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.