MATH 3007A Test 3 Solutions

[Marks]
MATH 3007A
Test 3 Solutions
November 4, 2011
Note: Unless stated otherwise, you may employ any theorem given in class in order to
evaluate integrals.
1. Let γ1(t) =2eπit [6]
, 0 ≤ t ≤ 1, and γ2(t) =2− 4t, 0 ≤ t ≤ 1. Is it necessarily true that
f(z) dz = f(z) dz? Justify your answer. (You need not evaluate these integrals.)
[3]
[3]
[3]
[4]
γ1
γ2
(a) f(z) = 1
(b) f(z) =
z − i
1
z +3
Solution:
(a) f(z) = 1
z − i is analytic on A = C \{i}. Sinceγ1is not homotopic to γ2 in A, the
integrals need not be equal.
(b) f(z) = 1
z +3 is analytic on A = C \{−3}. Since γ1 is homotopic to γ2 in A, the
integrals are equal by the deformation theorem.
e
2. Evaluate
z
dz, where the image of γ is the square with vertices at (1, 1), (−1, 1),
γ
z 2 − 4
(−1, −1) and (1, −1), traversed counterclockwise, once.
Solution:
f(z) = ez
z2 is analytic on A = C \{±2}. Sinceγis homotopic to a point in A, the
− 4
integral is 0 by Cauchy’s theorem (homotopy version).
3. Let f(z) = 1
z − 2 .
(a) Evaluate f(z) dz, whereγ1(t) =e2πit , 0 ≤ t ≤ 1.
γ1
Solution:
f is analytic on A = C \{2}. Sinceγis homotopic to a point in A, the integral
is 0 by Cauchy’s theorem (homotopy version).
(b) Evaluate f(z) dz, whereγ2(t) =2+e2πit , 0 ≤ t ≤ 1.
γ2
Solution:
f(z) dz =
γ2
4. Evaluate
γ
1
0
1
e 2πit 2πie2πit =2πi.
f(z) dz, wheref(z) =e z − sin(z) andγ(t) =2t + i sin(4πt), 0 ≤ t ≤ 1.
Solution:
f is continuous and has an analytic antiderivative F (z) =ez +cos(z) onA = C hence,
by the fundamental theorem,
f(z) dz = F (γ(1)) − F (γ(0)) = F (2) − F (0) = e 2 +cos(2)−2. γ

[4]
[4]
[3]
Alternatively, γ is homotopic to γ1, whereγ1(t) =2t, 0≤ t ≤ 1, hence, by the defor-
1
2t 2
mation theorem, f(z) dz = f(z) dz = e − sin(2t) · 2 dt = e +cos(2)−2. γ
γ1
5. Let C denote the straight line from (0, 0) to (1, 1), followed by the straight line from
(1, 1) to (1, 0).
(a) Parametrize C, i.e., determine the curve γ with image C.
Solution:
γ = γ1 ∪ γ2, whereγ1(t) =t + it, 0≤ t ≤ 1, and γ2(t) =1+i(1 − t), 0 ≤ t ≤ 1.
(b) Evaluate zdz.
γ
γ
Solution:
1
1
zdz= zdz+ zdz= (t + it)(1 + i) dt + [1 + i(1 − t)](−i) dt
γ
γ1
γ2
0
0
1 1
1
= 2it dt + 1 − t − idt= i + − i =
0
0
2 1
2 .
Alternatively, f is analytic on A = C,andγis homotopic to γ3
on A, whereγ3(t)
=
1
t, 0≤ t ≤ 1, hence, by the deformation theorem, zdz= zdz= tdt=
γ
γ3
0
1
2 .
Alternatively, f(z) =z is continuous and has an analytic antiderivative F (z) = 1
2 z2
on A = C, hence, by the fundamental theorem, zdz= F (γ(1)) − F (γ(0)) =
γ
1
2 .
1
6. Evaluate dz, where the image of γ is the square with vertices at (0, 0), (1, 1),
γ z − 1
(2, 0) and (1, −1), traversed clockwise, once.
Solution:
f is analytic on A = C \{1}, andγis homotopic to γ1 on A, whereγ1(t) =1+e−2πit ,
0
≤ t ≤ 1, hence,
by the deformation
theorem,
1
1
1
1
dz = dz =
z − 1 z − 1 e−2πit(−2πie−2πit ) dt = −2πi.
γ1
0
0
2