﻿﻿ Fractional Calculus Computer Algebra System math software
+ + =

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acosh(x)

The hyperbolic cosine differs from the sine merely in the sign of the exponential. Solving for this gives

$coshw=ew +e−w 2 =z e2w -2zew +1 =0 ew =z ±z2-1 w=cosh−1z =ln(z ±z2-1 )$

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

$cosh−1z =ln(z ±z2-1 ) +2πni$

The individual branches look like this:

One half of each branch again comes from using the plus sign on the square root, the other half from the minus sign. The lines where clearly discordant colors meet are again where transitions to higher or lower branches occur.

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: