Complex Analysis - Maths KU

5
0
–5
–2
–2
0
0
2
2
Figure 4.16 A Riemann surface for
w = sin−1 z
4.4 Inverse Trigonometric and Hyperbolic Functions 221
(b) From (14) we have:
d
dz cosh−1 �
�
z�
� z= √ 2/2
1
1
= ��√2/2�2 � = .
1/2 1/2
− 1 (−1/2)
After using f2 to find the square root in this expression we obtain:
d
dz cosh−1 �
�
z�
= 1
√ = −
2i/2 √ 2i.
Remarks
� z= √ 2/2
The multiple-valued function F (z) =sin −1 z can be visualized using the
Riemann surface constructed for sin z in the Remarksin Section 4.3 and
shown in Figure 4.16. In order to see the image of a point z0 under
the multiple-valued mapping w =sin −1 z, we imagine that z0 islying
in the xy-plane in Figure 4.16. We then consider all points on the Riemann
surface lying directly over z0. Each of these points on the surface
corresponds to a unique point in one of the squares Sn described in the
Remarksin Section 4.3. Thus, thisinfinite set of pointsin the Riemann
surface represents the infinitely many images of z0 under w =sin −1 z.
EXERCISES 4.4 Answers to selected odd-numbered problems begin on page ANS-15.
In Problems 1–10, find all values of the given quantity.
1. cos −1 i 2. sin −1 1
3. sin −1 √ 2 4. cos −1 5
3
5. tan −1 1 6. tan −1 2i
7. sinh −1 i 8. cosh −1 9. tanh
1
2
−1 (1+2i) 10. tanh −1 �√ 2i �
In Problems 11–16, use the stated branch of the multiple-valued function z 1/2 and
principal branch of ln z to (a) find the value of the inverse trigonometric or hyperbolic
function at the given point and (b) compute the value of the derivative of the
function at the given point.
11. sin −1 z, z = 1
i; use the principal branch of z1/2
2
12. cos −1 z, z = 5
3 ; use the branch √ re iθ/2 ,0