MAT2700 - EXAM, FALL 2009 (DEC 16, 4 HOURS) - FASIT
Problem 1.
a. Solve the equation
z 2 + iz + 2 = 0.
Express its two solutions z1 , z2 in polar form: zj = rj eiθj with θj ∈ (−π, π], j = 1, 2.
π
π
Answer. The two solutions are z1 = i = ei 2 , z2 = −2i = 2e−i 2 .
b. Compute the cube root of −2i, i.e., (−2i)1/3 .
π
π
2
Answer. (−2i)1/3 = |(−2i)|1/3 ei(− 2 +2πk)/3 = 21/3 ei(− 6 + 3 πk) , k = 0, 1, 2. Hence

1/3 −i π

2 e 6 , k = 0,
π
(−2i)1/3 = 21/3 ei 2 ,
k = 1,

21/3 ei 7π6 , k = 2.
c. Determine the Laurent series for the function
1
2
z + iz + 2
in the annulus 1 < |z| < 2.
Answer. By a) and partial fractions,
z2
1
i
i
1
=
=
−
.
+ iz + 2
(z + i)(z + 2i)
3(z + 2i) 3(z − i)
First, for |z| < 2,
i
i 1
1
1 1
=
,
=
z
3(z + 2i)
3 2i 1 − (− 2i )
61−c
c=−
z
iz
= .
2i
2
Since |c| = | z2 | < 1, we can apply the results on geometric series:
j
∞
X
i
1 i
=
zj .
3(z + 2i)
6
2
j=0
Next, for |z| > 1,
i
i 1
=
,
3(z − i)
3z 1 − zi
1
2
MAT2700 - EXAM, FALL 2009 (DEC 16, 4 HOURS) - FASIT
and as before, since |i/z| < 1, the results on geometric series apply:
∞
∞
∞
i X j −j X 1 j+1 −(j+1) X 1 k −k
i
=
iz =
i z
=
i z .
3(z − i)
3z j=0
3
3
j=0
k=1
Hence, the Laurent series is
∞
X1
1
=
z 2 + iz + 2
6
j=0
j
∞
X
i
1 j −j
j
iz .
z −
2
3
j=1
Problem 2. Consider the contour integral
Z
1
I :=
dz,
3
2
C2 z − z
where C2 is the set {z ∈ C : |z| = 2} traversed in the counter-clockwise direction.
a. Draw a figure of the contour C2 and signify the poles of the integrand. Compute
the integral I without using the residue theorem (justify all steps
R in your calculations). [Hint: As a part of your computations, explain why C2 1/(z − 1) dz =
R
1/(z − 1) dz, with the circle |z − 1| = 1 being oriented in the counter|z−1|=1
clockwise direction.]
2
1
-2
1
-1
2
-1
-2
Figure 1. Contour and poles for Problem 2 a).
Answer. By partial fractions, we can write the integrand as
1
1
1
1
1
= 2
=− 2 − +
.
3
2
z −z
z (z − 1)
z
z z−1
We then compute three contour integrals.
First,
Z
1
dz = 0,
2
C2 z
since since 1/z 2 has an antiderivative (−1/z) in a domain containing C2 .
MAT2700 - EXAM, FALL 2009 (DEC 16, 4 HOURS) - FASIT
3
Next, parameterizing C2 by z(t) = 2eit for t ∈ [0, 2π], we compute
Z 2π
Z 2π
Z
1 0
1
i dt = 2πi.
dz =
z (t) dt =
z(t)
0
0
C2 z
Similarly,
Z
1
dz =
z−1
C2
Z
|z−1|=1
1
dz = 2πi.
z−1
where the first equality comes as a consequence of the Deformation Invariance
Theorem.
Summarizing,
I = −0 − 2πi + 2πi = 0.
b. For each R > 2, set
Z
I(R) :=
CR
1
dz,
z3 − z2
where CR denotes the set {z ∈ C : |z| = R} traversed in the counter-clockwise
direction. Explain why we have
I = I(R),
for any R > 2.
Answer. The singularities of the integrand are inside C2 , and thus the integrand
is analytic in a suitable domain D containing the contours C2 and CR . The contour
C2 can be continuously deformed in D into CR . The result then follows from the
Deformation Invariance Theorem.
c. Prove that
|I(R)| ≤
2π
,
−R
R2
for R > 2.
Answer. For z ∈ CR , by triangle inequality
|z 3 − z 2 | ≥ |z|3 − |z|2 = R3 − R2 ,
and hence
|z 3
The result now follows:
Z
Z
1
≤
dz
z3 − z2 CR
CR
|z 3
1
1
≤ 3
.
2
−z |
R − R2
1
length(CR )
2πR
2π
dz ≤
≤ 3
= 2
2
3
2
2
−z |
R −R
R −R
R −R
d. Compute I by utilizing b) and c) [hint: let R → ∞].
4
MAT2700 - EXAM, FALL 2009 (DEC 16, 4 HOURS) - FASIT
Answer. According to c),
|I(R)| ≤
2π
→ 0 as R → ∞,
R2 − R
so, by b),
I = lim I(R) = 0.
R→∞
e. Use the residue theorem to compute I.
1
1
Answer. The integrand f (z) = z3 −z
2 = z 2 (z−1) has singularities at z = 0 (pole of
order 2) and z = 1 (simple pole), elsewhere it is analytic. We compute
d 2
d
1
1
Res(0) = lim (z f (z) = lim
= − lim
