MA2325: SOLUTIONS TO ASSIGNMENT 4
1. Sketch the C 1 paths γ : [0, 1] → C, t 7→ t + it2 and [0; 1 + i]. Then
compute the following integrals.
R
(a) γ Re(z) dz
R
(b) [0; 1+i] Re(z) dz
R
(c) γ z dz
R
(d) [0; 1+i] z dz
Solution:
(a) and (b). Note that Re(z) is not complex differentiable at any
point in C. The integral of Re(z) along a path joining two points will
in general depend on the choice of path. Here the path γ describes
part of a parabola joining 0 to 1 + i while [0; 1 + i] denotes the C 1
path
γ̃ : [0, 1] → C,
t 7→ (1 − t)0 + t(1 + i)
which describes the line segment joining 0 to 1 + i. We have
Z
Re(z) dz =
Z
=
Z
γ
1
Re(γ(t)) γ 0 (t) dt
0
1
t(1 + 2ti) dt
0
=
Z
1
t dt + i
0
=
1 2
+ i
2 3
1
Z
0
1
2t2 dt
2
MA2325: SOLUTIONS TO ASSIGNMENT 4
and
Z
Re(z) dz =
Z
=
Z
[0; 1+i]
1
Re(γ̃(t)) γ̃ 0 (t) dt
0
1
t(1 + i) dt
0
=
1
Z
t dt + i
0
=
1
Z
t dt
0
1 1
+ i
2 2
(c) and (d). Note that z is complex differentiable at every point
in C. Hence the integral of z along a piecewise C 1 path joining two
points is independent of the choice of path. Since F (z) =
z2
2
primitive for z we have
Z
z dz =
Z
γ
z dz = F (1 + i) − F (0) =
[0; 1+i]
(1 + i)2
−0=i
2
Or computing directly we have
Z
z dz =
Z
=
Z
=
Z
γ
1
γ(t) γ 0 (t) dt
0
1
(t + it2 )(1 + 2ti) dt
0
0
= i
1
3
t − 2t dt + i
Z
0
1
3t2 dt
is a
MA2325: SOLUTIONS TO ASSIGNMENT 4
3
and
Z
z dz =
Z
=
Z
[0; 1+i]
1
γ̃(t) γ̃ 0 (t) dt
0
1
(t + it)(1 + i) dt
0
Z
=
1
0 dt + i
Z
0
1
2t dt
0
= i
2. Let γ : [0, 2] → C be the piecewise C 1 path

 2eπit
t 7→
 4t − 6
for 0 ≤ t < 1
for 1 ≤ t ≤ 2
(a) Sketch the path γ.
(b) Compute the length of γ.
(c) What are the winding numbers for γ about the points i and
−i?
Solution:
(a) γ describes a semi-circular closed path in the upper half-plane
traversed once anticlockwise.
(b) The length of the arc is 2π and the length of the line segment
is 4. Or computing directly we have
Length(γ) =
Z
=
Z
2
|γ 0 (t)| dt
0
1
πit
|2πie | dt +
0
=
Z
Z
1
1
2π dt +
0
= 2π + 4
Z
1
2
4 dt
2
|4| dt
4
MA2325: SOLUTIONS TO ASSIGNMENT 4
(c) We answer this by inspecting the path. The winding number for
γ about the point i is 1 since γ winds around i exactly once and
in an anticlockwise direction. The winding number for γ about
the point −i is 0 since −i lies in the unbounded component of
C\γ.
3. Let z0 ∈ C. Prove that for k ∈ Z and any r > 0

 2πi
k
(z − z0 ) dz =
 0
|z−z0 |=r
Z
if k = −1
if k 6= −1
Solution: Here |z − z0 | = r denotes the C 1 path
t 7→ z0 + reit
γ : [0, 2π] → C,
which describes the circle with centre z0 and radius r. If k = −1 we
have
Z
−1
(z − z0 )
dz =
Z
=
Z
=
Z
|z−z0 |=r
2π
1
γ 0 (t) dt
γ(t) − z0
2π
1
rieit dt
reit
0
0
2π
i dt
0
= 2πi
If k 6= −1 then F (z) =
Z
(z−z0 )k+1
k+1
is a primitive for (z − z0 )k . Hence
(z − z0 )k dz = F (γ(2π)) − F (γ(0))
|z−z0 |=r
= 0
4. Let T be the triangular path determined by the points 0, 1 + i and
i. Compute the following integrals.
MA2325: SOLUTIONS TO ASSIGNMENT 4
(a)
R
T
(z 7 + 3z 4 − z + 2) dz
(b)
R
T
z̄ dz
5
Solution:
R
(a) T z 7 dz Note that the polynomial z 7 + 3z 4 − z + 2 is analytic
in C. Hence by Cauchy’s theorem for a triangle
Z
(z 7 + 3z 4 − z + 2) dz = 0
T
(b) Note that z̄ is not analytic on any open set containing T and
so we cannot apply Cauchy’s theorem as in (a). Instead we
compute directly
Z
z̄ dz =
T
Z
z̄ dz +
[0; 1+i]
Z
z̄ dz +
[1+i;i]
Z
z̄ dz
[i; 0]
[0; 1 + i] denotes the line segment [0, 1] → C, t 7→ t(1 + i).
Z
z̄ dz =
Z
[0; 1+i]
1
t(1 + i) (1 + i) dt
0
=
Z
1
(t − it) (1 + i) dt
0
=
Z
1
2t dt
0
= 1
[1+i; i] denotes the line segment [0, 1] → C, t 7→ (1−t)(1+i)+ti.
Z
z̄ dz =
[1+i; i]
Z
1
(1 − t)(1 + i) + ti (−1) dt
0
=
Z
1
(1 − t − i) (−1) dt
0
1
= − +i
2
6
MA2325: SOLUTIONS TO ASSIGNMENT 4
[i; 0] denotes the line segment [0, 1] → C, t 7→ (1 − t)i.
Z
z̄ dz =
[i; 0]
1
Z
(1 − t)i (−i) dt
0
=
1
Z
(t − 1)i (−i) dt
0
= −
1
2
Combining these three integrals we have
Z
T
z̄ dz = 1 −
1
1
+i− =i
2
2