# How to find a branch of $\mathrm{log}(iz)$ such that it's analytic in a fixed region

The question as a whole is given as (I'll remove part a since it's not relevant I think)

Define $f$ and $g$ by $f(z) = \mathrm{Log}(iz)$ and $g(z) = z^{-1}\mathrm{Log}(z+1)$.
b. Find a branch of $\mathrm{log}(iz)$ which is analytic in the region $\{z:{\bf Im}(z)>0\}.$
where $\mathrm{log}$ is multi-valued.

I know how to first find the set of points in the plane so that the Log of some polynomial allows it to be analytic, by using the principal branch $\mathbb{C} \backslash (-\infty,0]$. But I'm not sure how I could do the reverse.

• math.stackexchange.com/questions/555020/… – user29418 Aug 29 '17 at 3:20
• Removing any ray through the origin and infinity as long as it's not above the real axis should probably work – user29418 Aug 29 '17 at 3:22
• If $U$ is a simply connected open set not containing $0$ then $\frac{1}{z}$ is analytic on $U$ and so is $L(z) = \int_a^z \frac{1}{s}ds$ – reuns Aug 29 '17 at 9:30

If $g$ is the principle branch of logarithm then $g(-iz)+i\dfrac{\pi}{2}$ is such a branch.