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Does this imply $f$ is zero on the whole of $U$? If $U$ is a disc and $f$ is zero near the center then from the Taylor series I see that $f$ is constant. If $U$ is any connected opens set, then since it is open, it must be path connected, so my intuition is that for any $w\in U$ I can draw a path from $z$ to $w$ and since the path is compact I should be able to cover it with finitely many discs and I just have to make sure that the centre of the $n$th disc lies in the $n-1$th disc and use induction to show that $f(w)=0$. I think my claim is true by the above "intuition" but it would be to cumbersome to write down properly as a proof. Does anyone know how to prove this "properly" or disprove it?

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Your claim is certainly true. This is called the "principle of analytic continuation." I think your proof is just fine, but another way is to note that the set of points $z\in U$ such that all the derivatives $f^{(n)}(z)$ for $n\geq 0$ vanish is both open and closed. If this set is nonempty, it must be all of $U$, and hence $f$ must be $\equiv 0$. – froggie Jan 17 '13 at 15:50
Thank you. I got what you mean. – Montez Jan 17 '13 at 18:10

There is a well-known result which states that if a holomorphic map $f: \mathbb{C} \to \mathbb{C}$ is constant on a neighbourhood of a point then it is constant on all of $\mathbb{C}$.

To see this, assume that $f(z)=w$ for all $z \in U \subset \mathbb{C}$. The set $f^{-1}(w)$ of points $z \in \mathbb{C}$ for which $f(z)=w$ is closed because holomorphic functions are continuous and the set $\{w\} \subset \mathbb{C}$ is closed.

Let $V$ denote the interior of $U$, and assume that there exists a point $p$ on the boundary of $V$. It follows that $f$ is not constant in a neighbourhood of $p$. However, we can easily construct a sequence of points $(p_k)$, with each $p_i \in U$ such that $p_k \to p$ as $k \to \infty$. Since $p_k \in U$ for all $k$ we must have $f(p_k) = w$ for all $k$. This is a contradiction, and so the boundary of $V$ must be empty.

Since $U$ is closed without boundary it must be both open and closed and hence $U = \mathbb{C}$.

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