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The following two scenarios I hope will help me fill in gaps in my understanding of analytic continuation.

Scenario 1

We have a function $f(z)$ of a complex variable $z$. It is known to be zero at $z=a$, where $a$ is real for simplicity, but no loss of generality.

$$f(a)=0$$

We also know $f(z)$ is analytic in the domain $D$, and $a\in D$.

Question 1: There is not enough information to say $f(z)=0$ anywhere else in $D$. Is this correct?

Scenario 2:

We have a function $f(z)$ of a complex variable $z$. It is known to be zero for all values of $z$ in the range $[a,b]$, where $a,b$ are real, with no loss of generality.

$$f(z)=0 \text{ for } z \in [a,b]$$

We also know $f(z)$ is analytic in the domain $D$, and all values $[a,b]\in D$.

Question 2: This is enough information to say $f(z)=0$ everywhere in the region of analyticity $D$.

The reason, I suggest, there is enough information is that the gradient of $f(z)$ is zero along the line from $a$ to $b$ and therefore continues beyond it.

Additional Questions:

Question 3: If $f(z)$ is analytic in a subset $D$ of the complex plane $\mathbb{C}$, then it is analytic everywhere in $\mathbb{C}$, except at poles. Is this correct?

Question 4: By analytic, I mean the function is complex differentiable. That is, the cauchy-riemann conditions apply (wiki link). Is this what analyticity is?

Penelope
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    For scenario 1, correct - consider that if you have a holomorphic function, and subtract a constant, it will give you another holomorphic function. – Joe Oct 13 '21 at 00:47
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    For #2, correct, assuming $a \ne b$. In fact, if the function is zero on a set with a limit point *in* the domain, then it has to be the constant zero function. Richard Borcherds has a great video series on complex analysis in which he proves that here https://m.youtube.com/watch?v=Oh-IG_igGag&list=PL8yHsr3EFj537_iYA5QrvwhvMlpkJ1yGN&index=10 – Joe Oct 13 '21 at 00:56
  • For question #3, no, you can define a function to be identically equal to an analytic function on some open set $D$, but define it to be whatever else you want outside of $D$. Question #4 doesn't appear to be a question. – Joe Oct 13 '21 at 01:06
  • thanks @Joe - I have struggled with the idea of a "limit point" .. what is it? – Penelope Oct 13 '21 at 01:37
  • regarding q3 .. is there an easy example to illustrate a function that is not analytically continuable beyond a non-infinite domain $D \in \mathbb{C}$? – Penelope Oct 13 '21 at 01:43
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    A limit point is a concept from topology, but the simplest explanation in terms of $\mathbb{C}$ would be that, if you have a set of points $S$, and every punctured disk centered at a point $x$ contains points from $S$, then $x$ is a limit point of $S$. For example, if $S = { 1/n : n \in \mathbb{Z}^+ }$, then zero is a limit point of $S$. – Joe Oct 13 '21 at 02:35
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    I'm not familiar with easy examples to illustrate, but you may look at https://mathoverflow.net/a/10837 or https://math.stackexchange.com/questions/1749560/show-that-fz-sum-n-0-inftyz2n-cant-be-analytically-continued-past – Joe Oct 13 '21 at 02:40
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    For question 4, there is some slight ambiguity on the definition. Some authors use analytic to describe a function that can be expressed locally as a convergent power series and others that the function is locally differentiable in an open set. However, apart from possible confusion about the domain, these definitions are broadly equivalent, and any convergent power series extends to a complex differentiable function and every function differentiable in an open region domain has a local power series. – WA Don Oct 13 '21 at 08:35

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