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I do not know how to reconcile this answer with my understanding of the maximum modulus principle.

Is it correct to infer that if a holomorphic function is constant in some open disk then it is constant everywhere?

I believe that the whole of $\mathbb{C}$ is a connected open set containing any disk, right? If so, how on Earth does this answer work? In particular, the set $K_n := \{z∈\mathbb{C}:|z|≤1\ \&\ \operatorname{dist}⁡(z,R+)≥1/n\}$ surely contains a disk; so a function defined to be zero in $K_n$ must be zero everywhere.

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    $\begingroup$ constant everywhere on a connected open set containing that disk $-$ yes, that's correct. See the identity theorem for holomorphic functions. $\endgroup$
    – Targon
    Mar 19, 2020 at 9:07
  • $\begingroup$ @Targon, thank you, your comment helped me define my question better. $\endgroup$ Mar 19, 2020 at 9:16

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The set $L_n$ (using the notation from that answer you quoted) is not connected and the (open) neighbourhood of $L_n$ is explicitly also not connected, therefore the identity theorem (for which I gave the link in a comment) doesn't apply. If it was connected, you could not argue like in that answer. To illustrate this, consider for example the following function:$$f \colon z \longmapsto \begin{cases}e^{z} & \text{for } z \in B_1 (0)\\0 & \text{for } z \in B_1 (1)\end{cases}$$Now $f$ is holomorphic (because it is complex-differentiable everywhere it is defined) on an open subset of $\mathbb C$ $\big(\text{namely }B_1(0) \cup B_1 (1)\big)$ and it is constant on an open subset. But you can't infer, that $f$ is constant everywhere and there is no reason why it should be.

The identity therorem doesn't apply here, because the domain where $f$ is defined is not connected. If you have a connected, open subset of $\mathbb C$ the behaviour of a holomorphic function on any open disc, no matter how small, determines the function everywhere on that connected set. If the function is defined on a set which is not connected, then the parts don't have to have anything to do with each other at all (as you can see in my example above).
That is also why most of function theory is done and formulated for open, connected subsets of $\mathbb C$, i.e. domains.

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    $\begingroup$ I think my conceptual problem was with analytic continuation. I didn't realize that analytic continuation must (?) begin with a single domain. Doing analytic continuation on holomorphic functions with disconnected domains is not guaranteed to be consistent, right? $\endgroup$ Mar 19, 2020 at 10:13
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    $\begingroup$ correct, you could extend either $z \mapsto e^z$ from $B_1(0)$ to all of $\mathbb C$ and also $z \mapsto 0$ from $B_1(1)$ to all of $\mathbb C$. No problem there. But they will not agree. And so if you just say you extend $f$ (from my answer) you run into problems. $\endgroup$
    – Targon
    Mar 19, 2020 at 10:18
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    $\begingroup$ THANK YOU! I have been scratching my head all night on that one. I wish I paid more attention in undergrad. By the way, I did up vote your answer. $\endgroup$ Mar 19, 2020 at 10:20

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