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We are required to prove the following problem using the Inverse Function Theorem:

Let $X \subset \mathbb{C}$ open and connected. Let $f: X \to \mathbb{C}$ holomorphic. Suppose that $f(X) \subset \{z \in \mathbb{C}: |z| = 3\}$.

We want to show that $f$ is constant.

Inverse Function Theorem (for Complex Variable Functions). Let $f: X \subset \mathbb{C} \to \mathbb{C}$ holomorphic in $X$, with $X$ open set. Let $z_0 \in X$ such that $f'(z_0) \neq 0$. Then, exists $V, W \subseteq \mathbb{C}$ open sets, with $z_0 \in V \subseteq X$ and $f'(z_0) \in W$, such that, for every $z \in V$, $f(z) \in W$ and $f: V \to W$ has inverse function $g: W \to V$, with $g$ is holomorphic in $W$. Furthermore, $$g'(f(z)) = \frac{1}{f'(z)}, \quad \forall z \in V$$

Proof

If $f'(z) = 0$, for every $z \in X$, $f$ is constant, since $X$ is connected.

Suppose there is some $z_0 \in X$ such that $f'(z_0) \neq 0$. Therefore, exists $V, W \subseteq \mathbb{C}$ open sets such that $$g'(f(z)) = \frac{1}{f'(z)}, \quad \forall z \in V$$

(We are trying to find a contradiction, but I cannot do so. Somehow, we would need to use the fact that $f(X) \subset \{z \in \mathbb{C}: |z| = 3\}$).

By noticing that $g$ is holomorphic in $W$ and $f(z_0) \in W$, we can make absolute values come into the equation:

$$g'(f(z_0)) = \lim_{f(z) \to f(z_0)} \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)} \Longrightarrow \Big| g'(f(z_0)) \Big| = \lim_{f(z) \to f(z_0)} \frac{|g(f(z)) - g(f(z_0))|}{|f(z) - f(z_0)|}$$

Any hint/help is appreciated.

Blue Tomato
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    Hint: open mapping theorem for holomorphic functions – Didier Oct 16 '23 at 22:09
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    Actually, this statement doesn't require using advanced tools like the inverse function theorem or open mapping theorem. There is an elementary proof using just the Cauchy-Riemann equations. See here: https://math.stackexchange.com/questions/474723/holomorphic-function-with-constant-absolute-value – Mark Oct 16 '23 at 22:17
  • @Mark Yeah, I saw that. But at this one, I have the strictly restriction of using the I.F.T. – Blue Tomato Oct 16 '23 at 22:19
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    @BlueTomato Yeah, sorry, I didn't read the first line of your question. – Mark Oct 16 '23 at 22:21
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    The point is that the inverse function theorem tells you that the image has non-empty interior (unless $f$ is locally constant), however, this contradicts the assumption that the image is contained in a sphere. It is just an annoying way of using the open mapping theorem. – Severin Schraven Oct 17 '23 at 00:57
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    if you know that $f$ is locally invertible on $B(z_0,\delta)$ with holomorphic (i.e. continuous) inverse $g$ then $B(z_0,\delta)$ and $f\big(B(z_0,\delta)\big)$ are homeomorphic which obviously isn't true, e.g. delete two points from $B(z_0,\delta)$ and it is still connected but its image under $f$ is not (or delete one point and compare contractability). – user8675309 Oct 17 '23 at 01:08
  • Thanks to everyone! I got both ideas, the one with the mapping theorem and the one with the connected property of $B(z_0, \delta)$. – Blue Tomato Oct 17 '23 at 02:45

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