﻿﻿ Fractional Calculus Computer Algebra System math software
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# asech(x)

The hyperbolic secant is the algebraic inverse of the hyperbolic cosine. Solving for the exponential gives

$sechw=2ew +e−w =z ze2w -2ew +z =0 ew =1 ±1-z2 z w=sech−1z =ln(1 ±1-z2 z)$

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

$sech−1z =ln(1 ±1-z2 z) +2πni$

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: