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Let $f(z)$ be an analytic function on an open set of the complex plane containing the closed unit disc $D=\{z\in \mathbb{C}:|z|\leq 1\}$. Let $m$ be the minimum of $\{f(z)\ |\ z\in D\}$ and $M$ be the minimum of $\{|f(z)|\ |\ z\in C\}$ where $C=\{z\in \mathbb{C}:|z|=1\}$ of $D$. Assume that $m<M$. Then state whether the following are true or false?

(i) $f(z)$ admits a zero on $D$.

$(ii)$ $f(z)$ attains every complex number $w$ on $D$ such that $|w|<M$.

My attempt: I know that statement $(i)$ is true because of the Minimum modulus principle. But I am not able to find the explanation of statement $(ii)$. Please help.

PAMG
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  • Do you know the open mapping theorem for holomorphic functions? – Didier Oct 10 '22 at 06:24
  • Yes. Suppose that f is analytic and non-constant in an open set $G$. Then $f(G)$ is open. – PAMG Oct 10 '22 at 06:25
  • Can you please explain it. As $f(z)$ is non-constant, it can not be minimised inside the unit disc. However, as $m<M$, so it must have $m=0$. At what point, I am wrong? – PAMG Oct 10 '22 at 06:28
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    You're right. What I said is in fact used in the proof of this result, in this way: since $|f|$ is continuous on a compact, it has a minimum. From the assumption, it is on the interior. If it were not zero, from the open mapping principle, $f$ should be open in a neighbourhood of that point, and thus one could find a close enough point with strictly smaller modulus, which is a contradiction. What you said is a shortcut and I did not realized it. Sorry about that – Didier Oct 10 '22 at 06:31
  • Ok. I am sorry too for not giving the full information. – PAMG Oct 10 '22 at 06:32

2 Answers2

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For $\vert w\vert \lt M$ the line segment $[0,w]$ does not meet the image of $f\circ \gamma$ -- where $\gamma(t) = \exp\big(2\pi i\cdot t\big)$ for $t\in [0,1]$--, so they have the same winding numbers, i.e. $n\big(f\circ \gamma,w\big)=n\big(f\circ \gamma,0\big)\neq 0$ where the right hand side follows by (i) combined with the Argument Principle. Conclude $w\in f\big(D\big)$ again by the Argument Principle.

user8675309
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(i) is true, as you correctly said. If $f$ had no zero in $D$ then the maximum modulus principle applied to $1/f$ would give a contradiction.

(ii) is true as well, this can be seen as follows:

It follows from the open mapping theorem (and the fact that $D$ is bounded) that $$ \partial f(D) \subseteq f(\partial D) \, , $$ see for example Is it always true that $\partial f(U)=f(\partial U)$ when $f$ is holomorphic?.

It follows that the disk $B_M(0)$ does not contain a boundary point of $f(D)$, and since $0 \in f(D)$, as shown in part (i), the disk $B_M(0)$ is completely contained in $f(D)$.

In other words, $f$ takes every complex value $w$ with $|w| < M$.

Martin R
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