Complex Analysis - Maths KU

216 Chapter 4 Elementary Functions
The two square roots (−4) 1/2 of –4 are found to be ±2i using (4) in Section
1.4, and so:
sin −1 √ �
5=−i ln i √ � ��√ � �
5 ± 2i = −i ln 5 ± 2 i .
Because �√ 5 ± 2 � i isa pure imaginary number with positive imaginary part
(both √ 5+2 and √ 5 − 2 are positive), we have � � �√ 5 ± 2 � i � � = √ 5 ± 2 and
arg ��√ 5 ± 2 � i � = π/2. Thus, from (11) in Section 4.1 we have
��√ � � �√ � �
π
ln 5 ± 2 i = loge 5 ± 2 + i
2 +2nπ
�
for n =0,±1, ±2, ... . This expression can be simplified by observing that
�√ �
1
�√ � �√ �
loge 5 − 2 = loge √ = loge 1 − loge 5+2 =0− loge 5+2 ,
5+2
�√ � �√ �
and so loge 5 ± 2 = ± loge 5+2 . Therefore,
��√ � � � �√ � �
π
−i ln 5 ± 2 i = −i loge 5 ± 2 + i
2 +2nπ
��
and so
sin −1 √ 5=
for n =0,±1, ±2, ... .
� �√ �
= −i ± loge 5+2 + i
(4n +1)π
2
± i log e
�√ �
5+2
(4n +1)π
2
Inverse Cosine and Tangent We can easily modify the procedure
used on page 215 to solve the equations cos w = z and tan w = z. Thisleads
to definitions of the inverse cosine and the inverse tangent, which we now
state.
Definition 4.9 Inverse Cosine and Inverse Tangent
The multiple-valued function cos−1 z defined by:
cos −1 �
z = −i ln z + i � 1 − z 2�1/2 �
iscalled the inverse cosine. The multiple-valued function tan−1 z defined
by:
tan −1 z = i
2 ln
� �
i + z
(5)
i − z
iscalled the inverse tangent.
�
,
(4)