Prove that complex Log is holomorphic by using a weird function?

I want to prove that the complex $$Log(z)$$ function is holomorphic, when not negative.

To prove it, I am asked to use a "helper" function, which I do not understand how it helps. The function is:

$$F(\omega)=\frac{w-w_0}{e^w-e^{w_0}}$$ When $$w\neq w_0$$. $$F(w)=e^{-w_0}$$

Otherwise. And in the function $$w_0=Log(z_0)$$.

How can I use this function to prove that the Log function is holomorphic. As previous knowledge there is a theorem saying that combinations of functions have a derivative $$D(f(g(z_0))=g'(f(z_0))f'(z_0)$$ Provided certain conditions are met. I also know that the derivative of $$e$$ is $$e$$.

We have $$\frac {Log\, (z_0+h)-Log\, z_0} h=\frac {Log\, (z_0+h)-Log\, z_0} {e^{Log\, (z_0+h)} -e^{Log\, z_0}}$$. If you prove that $$F$$ is continuous, which is easy you see that above quantity tends to $$e^{-Log \, z_0}=\frac 1 {z_0}$$ which prove that $$Log$$ is differentiable at $$z_0$$ with derivative $$\frac 1 {z_0}$$. For this to work you have to make sure that Log is continuous at $$z_0$$ and for this you have to avoid $$(-\infty,0]$$.