Analytic function vanishes in an open unit disk. Show it vanishes identically. $f$ is an analytic function that vanishes in a unit disk which is a subset of domain $D$. Show that it vanishes all over $D$.
This was one of our quiz questions and I think we need to assume that $D$ is connected for it to be true. I couldn't find counterexample.
 A: You can write the domain $D$ as a disjoint union of two open sets:
$$U = \{z_0 \in D : f^{(k)}(z_0) = 0 \text{ for all } k\ge 0\}$$
and
$$V = \{z_0 \in D : f^{(k)}(z_0) \ne 0 \text{ for some } k\ge 0\}.$$
One of them is nonempty. Since $D$ is connected, the other must be empty.
The openness of $U$ is slightly subtle, hinging upon the fact that for any $z_0 \in D$, there is an open disk $B_r(z_0)$ with $r>0$ on which
$$f(z) = \sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!} (z-z_0)^k.$$
A: Typically, a domain is open and arcwise connected. There is a theorem of Weierstass that says that if a function is analytic in a domain, and there is a sequence of points with at least one limit point in $D,$ take any sequence of distinct points in the disk converging to the center of the disk here, and $f$ takes value $0$ at each term of the sequence, which your sequence does since it vanishes on the entire disk, then $f$ is identically $0$ on the domain. 
I always found this theorem pleasing. As a reference, you can find it in Hille, "Analytic Function Theory, Theorem 8.1.3 (I'm using the fourth printing of the first edition since I'm ancient. The text has been expropriated by Springer in its Graduate Texts series. In my index, the theorem is sourced under Zeros, frequency of.) I hope this helps. 
