Let $g(z)$ denote an analytic branch of $log z$ in a domain $D \subset\mathbb{C}$. Show that $g'(z) = 1/z$. Show also that if $h(z)$ is another analytic branch of $log z$ in $D$, then the function $$\frac{g(z)-h(z)}{\pi i}$$ is constant in $D$ and equal to an even integer.
Here is how I approached the problem:
$$z = exp(log\ z)$$
By differentiating both sides of the equation we get:
$$1 = exp(log\ z) * (log\ z)' = z*(log\ z)$$
By dividing both sides by $z$ we get:
$$(log\ z)' = \frac{1}{z}$$
For the second part of the problem:
$$\frac{g(z) - h(z)}{\pi i} = \frac{(log|z| + iArg\ z + 2k_1i\pi) - (log|z| + iArg\ z + 2k_2i\pi)}{\pi i} = \frac{2k_1i\pi - 2k_2i\pi}{\pi i}$$ $$= 2(k_1 - k_2)$$
I have been struggling with wrapping my head around branches and whatnot so I am a little unsure if I have done what was asked of me here. Does it look correct?
Note that\begin{align}\exp\bigl(g(z)-h(z)\bigr)&=\frac{\exp\bigl(g(z)\bigr)}{\exp\bigl(h(z)\bigr)}\\&=\frac zz\\&=1.\end{align}So, for each $z\in D$, $g(z)-h(z)=2\pi in$, for som integer $n$. Since $D$ is connected, the range of $g-h$ must be connected too, and therefore there is some $n\in\mathbb Z$ such that, for each $z\in D$, $g(z)-h(z)=2\pi i n$.