Why do injective holomorphic functions have nonzero derivative

Question

For some open sets $U$, $V$ in the complex plane, let $f:U\rightarrow V$ be an injective holomorphic function. Then $f'(z) \ne 0$ for $z \in U$.

Now I don't understand the proof, but here it is from my text. My comments are in italics. Suppose $f(z_0) = 0$ for some $z_0 \in U$.
$f(z) - f(z_0) = a(z - z_0)^k + G(z)$ for all $z$ near $z_0$, with $a \ne 0, k \ge 2.$ Also, $G$ vanishes to order $k+1$ at $z_0$.
I'm not clear on what this "vanishing" thing means. Maybe it means that $G$ can be expressed as a power series of order $k+1$ around $z_0$.

For sufficiently small $w$ we can write $f(z) - f(z_0) - w = F(z) + G(z)$, where $F(z) = a(z - z_0)^k - w$.
I'm not sure why we need to have $w$ small. This equation will work for any $w$.

Since $|G(z)| \lt |F(z)|$ on a small circle centered at $z_0$, and $F$ has at least two zeros inside that circle, Rouche's theorem implies that $f(z) - f(z_0) - w$ has a least two zeros there.

Now I think that $|G(z)| \lt |F(z)|$ can follow simply from the fact that $F$ is a polynomial of degree $k$ while $G$ has degree $k+1$. And the remark about the two zeros can follow from the fact that $F$ must have $k$ zeros in the complex plane. But the first part requires that we consider $z$ only on a small circle. The second part requires that our circle be big enough to capture two zeros. How do we know that we can satisfy both?

Since $f'(z) \ne 0$ for for all $z \ne z_0$ sufficiently close to $z_0$, the roots of $f(z) - f(z_0) - w$ are distinct, so $f$ is not injective - a contradiction.

I think that the derivative is never zero for values of $z$ other than $z_0$ because otherwise we would have a sequence of zeros limiting towards $z_0$ which would cause our function to be constant which is a contradiction. But again we have the same problem - we can only consider a small circle. The roots of $f$ may lie outside this circle.

Answer 1

First comment: Yes. I prefer to write $(z-z_0)^k\cdot(a+(z-z_0)H(z))$ in such proofs.

Second comment: $w$ small is not needed immediately, but we can only use a small circle (as guaranted by openness of $U$) and want to have $(z-z_0)^k=w$ at least once (end hence $k$ times) inside that circle. This is what forces $w$ to be small.

Third comment: We make our circle even smaller (and may revise our choice of $w$) in order to make $|G|<| F|$. By the vanishing order of $G$, we have $G(z)\le c\cdot |z-z_0|^{k+1}$ for some $c$ as long as $z\approx z_0$ (with my notation above, you can take any $c>|H(z_0)|$). Then what we need here for $|G(z)|<|F(z)|$ if $|z-z_0|=r$ is to chose $r\le \frac ac$. The very simple polynomial $F$ has $k$ zeroes in the circle because we choose our $w$ small enough (smaller than $ar^k$ if $r$ is the radius of our small circle).

Fourth comment: The zeroes of a holomorphic function are isolated unless the function is (locally) constant. If $f$ is constant on a small open disk, it is already far from injective.

Answer 2

I guess the text you were using was Stein and Shakarchi's Complex Analysis.

Actually, the proof is a bit problematic and should be modified in the following way.

We can first choose a small circle C centered at $z_0$ such that $|a(z-z_0)^k| > |G(z)|$ for any $z$ on this circle. This can be done simply because $G(z)$ is of order $k+1$ or higher.

Now we can choose a $w < \inf_{z\in C} \left(|G(z)| - |a(z-z_0)^k|\right) $. This guarantees that $|F(z)| = |a(z-z_0)^k-w| > |G(z)|$ on the circle.

Of course you can also require $w$ to be small enough to make sure the roots of $a(z-z_0)^k - w$ are inside the circle. (In fact, all those roots are on a circle centered at $z_0$ of radius $ {|w/a|}^{1/k}$.)