The inverse of a bijective holomorphic function is also holomorphic

Solution 1:

Step 1: $f'$ is never zero.

Indeed, if $f'(a)=0$ for some $a\in U$, then the Taylor expansion at $a$ is of the form $f(z)=f(a) + c_n (z-a)^n+\dots$ with $n\ge 2$, $c_n\ne 0$. This implies that $g(z) = (f(z)-f(a))/(z-a)^n$ is a nonzero holomorphic function near $a$, hence admits an $n$th degree root (a function $h$ such that $h^n=g$). Hence $$f(z) = f(a) + [(z-a)h(z)]^n$$ Since $z\mapsto (z-a) h(z) $ is an open map, its image contains a neighborhood of $0$; in particular it contains the points $\epsilon$ and $\epsilon \exp(2\pi i /n)$ for small $\epsilon$. These two points are sent into one, contradicting the injectivity of $f$.

Step 2: Inverse is smooth

This is just the Inverse function theorem: writing out $f=u+iv$, one can see that the Jacobian determinant of $(x,y)\mapsto (u,v)$ is $|f'(z)|^2\ne 0$.

Step 3: Inverse is holomorphic

Also the Inverse function theorem. Writing the derivative of $f$ as a $2\times 2$ real matrix, we get something of the form $$\begin{pmatrix} a & b \\ -b& a\end{pmatrix}$$ due to the Cauchy-Riemann equations. The inverse of such a matrix is also of this form: hence, $f^{-1}$ satisfies the Cauchy-Riemann equations.

Solution 2:

Hint: Prove that $f^{\prime}(z)\neq 0$ for all $z\in U$. Then use the inverse function theorem for analytic functions.

Solution 3:

Here is an approach using differential forms. Once you establish $f'$ has no roots and is therefore a local diffeomorphism, and hence a diffeomorphism $U\to V$, put $w = f(z)$ and $g(w) = f^{-1}(w)$. Then if $\tilde\gamma = f\circ \gamma$ is a closed, piecewise $C^1$ curve, \begin{align*} \oint_{\tilde\gamma} g(w)\,dw &= \oint_\gamma f^*(g\,dw)\\ &= \oint_\gamma z\,d(fz) \\ &= \oint_\gamma z\,\big(\partial_z f\,dz + \underbrace{\partial_{\bar z}f}_{=\ 0\ \text{because $f$ is holomorphic}}\,d\bar z\big) \\ &= \oint_\gamma z\,\partial_zf\,dz = 0, \end{align*} where the last integral is $0$ because $f$ is holomorphic. Hence $g$ is holomorphic by Morera's theorem.

Solution 4:

Another proof:

Without loss of generality we may assume that $U$ is connected. Obviously $f'$ does not vanish identically on $U$, so the set $Z$ of zeros of $f'$ is closed and discrete in $U$. By the open mapping theorem, $f:U\to V$ is a homeomorphism, so $f(Z)$ must also be closed and discrete. It is easy to check that $g=f^{-1}$ is holomorphic on $V\setminus f(Z)$, and the Riemann's theorem on removable singularities now shows that $g$ must be holomorphic on all of $V$.