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Solutions of Problems of
Math 113 Final Examination
May 20, 2009, 9:15 a.m. - 12:15 p.m.
Science Center A
Total Number of Points = 200
Problem 1. Evaluate the following two definite integrals by using the theory
of residues.
(a) (15 points)
∫ (^2) π
θ=
dθ
(5 + 4 cos θ)
2
(b) (15 points)
∞
x=−∞
x sin(2x)
x
2
dx.
Solution. (a) Rewrite the integral as
∫ (^2) π
θ=
dθ
(5 + 4 cos θ)
2 =^
∫ (^2) π
θ=
dθ ( 5 4
This is the same as Exercise #7 from Stein & Shakarchi, p.104, asking for
the computation of
∫ (^2) π
0
dθ
(a + cos θ)
2
2 πa
(a^2 − 1)
3 2
with a > 1 when a =
5 4
. To evaluate the integral (†) with a general a > 1,
we use the parametrization a = e
iθ for 0 ≤ θ ≤ 2 π so that
cos θ =
z +
z
, dθ =
dz
iz
Hence
∫ (^2) π
θ=
dθ
(a + cos θ)
2
= −i
|z|=
dz
z
a +
1 2
z +
1 z
= −i
|z|=
4 zdz
(2az + z
2
2
= −i
|z|=
4 zdz ( (z + a)
2 − (a^2 − 1)
= − 4 i
|z|=
zdz ( z + a +
a^2 − 1
z + a −
a^2 − 1
For the meromorphic function
f (z) =
z ( z + a +
a^2 − 1
z + a −
a^2 − 1
the only poles are at −a −
a
2 − 1 and −a +
a
2 − 1, both double poles.
Since a > 1, clearly the pole −a −
a^2 − 1 is outside the unit circle. The
point −a +
a
2 − 1 has absolute value < 1, because we can look at the right-
angled triangle with two sides of length 1 and
a^2 − 1 and with hypothenuse
a. The difference of the length of the hypothenuse a and the length of one
side abutting the right angle is less than the length of the other side which
is 1. The residue of f at the double pole −a +
a
2 − 1 is computed by
res−a+
√ a^2 − 1 f^ =^
d
dz
z + a −
a
2 − 1
f (z)
z=−a+
√ a^2 − 1
d
dz
z ( z + a +
a
2 − 1
z=−a+
√ a^2 − 1
−z + a +
a^2 − 1 ( z + a +
a
2 − 1
z=−a+
√ a^2 − 1
a
4 (a^2 − 1)
3 2
Thus
∫ (^2) π
0
dθ
(a + cos θ)
2
= (− 4 i)2πi
res−a+√a (^2) − 1 f
2 πa
(a
2 − 1)
3 2
With a =
5 4
we get
∫ (^2) π
θ=
dθ
(5 + 4 cos θ)
2
∫ (^2) π
θ=
dθ ( 5 4
) 2 =^
2 π
5 4 ( ( 5 4
2
10 π
and (^) ∣ ∣ ∣ ∣
CR
e
iz a
2 − z
2
i (z
2
2 )
2
dz
(a
2
2 ) πR
(R
2 − a
2 )
2
for R > a.
The residue resai f of f at the simple pole ai of f is given by
resai f = lim z→ai
(z − ai)
ze
iz
z
2
2
ze
iz
z + ai
z=ai
e
−a
By applying the theory of residues to the integration of f (z)dz over the
boundary of the upper half-disk of radius R centered at the origin and letting
R → ∞, we get
x=−∞
x sin xdx
x
2
2
= Im
∞
x=−∞
xe
ix dx
x
2
2
= Im
R
f (z)dz
= Im (2πi resai f )
= Im
2 πi
e
−a
= π e
−a .
With a = 4 we get (^) ∫ ∞
x=−∞
x sin x
x
2
dx =
π
e
4
Problem 2. Evaluate the following two definite integrals by using the theory
of residues and branch-cuts.
(a) (16 points)
x=
log x
x^2 + 4
dx.
(b) (16 points)
x=
x
α
x^2 + 9
dx for − 1 < α < 1.
Solution. (a) This is the same as Exercise #10 from Stein & Shakarchi,
p.104, asking for the computation of
0
log x
x^2 + a^2
dx
for a > 0 when a = 2. To evaluate (‡), we choose a branch of log z =
log |z| + i arg z by specifying the range −
π 2
< arg z <
3 2
and choose
f (z) =
log z
z
2
2
The contour is the boundary, in the counter-clockwise sense, of the set which
is equal to the upper half-disk of radius R > 0 minus the upper disk of radius
ε with 0 < ε < R. The integral over the upper half-circle CR of radius R
centered at the origin approaches 0 as R → ∞, because
∣ ∣ ∣ ∣
CR
f (z)dz
log R + π
R
2 − a
2
πR → 0 as R → ∞.
The integral over the upper half-circle Cε of radius ε centered at the origin
approaches 0 as r → 0, because
∣ ∣ ∣ ∣
Cε
f (z)dz
− log ε + π
a
2 − ε
2
πε → 0 as ε → 0.
