Can one extend/derive definition of holomorphic complex function on a fragment of domain uniquely by analytic continuation

Question

I can’t be sure if I learned this or am making up a false memory.

I remember a lecture in which I learned: if you see a fragment of a parabola, or a polynomial, you can define the function from it for all values of x. And analytic continuation is somewhat like that.

If true, I would like to find out how one can extend the following definition: $ f(z) = 2 $ where $ \Re(z) = 2 $ and $ -2 < \Im(z) < 2 $

Possible answers are:

(a) this cannot be done - period!

(b) yes it can be done, the function is $ f(z) = 2 $ for $ z \in \mathbb{C} $

(c) yes it can be done, the function is $ f(z) = \Re(z) $ for $ z \in \mathbb{C} $

(d) it can’t be done because we can’t choose between b and c above.

(e) other - please elaborate

Hope the question is clear enough.

Thanks!

Answer 1

The "Identity Principle" in complex analysis says that (on a EDIT: connected [not necessarily simply-connected, as @zhw. notes] region) there is at most a single holomorphic function assuming assigned values on a subset that has a limit point inside the given set. (So $f(z)=2$ is the current conclusion.)

There is no choice: $f(z)=2$ is holomorphic, and agrees on a set with accumulation point. Done. Also, $f(z)=\Re(z)$ is not holomorphic, although by chance it agrees with a holomorphic function ($f(z)=2$) on that set.