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The circular tangent differs from the hyperbolic tangent in having imaginary units in the exponential, plus an extra factor from the sine. Solving for the exponential gives

tanw=1i eiw -eiw eiw +eiw =z (1-iz) e2iw -1-iz =0 eiw =1+iz 1-iz w=tan1z =12i ln(1+iz 1-iz)

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

tan1z =12i ln(1+iz 1-iz) +πn

The individual branches look like this:

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