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acos(x)

The circular cosine differs from the hyperbolic cosine in having imaginary units in the exponential. Solving for this gives

cosw=eiw +eiw 2 =z e2iw -2zeiw +1 =0 eiw =z ±z2-1 w=cos1z =1iln(z ±z2-1 )

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

cos1z =1iln(z ±z2-1 ) +2πn

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: