Showing that choice of point on a curve is irrelevant to finding $f(z_0)$ in the curve Before I start I apologize for the horrible title but I have no idea how to title this.
So the problem is as follows:
Let $f(z)$ be analytic in and on a simple closed curve $\Gamma$, and let $f(z)$ have no zeros in or on $\Gamma$.
Now let $z_0$ be a point in $\Gamma$, and $z_1$, $z_2$ be two points on $\Gamma$. Let $\gamma_1$ & $\gamma_2$ be the two curves connecting $z_1$ & $z_2$ to $z_0$ respectively.  Where the integraion is towards $z_0$ both times show that 
$$f(z_1)e^{\int_{\gamma_1}\frac{f'(x)}{f(x)}dx}=f(z_2)e^{\int_{\gamma_2}\frac{f'(x)}{f(x)}dx}$$
And then show that both sides are equal to $f(z_0)$.
I'm given a hint to represent $f(z_2)/f(z_1)$ as an integral along the part of $\Gamma$ connecting these two points.
This is where the problem begins and thats where I would start, but I'm struggling with that.
One idea I had is that since $f(z_0)$ doesn't equal $0$, I could use that I could make a disk large enough centered at $z_0$ such that $e^{w(z)}=f(z)$, where $w(z)$ is essentially the two integral above, just flip the sign.
Another idea I think is better.  If I solve the hint, then I could make a new curve $\gamma_3$ using the arc connecting $z_1$ and $z_2$ and $\gamma_1$ and $\gamma_2$, and then use Cauchy's theorem, so that $\int_{\gamma_3}f(z)=0$, and I would split this integral up into the three parts forming this curve if I could figure it out.
Any help is appreciated, thanks in advance.
 A: This may be naive of me, but $f'/f$ has an anti derivative (some branch of the logarithm, call it $\log f$) defined everywhere in and on the curve, so we could just evaluate each expression at the endpoints:
\begin{align*}
f(z_1)e^{\int_{\gamma_1}\frac{f'(z)}{f(z)}dz}&=f(z_1)e^{\log f(z_0)-\log f(z_1)}=f(z_1)\big(f(z_0)/f(z_1)\big)=f(z_0),\\
f(z_2)e^{\int_{\gamma_2}\frac{f'(z)}{f(z)}dz}&=f(z_2)e^{\log f(z_0)-\log f(z_2)}=f(z_2)\big(f(z_0)/f(z_2)\big)=f(z_0).
\end{align*}
Is there some technical point I'm missing here?
Edit: The previous procedure can be used without having to assume a branch of the logarithm exists on the entire curve--just assume it exists in simply connected neighborhoods of $\gamma_1,\gamma_2$. Also, Here's a method that uses the hint:
Let $\gamma_3$ is a simple curve beginning at $z_1$ and travelling to $z_2$. $f'/f$ is analytic ($f\neq 0$) in neighborhood (which we can assume--through shrinking if necessary--is simply-connected) of the image of $\gamma_3$, so there exists a holomorphic branch of the logarithm, $\log f$ in just this neighborhood. Thus
$$
\frac{f(z_2)}{f(z_1)}=e^{\int_{\gamma_3}\frac{f'(z)}{f(z)}dz}.
$$ 
Then
$$
f(z_1)e^{\int_{\gamma_{1}}\frac{f'(z)}{f(z)}\,dz}=f(z_2)e^{\int_{\gamma_{1}}\frac{f'(z)}{f(z)}\,dz-\int_{\gamma{3}}\frac{f'(z)}{f(z)}\,dz}=f(z_2)e^{\int_{\gamma_2}\frac{f'(z)}{f(z)}\,dz}
$$
This is true because integrating over "$\gamma_1-\gamma_2-\gamma_3$" gives zero by by Cauchy's theorem since $f(z)\neq 0$ everywhere (draw a picture) so it follows that integrating over "$\gamma_1-\gamma_3$" is the same as integrating over $\gamma_2$. Hopefully this (not really that different) method will lend some insight.
A: Notice that:
$\displaystyle\frac{f(z_{2})}{f(z_{1})}=\exp({\int_{C}\frac{f'(z)}{f(z)}d\mathrm{z}})$
Where $C$ is the arc going from $z_{1}$ to $z_{2}$ in the curve $\Gamma$. You can now expand the terms and use the theorem of Cauchy to get the result.
