SOLUTION OF TEST 3
MINGFENG ZHAO
November 08, 2012
1. [10 Points] Find the integral
Z
∞
−∞
1
dx
5 − 2x + x2
using the Residue Theorem. Show all steps.
2
Proof. Consider z − 2z + 5 = 0, then z =
2±
√
4 − 20
= 1 ± 2i, that is, z 2 − 2z + 5 = [z − (1 +
2
2i)][z − (1 − 2i)].
For any R >
Z
√
R
−R
Z
For
CR
5, let CR be the upper semi-cicle of |z| = R, then by the Residue Theorem, then
1
dx +
5 − 2x + x2
Z
CR
1
1
π
dz = 2πi ·
=
5 − 2z + z 2
(1 + 2i) − (1 − 2i)
2
1
dz, then
5 − 2z + z 2
Z
CR
1
dz
2
5 − 2z + z
Z
≤
CR
Z
≤
CR
1
1
|dz|
|5 − 2z + z 2 |
1
|dz|
|z|2 − 2|z| − 5
2
MINGFENG ZHAO
Z
=
CR
=
R2
→ 0,
1
|dz|
R2 − 2R − 5
πR
− 2R − 5
as R → ∞
So we know that
Z
∞
−∞
π
1
dx =
2
5 − 2x + x
2
SOLUTION OF TEST 3
2. [10 Points] Write the series for the function
3
1
around z1 , which is its pole in the upper
5 − 2x + x2
half plane. Also, find its radius of convergence.
Proof. By the proof of the Problem 2, then z1 = 1 + 2i and
1
5 − 2z + z 2
=
1
[z − (1 + 2i)][z − (1 − 2i)]
=
1
1
·
z − (1 + 2i) z − (1 − 2i)
=
1
1
·
z − (1 + 2i) z − (1 + 2i) + 4i
=
=
1
1
1
·
·
z − (1 + 2i) 4i 1 − z−(1+2i)
4i
k
∞
X
1
z − (1 + 2i)
1
−
·
4i z − (1 + 2i)
4i
If |z − (1 + 2i)| < 4
k=0
=
∞
X
(4i)−k−1 (−1)k [z − (1 + 2i)]k−1
k=0
The radius of convergence is 4.
4
MINGFENG ZHAO
3. [10 Points] Find the integral
∞
Z
−∞
cos(x)
dx
4 + x2
using the Residue Theorem. Show all steps.
Proof. Notice that eiz = cos(z) + i sin(z). Consider z 2 + 4 = 0, then z = 2i, that is, z 2 + 4 =
(z + 2i)(z − 2i).
For any R > 2, let CR be the upper semi-cicle of |z| = R, then by the Residue Theorem, then
Z
R
−R
Z
R
For
−R
eix
dx +
4 + x2
eiz
ei(2i)
π
dz = 2πi ·
= 2
2
4+z
2i + 2i
2e
Z
CR
eix
dx, then
4 + x2
Z
R
−R
eix
dx
4 + x2
Z
R
=
−R
Z
R
=
−R
Z
R
=
−R
Z
For
CR
eiz
dz, then
4 + z2
Z
eiz
dz
2
CR 4 + z
Z
≤
CR
Z
=
CR
cos(x) + i sin(x)
dx
4 + x2
Z R
cos(x)
sin(x)
dx
+
i
dx
2
2
4+x
−R 4 + x
cos(x)
dx
4 + x2
iz e
4 + z 2 |dz|
|eiz |
|dz|
|4 + z 2 |
SOLUTION OF TEST 3
5
|eiz |
|dz|
2
CR |z| − 4
Z π
1
iReiθ dReiθ e
2
R −4 0
Z π
R
−R sin(θ)+iR cos(θ) dθ
e
R2 − 4 0
Z π
R
e−R sin(θ) dθ
R2 − 4 0
Z
≤
=
=
=
Rπ
R2 − 4
≤
→ 0,
Since sin(θ) ≥ 0 for all θ ∈ [0, π]
as R → ∞
Hence we know that
Z
∞
−∞
π
cos(x)
dx = 2
2
4+x
2e
6
MINGFENG ZHAO
cos(x)
around z1 , which is its pole in the upper half
4 + x2
4. [10 Points] Write the series for the function
plane. Also, find its radius of convergence.
Proof. By the proof of the Problem 3, then z1 = 2i and
eiz
=
ei(z−2i)−2
=
e−2
∞
X
(z − 2i)k
k!
k=0
e−iz
=
e−i(z−2i)+2
=
e2
∞
X
(−i)k (z − 2i)k
k!
k=0
cos(z)
=
=
eiz + e−iz
2
∞
X
e−2 + e2 (−i)k
k=0
1
4 + z2
2k!
(z − 2i)k
=
1
(z − 2i)(z + 2i)
=
1
1
·
z − 2i z − 2i + 4i
=
=
1
1
1
·
·
4i z − 2i 1 + z−2i
4i
k
∞
X
1
1
z − 2i
·
·
−
4i z − 2i
4i
If |z − 2i| < 4
k=0
Hence we have
cos(z)
4 + z2
=
=
n−k
∞ X
n
X
1
1
e−2 + e2 (−i)k
z − 2i
·
·
· −
4i z − 2i n=0
2k!
4i
k=0
" n
#
∞
X
X e−2 + e2 (−i)k
1
1
·
·
(z − 2i)n
4i z − 2i n=0
2(−4i)n−k k!
k=0
The radius of convergence is 4.
SOLUTION OF TEST 3
7
5. [10 Points]Find the integral
Z
2π
0
1
dx
2 + cos(x)
using the Residue Theorem. Show all steps.
1
2z
= 2
. Consider z 2 + 4z + 1 = 0,
z + 4z + 1
z + z1
√
√
√
then z = −2 ± 3, that is, z 2 + 4z + 1 = [z − (−2 + 3)][z − (−2 − 3)].
Proof. Notice that cos(x) = eix + e−ix and
2+
1
2
By Residue Theorem, then
2
√
√
2πi ·
−2 + 3 − (−2 − 3)
Z
=
|z|=1
Z
=
|z|=1
Z
2
dz
z 2 + 4z + 1
1
·
z 2+
1
dz
z + z1
2π
e−ix ·
=
0
Z
=
1
2
i
0
2π
2+
1
2
1
deix
(eix + e−ix )
Let z = eix
1
dx
2 + cos(x)
Hence we have
Z
0
2π
1
dx
2 + cos(x)
=
=
2
2π · √
2 3
√
2π 3
3
8
MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu