If $f(z)$ analytic on $E(L)$, and $\lim_{z \to \infty}f(z)=A$, prove that $\frac{1}{2\pi i} \int_{L}\frac{f(\zeta)}{\zeta-z}dz = A$ or $-f(z)+A$ I am working on the following problem:

Let $L$ be a closed, rectifiable simple curve, traversed
  counterclockwise.  Let $f(z)$ be a differentiable function on a domain
  $G$ where $L \cup E(L)$ ($E(L)$ is the exterior of $L$, $I(L)$ is
  the interior of $L$) is contained in $G$. (That is, $f$ is
  differentiable on a neighborhood of $L$ and outside of $L$, but not
  necessarily inside.) 
Suppose that $\displaystyle \lim_{z \to \infty}f(z) = A$.
Prove that $\displaystyle \frac{1}{2\pi i}\int_{L}
\frac{f(\zeta)}{\zeta-z}d \zeta = A$ if $z \in I(L)$ and that
  $\displaystyle \frac{1}{2\pi i}\int_{L} \frac{f(\zeta)}{\zeta-z}d
\zeta = -f(z)+A$ if $z \in E(L)$

Our professor gave us a hint: Pass to integration over a large circle $|z|=R$ (which, I assume must contain $L$?) using Cauchy Theorem for multiple contours. Then, use the following result:

If $f(z)$ is continuous in the closed domain $|z| \geq R_{0}$, $0 \leq arg z \leq \alpha$ ($0 \leq \alpha \leq 2\pi$), and if the limit $\displaystyle \lim_{z \to \infty}zf(z) = A$ exists, then $\displaystyle \lim_{R \to \infty} \int_{\displaystyle \Gamma_{R}} f(z) dz = iA\alpha$ where $\displaystyle \Gamma_{R}$ is the arc of the circle $|z| = R$ lying in the given domain.

The professor's hint is confusing me a bit, and I'm not sure how to apply it here. I.e., how do limits come into play here? Also, how do I use this circle to split the domain into multiple contours? 
I tried applying it, and this is all I have so far: Let $\zeta = Re^{i \theta}$, then $d \zeta = i Re^{i \theta} d \theta$, so the integral becomes: $\displaystyle \frac{1}{2 \pi i} \int_{0}^{2\pi} \frac{f(Re^{i \theta})}{Re^{i \theta} - z}d\theta $, and from that point, I am completely lost...
 A: METHODOLOGY $1$: Using the Residue Theorem and the Residue at Infinity
From the residue theorem we have
$$\begin{align}
\oint_{L}\frac{f(z')}{z'-z}\,dz'&=-2\pi i\begin{cases}
\text{Res}\left(\frac{f(z')}{z'-z},z'=z\right)+\text{Res}\left(\frac{f(z')}{z'-z},z'=\infty\right)&,z\in E(L)\\\\
\text{Res}\left(\frac{f(z')}{z'-z},z'=\infty\right)&,z\in I(L)
\end{cases}
\end{align}$$

Note that the minus sign in the term $-2\pi i$ is a consequence of the orientation of $L$.  Although $L$ is traversed counter-clockwise, the contour $L$ encloses the point at $\infty$ with clockwise orientation.  And if $z\in I(L)$, then $L$ also encloses $z$ with clockwise orientation. 


Evaluating the Residues
The residue at $z'=z$ is trivial to evaluate with $\text{Res}\left(\frac{f(z')}{z'-z},z'=z\right)=f(z)$.  
The residue at infinity can be evaluated by transforming $z' =1/w$ and evaluating the residue at $w=0$.  Proceeding we find 
$$\begin{align}
\text{Res}\left(\frac{f(z')}{z'-z},z'=\infty\right)&=\text{Res}\left(-\frac{1}{w^2}\frac{f(1/w)}{1/w-z},w=0\right)\\\\
&=-A
\end{align}$$

Result
Putting it all together we find 
$$\frac{1}{2\pi i}\oint_{L}\frac{f(z')}{z'-z}\,dz'=
\begin{cases}
A-f(z)&,z\in E(L)\\\\
A&,z\in I(L)
\end{cases}$$

METHODOLOGY $2$: Using the Hint
Note that for $z\in I(L)$, and $R$ sufficiently large we can deform the contour $L$ to a circle of radius $R$, centered at the origin, and enclosing $L$.  Then, we can write
$$\begin{align}
\oint_{L}\frac{f(z')}{z'-z}\,dz'&=\oint_{|z'|=R}\frac{f(z')}{z'-z}\,dz'\\\\
&=\int_0^{2\pi}\frac{f(Re^{i\phi})}{Re^{i\phi}-z}\,(iRe^{i\phi})\,d\phi\\\\
&\to 2\pi i A\,\,\text{as}\,\,R\to \infty
\end{align}$$
For $z\in E(L)$, if we deform the contour $L$ as in the previous development, we need to account for the introduction of the simple pole of $\frac{1}{z'-z}$ at $z'=z$.  Therefore,  
$$\begin{align}
\oint_{L}\frac{f(z')}{z'-z}\,dz'&=\oint_{|z'|=R}\frac{f(z')}{z'-z}\,dz'-2\pi i \text{Res}\left(\frac{f(z')}{z'-z},z'=z\right)\\\\
&=\int_0^{2\pi}\frac{f(Re^{i\phi})}{Re^{i\phi}-z}\,(iRe^{i\phi})\,d\phi-2\pi i f(z)\\\\
&\to 2\pi i (A-f(z))\,\,\text{as}\,\,R\to \infty
\end{align}$$
as was to be shown!
