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# asec(x)

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

$secw=2 eiw +e−iw =z ze2iw -2eiw +z =0 eiw =1 ±1-z2 z w=sec−1z =1iln(1 ±1-z2 z)$

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

$sec−1z =1iln(1 ±1-z2 z) +2πn$

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: