# Analytic branch of log z

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 $$z\in D$$, $$g(z)-h(z)=2\pi i n$$.

• Sorry, I don't really understand what you mean. How did you get to $exp(g(z) - h(z))$?
– Max
Apr 1, 2020 at 9:30
• I wanted to prove that $g(z)-h(z)$ is constant and my first step was to prove that $\exp\bigl(g(z)-h(z)\bigr)$ is constant. Apr 1, 2020 at 9:33
• Oh, I see. So the way I tried proving it is wrong?
– Max
Apr 1, 2020 at 9:39
• I didn't understand it. There is no $\log$ function. So, what is $\log'$? And what is $\operatorname{Arg}$? Apr 1, 2020 at 9:46
• $g(z), h(z)$ are two analytical branches of $logz$. $log'(z)$ is how I noted it as a derivative (i.e. $g'(z)$). Arg is the principal argument
– Max
Apr 1, 2020 at 9:51