cute branch cut example from paper by Fateman and Dingle ..

r(z):=(z+1/z)/2$
s(w):=w+sqrt(w+1)*sqrt(w-1)$

radcan(s(r(z))  ==>  z
radcan(r(s(z))  ==> z  ..  so r and s seem to be inverses
s(r(12)) is 12,
s(r(-12)) is -12
s(r(%i))  is %i
s(r(-%i)) is %i.   surprise.

actually s(r(z))  is either z  or 1/z.  depending. Feel free to explore
r(s(z)).
Using rectform() and numer() and radcan()  do different things.

Here's a link to the paper

https://dl.acm.org/doi/pdf/10.1145/190347.190424

(This paper describes some programs we wrote (c 1994) using Mathematica,
pointing out some
difficulties. I expect that some of the shortfalls we had to program around
were fixed in subsequent Mathematica versions.).
A Google search will show some later papers on related topics, e.g. by fans
of Maple ( M. England, 2013)

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