Background: I'm trying to solve a previous question that people tried to help with. I've got almost what I need from good posts like 1, 2. When you look at functions like $\sqrt{(z-a)(z-b)}$ You can recognize the branch cuts graphically around the circles of the roots $\sqrt{|z-a|e^{i \arg(z-a)} |z-b|e^{i \arg(z-b)}}$ and figure out the interval where the two jumps cancel. Is there a comprehensible way to visualize the inverse trig function's branch cut in the same way?
Plot3D[Re[Sinh[x + I y]], {x, -20, 20}, {y, -20, 20}, AxesLabel -> Automatic]
As you can see the real part sinh $\frac{e^z-e^{-z}}{2i}$ traces out a sin wave along the y axis, and an exponential $e^{|x|}$ along the x-axis. The Im[sinh] traces out a cos wave along the y axis with an similar looking graph.
The Real part of the inverse $\sinh^{-1}=\ln|z+(z^2+1)^{1/2}|$ is taking the infinite positive and negative z axis above and mapping it to a restricted x valued z-height. Branch cut $(-i\infty,-i)$,$(i,i\infty)$
The imaginary inverse sinh =$iArg(z+(z^2+1)^{1/2}$ appears as follows with the branch cut :
Plot3D[Im[ArcSinh[(x + I y)]], {x, -20, 20}, {y, -20, 20}, AxesLabel -> Automatic]
Is there any intuitive way to understand the branch cut?
