[SOLVED] Amath 567, problem set 2

1. Problem 1: Evaluate H
C
f(z)dz where C is the unit circle centered at the
origin and f(z) is given by the following:
(a) e
iz
(b) e
z
2
(c) 1
z−1/2
(d) 1
z
2−4
(e) 1
2z
2+1
(f) √
z − 4, with 0 ≤ z − 4 < 2π2. Problem 2: We wish to evaluate the integral
Z ∞
0
e
ix2
dx
Consider the contour
IR =
I
C(R)
e
iz2
dz
where C(R)
is the closed circular sector in the upper half plane with boundary points 0, R, and Reiπ/4
. Show that IR = 0 and that
lim
R→∞ Z
C1(R)
e
iz2
dz = 0
where C1(R)
is the line integral along the circular sector from R to Reiπ/4
.
Hint: Use sin(x) ≥
2x
π
on 0 ≤ x ≤ π/2.
Then, breaking up the contour C(R)
into three component parts, deduce
lim
R→∞ Z R
0
e
ix2
dx − e
iπ/4
Z R
0
e
−r
2
dr!
= 0
and from the well-known result of real integration:
Z ∞
0
e
−x
2
dx =
√
π
2
deduce that I = e
iπ/4√
π/2.3. Problem 3: Consider the integral
I =
Z ∞
−∞
dx
x
2 + 1
Show how to evaluate this integral by considering
I
C(R)
dz
z
2 + 1
where C(R)
is the closed semicircle in the upper half plane with endpoints
at (−R, 0) and (R, 0) plus the x axis. Hint: use
1
z
2 + 1
=
−1
2i

1
z + i
−
1
z − i

and show that the integral along the open semicircle in the upper half
plane vanishes as R → ∞. Verify your answer by usual integration in real
variables.4. Problem 4: Let
f(z) = e
t
2
(z−1/z) =
X∞
n=−∞
Jn(t)z
n
Show from the definition of Laurent series and using properties of integration that
Jn(t) = 1
2π
Z π
−π
e
−i(nθ−t sin θ)
dθ =
1
π
Z π
0
cos (nθ − tsin θ)dθ
The functions Jn(t) are called Bessel functions, which are well-known special functions in mathematics and physics.