Problem 1
Let u(x, y) = Re f (x, y) and v(x, y) = Im f (x, y). Then
2
u(x, y) = ex
−y 2
v(x, y) = ex
cos(2xy),
2
−y 2
sin(2xy),
and
2
2
∂u
∂v
= ex −y (2x cos(2xy) − 2y sin(2xy)) =
,
∂x
∂y
2
2
∂u
∂v
= ex −y (−2y cos(2xy) − 2x sin(2xy)) = − .
∂y
∂x
Hence f (z) satisfies the Caychy-Riemann equations for all z ∈ C, so f (z) is entire.
Problem 2
We calculate
∂u
= 3x2 − 3y 2 − 2x
∂x
∂u
= −6xy + 2y
∂y
so
v(x, y) = 3yx2 − y 3 − 2xy
is an harmonic conjugate. (The others only differ by a constant.)
Problem 3
i. Domain of analyticity is C \ {0, −1} while f 0 (z) =
−3z 2 +4z+2
(z 2 +z)2 .
ii. Domain of analyticity is C \ {iπ(k + 1/2) : k ∈ Z}, and f 0 (z) = 1/ cosh2 z.
Problem 4
Use the branch Log(15z 2 − 227z + 23) with domain D∗ = {z : |z| < 1/10}.
Then the principal branch of the logarithm is defined on the domain. Here f 0 (z) =
30z−227
15z 2 −227z+23 .
Problem 5
i. Write (z 2 + z)2/7 = exp 27 log(z 2 + z). For z = −1/2, z 2 + z = −1/4, so we
can use a branch of the logarithm which is defined for z 6∈ R+ to define f (z) =
exp 72 (Log(−(z 2 + z)) + πi) with domain D∗ = {z : |z + 12 | < 21 }.
ii. As before (z 2 +z)2/7 = exp 27 log(z 2 +z). Note that z 2 +z does not take negative
real values unless x ≤ 0. Hence we can use the principal branch of the logarithm
to define f (z) = exp 72 Log(z 2 + z) with domain D∗ = {z : |z − 12 | < 21 }.
In both cases f 0 (z) = 27 (z 2 + z)2/7 (2z + 1)(z 2 + z)−1 , with the branch used in
f (z) being the one from the respective problem.
0
1
2
Problem 6
Problem 7
We can write
Z
sin2 zdz =
Γ
Z
0
sin2 z dz +
−π
Z
sin2 z dz,
Γ0
where Γ0 is the circle part of the contour. Since sin2 z is entire, the second integral
vanishes, while standard calculus results give that the first integral equals π/2.
Problem 8
Using the branch F (z) = Lπ/2 (z − i) as defined on p. 121 in the book, we see
that F (z) is an antiderivative of (z − i)−1 defined on D∗ = C \ i[1, ∞). Hence by
theorem 7 p. 176 we get
Z
1
dz = F (1 + i/2) − F (−1 + i/2)
Γ z−i
= Log |1 + 3i/2| + i Argπ/2 (1 − i/2) − Log | − 1 − i/2| − i Argπ/2 (−1 + 3i/2)
= iπ − 2i arctan 1/2.
Problem 9
Using the branch F (z) = Lπ/2 (z + i) as defined on p. 121 in the book, we see
that F (z) is an antiderivative of (z + i)−1 defined on D∗ = C \ i[−1, ∞). Hence by
theorem 7 p. 176 we get
Z
1
dz = F (1 + i/2) − F (−1 + i/2)
Γ z−i
= Log |1 + 3i/2| + i Argπ/2 (1 + 3i/2) − Log | − 1 + 3i/2| − i Argπ/2 (−1 + 3i/2)
= iπ + 2i arctan 3/2.
Problem 10
Decomposing into partial fractions, we get
Z
Z
Z
i
2−i
2z − 1
dz =
dz +
dz,
2 + iz
z
z
z
+i
Γ
Γ
Γ
where the first integral vanishes since it is analytic on and inside the contour Γ,
while for the second integral we have
Z
2−i
dz = 2πi(2 − i)
Γ z+i
so our integral equals 2π(1 + 2i).