MATH4555 - HW 7 Solutions
Janis lai
1
Section 7.4
#1 (2 points each)
(a) f (0) 0 , therefore f (z) has zero of order 2 at origin.
f ( n, i n) 0 , therefore f (z) has simple zeros at n and i n.
(h) f (
MATH4555 - HW 3 Solutions
Janis lai
1
Section 3.2
#7 (3 points each)
(a)h(z) = ey cos x + iey sin x
u x = ey sin x
vy = ey sin x
uy = ey cos x
only when x = n
1
only when x = (n + )
2
v x = ey cos x