# A holomorphic function with non-vanishing derivative

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:

1. 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?

## 1 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.

• Thank you very much for such a clear explanation! – L.G Oct 21 '14 at 5:42
• I'm kind of curious what the book's proof is now, since I can't really see any other way of proving this. – Daniel McLaury Oct 21 '14 at 5:51
• I'm guessing that the book really proves that $f$ is injective on some neighbourhood of a point where $f'$ is non-zero. This is a little more complicated, and usually done via Rouché's theorem. For the result above, we don't really need that $f'$ is non-vanishing. (Just that $f'$ is not identically zero.) – mrf Oct 21 '14 at 6:34
• The proof in the book goes "Without loss of generality, let $z_0 = 0$ and $f(z_0) = 0$. Claim $i$ is almost trivial: Since $f$ is holomorphic, it can be decomposed as $$f(z) = \Delta(z)z,$$ where $\Delta$ is continuous at $0$ and takes the value $$\Delta(0) = f'(0) \neq 0$$ there. Thus $\Delta$ is nonzero in a neighbourhood of $0$, and $0 = \Delta(z)z$ implies $z = 0$." (@mrf This is at the very start of the book, long before the Cauchy formula, identity theorem and so on.) – Daniel Fischer Oct 21 '14 at 10:00
• Oh, so the point is that they're proving a bigger result and are just observing that this falls out of the proof at the beginning? – Daniel McLaury Oct 21 '14 at 13:33