by making the radius $r$ small, we can get the angle of the arc as close to $\pi$ as we wish. Thus unless $b$ is a negative real number, by choosing $r$ small enough we can find a $w$ with $\lvert w\rvert = r$, $\lvert w-1\rvert < 1$ and $\lvert f(w)\rvert = r^n + \lvert b\rvert$. If $b < 0$, we find

$$\lvert f(re^{i\varphi})\rvert^2 = \lvert r^2 \cos (2\varphi) + b + ir^2\sin (2\varphi)\rvert^2 = r^4 + b^2 - 2r^2\lvert b\rvert \cos (2\varphi),$$

and that is larger than $\bigl( \frac{r^2}{2} + \lvert b\rvert\bigr)^2 = \frac{r^4}{4} + b^2 + r^2\lvert b\rvert$ if and only if

$$\frac{3r^4}{4} > r^2\lvert b\rvert \bigl(1+\cos (2\varphi)\bigr) = 2r^2\lvert b\rvert \cos^2 \varphi,$$

which we can rearrange to

$$\frac{\cos^2 \varphi}{r^2} < \frac{3}{8\lvert b\rvert}.\tag{1}$$

Since $\lvert re^{i\psi} - 1\rvert^2 = (1-r\cos\psi)^2 + r^2\sin^2\psi = 1 + r^2 - 2r\cos\psi$, we have $\lvert re^{i\psi} - 1\rvert = 1$ if and only if $\cos \psi = \frac{r}{2}$, and so in $(1)$ the infimum of the possible left hand sides is $\frac{1}{4}$, and we can achieve $\lvert f(w)\rvert > \frac{r^2}{2} + \lvert b\rvert$ on the arc $\{w : \lvert w\rvert = r, \lvert w-1\rvert < 1\}$ only for $b \in \bigl(-\frac{3}{2},0\bigr)$ when $b < 0$. That implies that

$$\sup \{ \lvert f(w)\rvert : \lvert w\rvert \leqslant r, \lvert w-1\rvert < 1\} \leqslant \frac{r^2}{2} + \lvert b\rvert$$

for $b \in \bigl(-\infty, -\frac{3}{2}\bigr]$, since

$$\lvert f(w)\rvert \leqslant \frac{\lvert w\rvert^2}{2} + \lvert b\rvert \leqslant \frac{r^2}{2} + \lvert b\rvert$$

by the above.

Summarising, such a radius $r$ point $w$ exists if and only if $a\neq 0$ and either $n \geqslant 3$ or $n = 2$ and $\frac{b}{a} \in \mathbb{C}\setminus \bigl(-\infty, -\frac{3}{2}\bigr]$.