300 Chapter 5
5.25 Tiffi INVERSE OF THE CIRCULAR AND
HYPERBOLIC FUNCTIONS
Given any z E C there is a set of values of w satisfying the equation
sinw = z (5.25-1)Since sin(w + 2k7r) = sinw, where k is any integer, it is clear that if
the equation (5.25-1) has a solution at all, it has infinitely many, so that
equation defines implicitly a multiple-valued function called the inverse
sine, denoted
w = arcsinz or w=sin-^1 z
To show that equation (5.25-1) actually has solution for any given z f=
oo, and to give a formula for their computation, we begin by writing (5.25-1)
in the equivalent form
__ 2_i __ =z
which leads to the quadratic equation in eiw:
e^2 iw - 2izeiw - 1 = 0Hence
so
· w = arc sin z = %1f - i log( z + *.Jz2=1) (5.25-2)
If V z^2 - 1 represents the principal value of *v'z2="1, meaning the root
with argument in the interval (-%7r, 1/ 2 7r), the second square root can be
represented by -Jz^2 - 1. Since
(z - .Jz2=l)(z + Jz2 -1) = 1
we have
log(z - .Jz2=1) = - log(z + .Jz2=1)
modulus 27r. Hence formula (5.25-2) can be expressed in the form
w = arcsinz = %7r =i= ilog(z + .Jz2=1)
or