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acot(x)

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

cotw=i eiw +eiw eiw -eiw =z (i-z) e2iw +i+z =0 eiw =z+i z-i w=cot1z =12i ln(z+i z-i)

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

cot1z =12i ln(z+i z-i) +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: