I really want to understadn the proof of the following theorem from Lieb's Complex Analysis:
Let $f:U\rightarrow \mathbb{C} $ be a holomorphic function with non-vanishing derivative. Then:
- For every $z_0\in U$, there exists a neighborhood $U(z_0)$ such that $z=z_0$ isthe only solution of th eequation $f(z)=f(z_0)$.
I don't understand the proof given in the book.
Can someone give an alternative proof to this?
Best Answer
Suppose not. Then for any arbitrarily small neighborhood $V$ of $z_0$, we can find some $z \in V$ with $f(z) = f(z_0)$. In particular, we can find a sequence $z_1, z_2, \ldots$ converging to $z_0$ and such that $f(z_i) = f(z_0)$ for all $i$: just take smaller and smaller neighborhoods converging to $z_0$ and pick such a point in each.
Now, by definition, $$f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$$
Since $f$ is holomorphic, this limit exists. That means we can take the limit along any series converging to $z_0$ and get the right answer. So let's use the series we just built! The numerator, $f(z_i) - f(z_0)$, is always zero, so the limit is zero, so the derivative vanishes, contradicting our assumption.