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# acoth(x)

The hyperbolic cotangent differs from the tangent merely in the sign of the exponential. Solving for this gives

$cothw=ew +e−w ew -e−w =z (1-z) e2w +1+z =0 ew =z+1 z-1 w=coth−1z =12 ln(z+1 z-1)$

Applying the behavior of the logarithm, the inverse hyperbolic cotangent on an arbitrary branch is

$coth−1z =12 ln(z+1 z-1) +πni$

The individual branches look like this:

The real part of this function retains the same numerical value between branches, while the imaginary part moves up and down in value. Visualize the imaginary part of several branches simultaneously: