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atanh(x)

Consider first the hyperbolic tangent in terms of the exponential. Solving for this gives

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

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

$tanh−1z =12 ln(1+z 1-z) +πni$

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:

This graphic illustrates a feature of the logarithm that will apply to all of the functions considered here: its coloration by argument quickly becomes monochromatic on branches with an index of higher absolute value.

The gaps between adjacent branches here and in subsequent visualizations of multiple branches represent the removal of artifacts arising from the parametric representations employed. They could be removed using the healing method of this presentation, but will be left as is for simplicity of rendering. The analytic functions naturally continue across all such gaps.