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Let $D=\{z\in\mathbb{C}:|z|<1\}$.

Let $f:\overline{D}\to\mathbb{C}$ be a continuous function that is holomorphic in $D$.

Suppose that $f$ is injective.

Prove that $\forall w_0\in f(D)$

$$f^{-1}(w_0)=\frac{1}{2\pi i}\int_{\partial D}\frac{zf'(z)}{f(z)-w_0}dz$$

From injectivity $f'\neq 0$ in $D$, so $f^{-1}$ is holomorphic in $f(D)$.

Then, I wanted to say that $\partial f(D)=f(\partial D)$ and use

$$f^{-1}(w_0)=\frac{1}{2\pi i}\int_{\partial f(D)}\frac{f^{-1}(z)}{z-w_0}dz$$

And show the equivalence of the integrals using the parametrizations

$$\forall t\in[0,\pi],\gamma_1(t)=e^{it},\gamma_2(t)=f(e^{it})$$

of $\partial D,f(\partial D)$, respectively.

The problem is , of course, that for Cauchy's formula we need holomorphicity in a domain $\Omega$ that contains $\overline{f(D)}$.

Edit:

Can we define $\forall\ 0<R<1:D_R=\{z\in\mathbb{C}:|z|<R\}$, use Cauchy there and show the equivalence, and finally take $R\to 1$ ?

Don Fanucci
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1 Answers1

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Following your own suggestion, you can start by noting that$$f^{-1}(w_0)=\frac1{2\pi i}\int_{\partial f(D_r)}\frac{f^{-1}(z)}{z-w_0}\,\mathrm dz,$$assuming that $R>|w_0|$. And now you can do the substitution $z=f(w)$ and $\mathrm dz=f'(w)\,\mathrm dw$, thereby proving that the previous integral is equal to$$\frac1{2\pi i}\int_{\partial D_R}\frac{zf'(w)}{f(w)-w_0}\,\mathrm dw.$$

José Carlos Santos
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