MA2325: SOLUTIONS TO ASSIGNMENT 5
Compute the following integrals.
1. (a)
Z
|z−1|=4
1
dz
(z − 2)(z − 4)
|z−1|=1
1
dz
(z − 1)(z + 1)5
(b)
Z
Solution:
(a) The function f (z) =
1
(z−2)(z−4)
has isolated singularities at 2
and 4. The winding number of the circle |z − 1| = 4 about both
of these singularities is 1. Note that
1
z−4
has a local power series
representation centred at 2,
1
= a0 + a1 (z − 2) + a2 (z − 2)2 + · · ·
z−4
Hence the Laurent series representation for f (z) centred at 2 is
f (z) =
a0
+ a1 + a2 (z − 2) + · · ·
z−2
The residue of f (z) at 2 is
1
Res(f (z), 2) = a0 =
z−4
Similarly,
1
z−2
z=2
=−
1
2
has a local power series representation centred at
4,
1
= b0 + b1 (z − 4) + b2 (z − 4)2 + · · ·
z−2
1
2
MA2325: SOLUTIONS TO ASSIGNMENT 5
Hence the Laurent series representation for f (z) centred at 4 is
f (z) =
b0
+ b1 + b2 (z − 4) + · · ·
z−4
The residue of f (z) at 4 is
1
Res(f (z), 4) = b0 =
z−2
=
z=4
1
2
By the Residue Theorem,
Z
|z−1|=4
1
1 1
dz = 2πi − +
=0
(z − 2)(z − 4)
2 2
(b) The function f (z) =
1
(z−1)(z+1)5
has isolated singularities at 1
and −1. The winding number of the circle |z − 1| = 1 about 1
is 1. The winding number of the circle about −1 is 0. We only
need to compute the residue of f (z) at 1. Note that
1
(z+1)5
has
a local power series representation centred at 1,
1
= a0 + a1 (z − 1) + a2 (z − 1)2 + · · ·
(z + 1)5
Hence the Laurent series representation for f (z) centred at 1 is
f (z) =
a0
+ a1 + a2 (z − 1) + · · ·
z−1
The residue of f (z) at 1 is
1
Res(f (z), 1) = a0 =
(z + 1)5
=
1
32
=
z=1
By the Residue Theorem,
Z
|z−1|=1
1
dz = 2πi
(z − 1)(z + 1)5
1
32
πi
16
MA2325: SOLUTIONS TO ASSIGNMENT 5
3
2. (a)
Z
|z|=2
z4
1
dz
− z2
(b)
Z
γ
z2
1
dz
+1
where γ is composed of the semicircular arc C : [0, 1] → C,
t 7→ 2eπit followed by the line segment [−2; 2].
Solution:
(a) The function f (z) =
1
z 4 −z 2
=
1
z 2 (z+1)(z−1)
has isolated singulari-
ties at 0, 1 and −1. The winding number of the circle |z| = 2
about each of these singularities is 1. Note that
1
(z+1)(z−1)
has a
local power series representation centred at 0,
1
= a0 + a1 z + a2 z 2 + a3 z 3 + · · ·
(z + 1)(z − 1)
Hence the Laurent series representation for f (z) centred at 0 is
f (z) =
a0 a1
+
+ a2 + a3 z + · · ·
z2
z
The residue of f (z) at 0 is
d
Res(f (z), 0) = a1 =
dz
1
(z + 1)(z − 1)
z=0
The residues of f (z) at the other singularities are
Res(f (z), 1) =
1
2
Res(f (z), −1) = −
1
2
By the Residue Theorem,
Z
|z|=2
1
1 1
dz = 2πi 0 + −
=0
z4 − z2
2 2
=0
4
MA2325: SOLUTIONS TO ASSIGNMENT 5
1
z 2 +1
(b) The function f (z) =
=
1
(z+i)(z−i)
has isolated singularities
at i and −i. The winding number of the path γ about i is 1
and the winding number of γ about −i is 0. We only need to
compute the residue of f (z) at i.
Res(f (z), i) =
1
2i
By the Residue Theorem,
Z
γ
1
dz = 2πi
2
z +1
1
2i
=π
3. (a)
Z
(z 2
|z|=4
z
dz
+ 4)(z − 6)
(b)
Z
γ
z2
dz
(z 2 + 4)2
where γ is composed of the semicircular arc C : [0, 1] → C,
t 7→ 4eπit followed by the line segment [−4; 4].
Solution:
(a) The function f (z) =
z
(z 2 +4)(z−6)
=
z
(z−2i)(z+2i)(z−6)
has isolated
singularities at 2i, −2i and 6. The winding numbers of the circle
|z| = 4 about 2i and −2i are both 1. The winding number of
the circle about 6 is 0. We only need to compute the residue of
f (z) at 2i and −2i.
Res(f (z), 2i) =
Res(f (z), −2i) =
2i
1
12 + 4i
=−
=−
4i(2i − 6)
12 − 4i
160
2i
1
12 − 4i
=−
=−
4i(−2i − 6)
12 + 4i
160
MA2325: SOLUTIONS TO ASSIGNMENT 5
5
By the Residue Theorem,
Z
|z|=4
z
12 + 4i 12 − 4i
3πi
dz
=
2πi
−
−
=
−
(z 2 + 4)(z − 6)
160
160
10
z2
(z 2 +4)2
(b) The function f (z) =
=
z2
(z−2i)2 (z+2i)2
has isolated singu-
larities at 2i and −2i. The winding number of the path γ about
2i is 1 and the winding number about −2i is 0. We only need
to compute the residue of f (z) at 2i.
Res(f (z), 2i) = −
i
8
By the Residue Theorem,
z2
i
π
dz = 2πi −
=
2
2
(z + 4)
8
4
Z
γ
4. (a)
Z
1
z e z dz
|z|=1
(b)
Z
|z|=2
eiz
dz
z + z3
Solution:
1
(a) The function f (z) = z e z has an isolated singularity at 0. The
winding number of the circle |z| = 1 about this singularity is 1.
Recall that for all z ∈ C,
ez = 1 + z +
1 2 1 3
z + z + ···
2!
3!
Hence for all z 6= 0,
1
ez = 1 +
1
1 1
1 1
+
+
+ ···
2
z 2! z
3! z 3
6
MA2325: SOLUTIONS TO ASSIGNMENT 5
and so the Laurent series for f (z) at 0 is
f (z) = z + 1 +
11
1 1
+
+ ···
2! z 3! z 2
The residue of f (z) at 0 is
Res(f (z), 0) =
1
2
By the Residue Theorem,
Z
1
z e dz = 2πi
= πi
2
1
z
|z|=1
(b) The function f (z) =
eiz
z+z 3
=
eiz
z(z−i)(z+i)
has isolated singularities
at 0, i and −i. The winding number of the circle |z| = 2 about
each of these singularities is 1. The residues of f (z) at these
points are
Res(f (z), 0) = 1
1
2e
e
Res(f (z), −i) = −
2
Res(f (z), i) = −
By the Residue Theorem,
Z
|z|=2
eiz
1
e
dz = 2πi 1 −
−
z + z3
2e 2