I came across the following transition in a proof a complex analysis book that I was going through.
For $z \in \mathbb{C}$, let $h(z) = \int^z_{z_0} g'(z) / g(z) dz$ and suppose $h$ is analytic, then we have $h'(z) = g'(z) / g(z)$. Also, $\frac{d}{dz} (g(z) e^{-h(z)}) = 0$...
My attempt
I was trying to understand the last transition in the following way,
By the derivative of logarithm, I went;
$\frac{d}{dz} h(z) = \frac{d}{dz} \log (g(z))$, thus, $g'(z)/g(z) = e^{-h(z)}$ and $\frac{d}{dz} (g(z) e^{-h(z)}) = \frac{d}{dz} g'(z) = 0 (?)$.
I was not sure why this is supposed to lead to 0,, or maybe I'm completely going wrong. could you tell me about this?
$\frac d {dz} (g(z)e^{-h(z)})=g'(z)e^{-h(z)}-h'(z)g(z)e^{-h(z)}=e^{-h(z)}[g'(z)-h'(z)g(z)]=0$ since $h'(z)=\frac {g'(z)} {g(z)}$ (by FTC).
Incidentally, you forgot to state that $g(z)$ does not vanish at any point.