I'm an undergrad taking my first complex analysis course, and I ran into something I wasn't completely clear about in Stein & Shakarchi's complex analysis book right at the beginning of the section on conformal mappings.
If $f: U \rightarrow V$ is holomorphic and injective, then $f'(z) \neq 0$ for all $z \in U$. In particular, the inverse of $f$ defined on its range is holomorphic, and thus the inverse of a conformal map is also holomorphic.
The proof starts by supposing $f'(z_0) = 0$ for some $z_0 \in U$, and then says that
f(z) – f(z_0) = a(z-z_0)^k + G(z)
Where $k \geq 2$, and $G(z)$ vanishes of order $k+1$ near $z_0$. The authors don't really explain why this is true, but my best guess is that they're just taking the Taylor expansion of $f$ at $z_0$. Is that accurate?
Next, they say that for sufficiently small circle around $z_0$, and for sufficiently small $w$, we have the following two properties:
1] $|G(z)| < |a(z-z_0)^k – w|$ on the circle
2] $|a(z-z_0)^k – w|$ has at least two zeros inside the circle
Neither of these is explained either. But I'm wondering if the following is a valid explanation: because $G(z)$ vanishes of order greater than $k$, there is some circle $C$ around $z_0$ such that on $C$, $|G(z)| < |a(z-z_0)^k|$, so choosing $|w| < |a(z-z_0)^k| – |G(z)|$, we have $|a(z-z_0)^k – w| \geq |a(z-z_0)^k| – |w| > G(z)$. Also, because $|G(z)| > 0$, we also have $|w| < |a(z-z_0)^k|$ so by Rouche's theorem, $|a(z-z_0)^k – w|$ still has at least two zeros inside the circle.
Therefore, by Rouche's theorem, $f(z) – f(z_0) – w$ also has two zeros inside the circle.
So far, if my above explanations are correct, I'm not really running into trouble. However, now the authors assert
Since $f'(z_0) \neq 0$ for all $z \neq z_0$ in a small circle around $z_0$, the roots of $f(z) – f(z_0) – w$ must be distinct.
Here for some reason I don't follow at all why this is true, my brain just can't seem to pick up the connection. Would someone mind explaining why the roots have to be distinct?
Later on, Stein and Shakarchi say that if $f$ is bijective and holomorphic, and $g$ is its inverse, we have
$$\frac{g(w) – g(w_0)}{w-w_0} = \frac{1}{\frac{w-w_0}{g(w) – g(w_0)}} = \frac{1}{\frac{f(z) – f(z_0)}{z-z_0}}$$
And then they let $z \rightarrow z_0$ to get the derivative of $g$. But my question is, when we calculate the derivative of $g$, shouldn't we instead let $w \rightarrow w_0$? If we say that this is the same as letting $g(w) \rightarrow g(w_0)$, doesn't that require g to be continuous? How do we know continuity of $g$ from holomorphicity of $f$?
Would anyone mind verifying that my earlier explanations make sense, and also explaining my latter two questions? Edit: I bolded my questions to separate them from the rest of the proof. Thanks!
Best Answer
This is correct. Recall that since $f$ is holomorphic in $U$, it is also analytic in $U$. Therefore its power series must converge for some positive radius of convergence at all points in $U$ (recall the definition of analytic function found on page 18 of Chapter 1). Thus, the power series expansion around $z_0$ converges for some positive radius of convergence, i.e. $f(z) = f(z_0) + (z-z_0)f'(z_0) + \cdots$ but since $f'(z_0)=0$, it must be the case that $$f(z) = f(z_0) + a(z - z_0)^2 + G(z) \implies f(z) - f(z_0) = a(z - z_0)^k + G(z)$$ for some $k \geq 2$ and $G$ is a polynomial in $z-z_0$ with a factor of $(z-z_0)^{k+1}$.
This looks right to me.
Assume the roots are not distinct, then let the root be $r$ and it must have multiplicity 2. Then, since $r$ is a root of $f(z)-f(z_0)-w$, we can write $f(z) - f(z_0) - w = (z - r)^2 g(z)$ where $g(z)$ is non-vanishing in some neighborhood of $r$. Also, note that $r \neq z_0$, otherwise $w = 0$. But then taking the derivative of both sides, $f'(z) = 2(z-r)g(z) + (z-r)^2g'(z)\implies f'(r) = 0$. But $f'(r) \neq 0$ for all $r\neq z$ in that neighborhood, so we have a contradiction.
I believe the continuity of $g$ follows from the Inverse-function theorem. That is, since the derivative of $f$ is nonzero around $z_0$, $g$ must be differentiable, and therefore continuous.