# Analogue of Cauchy Integral Formula to compute $f(a)$ in terms of $f(z)$, $z$ on $\gamma$, when $a$ is outside $\gamma$

Let $$f:\Omega\to\mathbb C$$ be holomorphic, where $$\Omega$$ is a simply connected domain. Suppose we know the values of $$f$$ on a simple closed curve $$\gamma$$ contained in $$\Omega$$.

The Cauchy Integral Formula tells us how to calculate $$f(a)$$ for $$a$$ inside $$\gamma$$. But $$f(a)$$ is also uniquely determined for $$a\in\Omega$$ outside $$\gamma$$ (by the Identity Theorem). For such $$a$$, is there an explicit Cauchy-like formula for $$f(a)$$ in terms of the values of $$f$$ on $$\gamma$$?

I don't think there are easy ways to do this. It will require at least several steps for all points outside $$\gamma$$ covered. Certainly, we would not expect any easy formula for $$f(a_1)$$ in terms of values on $$\gamma$$ in the figure 1 below. Let $$R$$ be the region inside $$\gamma$$.

First, you have already that inside of $$\gamma$$ is covered.

Consider $$R_1=\{z\in\Omega | \exists a_0\in R, \exists \epsilon>0 \ \mathrm{such \ that} \ |z-a_0|<\epsilon \ \mathrm{ and }\ \overline{D}(a_0,\epsilon)\subseteq \Omega\}.$$ (Figure 2 below)

The function values on $$R_1$$ in terms of $$\gamma$$ is obtained as follows: $$f(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\gamma} \frac{f(w)}{(w-a_0)^{k+1}} dw\right)(z-a_0)^k$$ The function $$f$$ is holomorphic on $$\Omega$$, so the series is valid for any $$z\in R_1$$. The region $$R_1$$ properly contains $$R$$. We repeat this idea to obtain a larger region $$R_2$$ by using the boundary of $$R_1$$ as a new curve $$\gamma_1$$.

For any $$z\in\Omega$$, the above process is repeated finitely many times to reach $$z$$ from the inside of $$\gamma$$. (See figure 3).

Alternative approach

I present an alternative approach using Riemann mapping theorem. Fix $$a_0$$ inside $$\gamma$$. Let $$\phi:\Omega \rightarrow \mathbb{D}$$ such that $$\phi(a_0)=0$$ and $$\phi$$ is bijective holomorphic. See Figure 1.

Practically, this is almost as equally difficult as my previous approach, since $$\phi$$ encodes how to reach from $$a_0$$ to any point in $$\Omega$$. The presentation is more elegant than the previous approach.

Consider for $$z\in \mathbb{D}$$, the function $$F(z)=f(\phi^{-1}(z)).$$ This is a holomorphic function on $$\mathbb{D}$$ and $$\phi(\gamma)$$ encloses $$0\in\mathbb{D}$$. Thus, we are able to write $$F$$ as a power series about $$0$$ using Cauchy integral formula as before. $$F(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\phi(\gamma)} \frac{F(w)}{w^{k+1}} dw\right) z^k.$$

Since $$F$$ is holomorphic on $$\mathbb{D}$$, the power series is valid for any $$z\in \mathbb{D}$$.

Then for any $$z\in\Omega$$, we have $$f(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\phi(\gamma)} \frac{F(w)}{w^{k+1}} dw\right) (\phi(z))^k.$$ By a change of variable $$\phi^{-1}(w)$$->$$w$$, we have $$f(z)=\sum_{k=0}^{\infty} \left(\frac1{2\pi i} \int_{\gamma} \frac{f(w)\phi'(w)}{(\phi(w))^{k+1}} dw\right) (\phi(z))^k.$$