= −1
z→0 dz
z→0 dz
z→0 (z − 1)2
z−1
and
1
Res(1) = lim(z − 1)f (z) = lim 2 = 1.
z→1
z→1 z
Consequently,
I = 2πi (Res(0) + Res(1)) = 2πi(−1 + 1) = 0.
Problem 3. The Laplace equation is a second-order partial differential equation
of the form
∂ 2φ ∂ 2φ
+
= 0.
(1)
∂x2 ∂y 2
A real-valued function φ = φ(x, y) is said to be harmonic in a domain D ⊂ R2 if φ
2
2
2
2
is twice continuously differentiable (i.e., ∂φ
, ∂φ , ∂ φ , ∂ φ , ∂ φ , ∂ φ exist and are
∂x ∂y ∂x∂y ∂y∂x ∂x2 ∂y 2
continuous in D) and satisfies the Laplace equation (1) at each point (x, y) in D.
a. Suppose f (z) = u(x, y) + i v(x, y) is analytic in a domain D. Show that u(x, y)
and v(x, y) are harmonic in D. [Hint: Differentiate the Cauchy-Riemann equations.]
Answer. Since f is analytic, all partial derivatives of u and v up to order 2 exist
and are continuous in D, so it remains to show that the Laplace equation holds.
Since f is analytic, the Cauchy-Riemann equations demand that
∂v
∂u
∂v
∂u
=
,
=− .
∂x
∂y
∂y
∂x
Taking partial derivatives with respect to x yields
∂ 2u
∂ 2v
∂ 2u
∂ 2v
=
,
=
−
.
∂x2
∂y∂x
∂y∂x
∂x2
Similarly, taking partial derivatives with respect to y yields
∂ 2u
∂ 2v
∂ 2u
∂ 2v
= 2,
=
−
.
∂x∂y
∂y
∂y 2
∂x∂y
MAT2700 - EXAM, FALL 2009 (DEC 16, 4 HOURS) - FASIT
5
Since all the partial derivatives are continuous, it is a calculus fact that
∂ 2v
∂ 2v
=
,
∂y∂x
∂x∂y
∂ 2u
∂ 2u
=
.
∂x∂y
∂y∂x
Hence, it follows that
∂ 2v
∂ 2u
∂ 2v
∂ 2u ∂ 2u
∂ 2u
=
=
−
=
=⇒
+
=0
∂x2
∂y∂x
∂x∂y
∂y 2
∂x2 ∂y 2
and
−
∂ 2u
∂ 2v
∂ 2v
∂ 2u
∂ 2v ∂ 2v
=
=
=
=⇒
+
= 0;
∂x2
∂y∂x
∂x∂y
∂y 2
∂x2 ∂y 2
Therefore, u and v are both harmonic functions in D.
b. Suppose φ(x, y) is a harmonic function in a simply connected domain D. Set
g(z) :=
∂φ
∂φ
−i .
∂x
∂y
Show that g is analytic.
Answer. Let us show that the Cauchy-Riemann equations are satisfied for g.
Indeed, since φ is harmonic,
∂ ∂φ
∂
∂φ
=
−
.
∂x ∂x
∂y
∂y
Moreover, since
∂2φ
∂y∂x
=
∂2φ
,
∂x∂y
∂
∂y
∂φ
∂x
∂
=−
∂x
∂φ
−
∂y
.
(all the displayed second order partial derivatives exist and are continuous since φ
is harmonic).
c. Let g(z) be the function defined in b). Since g is analytic in the simply connected
domain D we know that there exists an antiderivative G(z) = u(x, y) + i v(x, y)
for some functions u and v, G0 = g in D. Define the function
f (z) := G(z) + c,
where c is a constant.
Show that c can be chosen so that the real part of f is φ: Re(f ) = φ.
6
MAT2700 - EXAM, FALL 2009 (DEC 16, 4 HOURS) - FASIT
Answer. We have
∂v
∂u
∂u
∂u
+i
=
−i ,
∂x
∂x
∂x
∂y
where the last equality comes from the Cauchy-Riemann equations. At the same
time we also have
∂φ
∂φ
G0 (z) = g(z) =
−i ,
∂x
∂y
so necessarily
∂u
∂φ
∂u
∂φ
=
,
=
∂x
∂x
∂y
∂y
Hence,
u = φ + constant,
and choosing c = −constant it follows that the real part of
f 0 (z) = G0 (z) =
f = G + c = u + i v + c = φ + constant + i v + c = φ + i v
is indeed φ.
d. The Maximum Modulus Principle (MMP) says that a complex-valued function
f (z) that is analytic in a bounded domain D and continuous up to and including
its boundary attains its maximum modulus |f | on the boundary.
Use the MMP to show the maximum principle for solutions of the Laplace
equation, namely that a function φ(x, y) that is harmonic in a bounded simply
connected domain D and continuous up to and including the boundary attains its
maximum on the boundary [Hint: Apply MMP to the function |ef | with f from
c).]
Answer. With
f = φ + i v,
it follows that
f
e = eφ ei v = eφ .
According to the MMP, ef attains its maximum at the boundary and consequently the same must be true of eφ , and thus also of φ (since the real-valued
exponential function is monotone).