Complex Analysis - Maths KU

4.4 Inverse Trigonometric and Hyperbolic Functions 215
Section 4.3. We begin by using Definition 4.6 to rewrite the equation sin w = z
as:
e iw − e −iw
2i
= z or e 2iw − 2ize iw − 1=0.
Because e 2iw − 2ize iw − 1 = 0 isa quadratic equation in e iw we can then use
the quadratic formula (3) in Section 1.6 to solve for e iw :
e iw = iz + � 1 − z 2� 1/2 . (1)
Since we are using the quadratic formula, we should keep in mind that the
expression � 1 − z 2� 1/2 in (1) represents the two square roots of 1−z 2 . Finally,
we solve for w using the complex logarithm:
�
iw =ln iz + � 1 − z 2�1/2 �
�
or w = −i ln iz + � 1 − z 2�1/2 �
. (2)
Each value of w obtained from the second equation in (2) satisfies the equation
sin w = z. Therefore, we call the multiple-valued function defined by the
second equation in (2) the inverse sine. We summarize this discussion in
the following definition.
Definition 4.8 Inverse Sine
The multiple-valued function sin −1 z defined by:
sin −1 �
z = −i ln iz + � 1 − z 2�1/2 �
iscalled the inverse sine.
At times, we will also call the inverse sine the arcsine and we will denote
it by arcsin z. It is clear from (3) that the inverse sine is multiple-valued since
it isdefined in termsof the complex logarithm ln z. It isalso worth repeating
that the expression � 1 − z 2� 1/2 in (3) represents the two square roots of 1−z 2 .
EXAMPLE 1 Values of Inverse Sine
Find all valuesof sin −1 √ 5.
Solution By setting z = √ 5 in (3) we obtain:
sin −1 √ �
5=−i ln i √ � �√ � � �
2
1/2
5+ 1 − 5
�
= −i ln i √ 5+(−4) 1/2�
.
(3)