# Shifting the branch cut of the complex logarithm

I have some troubles understanding branch cuts of the complex logarithm, or to be more precise, the shift thereof. So typically, the branch cut is along the negative real axis, connecting the branch points at $$0$$ and $$\infty$$. We know, that this choice is not unique - I could, for instance, also choose the positive real axis.

Now, this is where I'm puzzled: is it also possible to shift it to any different spot, say, to the negative imaginary axis? This would lead to a single-valued function on all of the real axis except at the origin, where we still have a singularity - right?

Surely you can do that.

Define $$\log \, z=\log|z|+i\theta$$ with $$-\pi /2 <\theta < 3\pi /2$$, where $$z=|z|e^{i\theta}$$, and you will get an analytic branch of logarithm in $$\mathbb C\smallsetminus {\{-it: t\geq 0\}}$$.

• Sorry for the nitpick: Don't you need a domain where (precisely) one of the endpoints ; either $\pi/2$ or $3\pi/2$ here, are included in order to be a local inverse of logz on $\mathbb C - \ \{ 0\}$ ?
– MSIS
Commented Mar 30 at 14:26

Essentially it is not possible to define the complex logarithm on a closed curve around the origin. To this end the cutting is usually done from the origin in a straight line (in any direction). However you could in theory also cut along a general curve (from the origin). See also the wikipedia article for information: https://en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts

$$\log z=\log |z|+i\angle z$$

and you can choose where you want the phase jump of $$\angle z$$ to occur.

You would have for positive $$x$$

$$\log(-x)=\log x-i\pi.$$