On the upper half-plane there is only one pole for f (z) which is a simple pole
at ai. The residue of f (z) at ai is computed by
resaif = lim z→ai
(z − ai)
log z
z
2
2
log z
z + ai
z=ai
log a +
π 2
i
2 ai
The theory of residues yields
∞
x=
log x dx
x^2 + a^2
0
x=−∞
(log(−x) + πi) dx
x^2 + a^2
= 2πi resaif =
π
log a +
π 2
i
a
Taking the real part of both sides, we get
∞
x=
log x dx
x
2
2
0
x=−∞
log(−x) dx
x
2
2
π log a
a
Changing the sign of the variable x from x to −x in the second integral on
the left-hand side, we get
x=
log x dx
x
2
2
π log a
a
By the residue theorem over the contour described above,
∫ R
r
x
α
(1 + x
2 )
2
dx +
CR
f (z)dz −
Cr
f (z)dz −
∫ R
r
e
iα 2 π x
α
1 + x^2
dx
= 2πi (Resz=if (z) + Resz=−if (z))
= 2πi
e
απi 2
2 i
e
3 απi 2
− 2 i
= π e
απi 2
1 − e
απi
The reason for the last integral on the left-hand side of the equation is that
the value of z
α is x
α e
iα 2 π at the real point x for that integrand, because for
that integrand at the real point x the angle θ for its polar representation is
2 π. As R → ∞,
CR
f (z)dz
≤ sup z∈CR
|f (z)| · (length of CR) ≤
R
α
1 + R^2
· 2 πR
approaches 0, because α < 1. As r → 0,
∣ ∣ ∣ ∣
Cr
f (z)dz
≤ sup z∈Cr
|f (z)| · (length of Cr) ≤
r
α
1 + r^2
· 2 πr
approaches 0, because α > −1. Hence
1 − e
iα 2 π
0
x
α
1 + x
2
dx = π e
απi 2
1 − e
απi
and
∞
0
x
α
1 + x
2
dx =
π e
απi (^2) (1 − eαπi)
1 − e
iα 2 π
π e
απi 2
1 − eiαπ^
π
e
−απi (^2) − e
απi 2
π 2
cos
απ 2
The final answer is (^) ∫ ∞
0
x
α
1 + x^2
dx =
(1 − a)π
4 cos
aπ 2
The final answer is (^) ∫ ∞
x=
x
α
x^2 + 9
dx =
α− 1 π 2
cos
πα 2
Problem 3 (20 points). Let D be a connected open subset of C. A holo-
morphic function g(z) on D is said to be univalent if g (z 1 ) 6 = g (z 2 ) for any
z 1 6 = z 2 in D. Suppose fn(z) is a sequence of univalent holomorphic functions
on D such that fn(z) approaches some nonconstant function f (z) uniformly
on D as n → ∞. Prove that f (z) is a univalent holomorphic function by
using Rouch´e’s theorem to show that if f (z 1 ) = f (z 2 ) for some z 1 6 = z 2 in
D, then there exist some positive integer N and some 0 < ε < |z 1 − z 2 | such
that for n ≥ N there exists ζ 2 ∈ D with |ζ 2 − z 2 | < ε and fn (z 1 ) = fn (ζ 2 ).
Solution. Since f (z) is the uniform limit D of the sequence of holomorphic
functions fn(z) on D, it follows for any closed solid triangle T in D
∂T
f (z)dz = lim n→∞
∂T
fn(z)dz = 0
(where ∂T is the boundary of T ) and by Morera’s Theorem f (z) is holomor-
phic on D.
Assume that f (z) is not univalent on D so that f (z)1 = f (z 2 ) for some
z 1 6 = z 2 in D and we are going to derive a contradiction. Since f (z) is
nonconstant on D and D is connected, f (z) − f (z 1 ) is not identically zero
and there exists some 0 < ε < |z 1 − z 2 | which is less than the distance from
z 2 to the boundary of D such that the holomorphic function f (z) − f (z 1 )
is nowhere zero on |z − z 2 | = ε. Let η > 0 such that |f (z) − f (z 1 )| ≥ η on
|z − z 2 | = ε. There exists some positive integer N such that |f (z) − fn(z)| <
η 2
on |z − z 2 | = ε for n ≥ N and |f (z 1 ) − fn(z 1 )| <
η 2
for n ≥ N. Then
|(f (z) − f (z 1 )) − (fn(z) − fn(z 1 ))|
≤ |f (z) − fn(z)| + |f (z 1 ) − fn(z 1 )|
η
η
= η ≤ |f (z) − f (z 1 )|
on |z − z 2 | = ε for n ≥ N. By Rouch´e’s theorem, fn(z) − fn(z 1 ) and f (z) −
f (z 1 ) have the same number of solutions in |z − z 2 | < ε for n ≥ N. Since z 2
is a zero of f (z) − f (z 1 ) on |z − z 2 | < ε, we conclude that for n ≥ N there
exists some ζ 2 ∈ D with |ζ 2 − z 2 | < ε such that ζ 2 is a root of fn(z) − fn(z 1 ).
This contradicts the univalent property of fn(z) and we conclude that f (z)
must be univalent on D.
Problem 4. Let τ be a complex number with Im(τ ) > 0.
which implies that, for ξ < 0,
lim R→∞
z=R+ηi, 0 ≤η≤b
e
− 2 πiξz
(τ + z)
2
dz = 0,
lim R→∞
z=−R+ηi, 0 ≤η≤b
e
− 2 πiξz
(τ + z)
2
dz = 0
and
fˆ (ξ) =
x∈R
f (x)e
− 2 πiξx dx =
z=x+ib, −∞<x<∞
e
− 2 πiξz
(τ + z)
2
dz.
On the other hand,
z=x+ib, −∞<x<∞
e
− 2 πiξz
(τ + z)
2
dz
≤ e
2 πξ b
x=−∞
dx
(x + Re τ )
2
2
approaches 0 as b → ∞, because ξ < 0. Hence fˆ (ξ) = 0 for ξ < 0.
For the computation of fˆ (ξ) for ξ > 0 we introduce the closed rectangle
Γ
− R,b for^ R, b >^ Im^ τ^ whose four vertices are^ −R, R, R^ −^ bi,^ −R^ −^ bi. Since
e
− 2 πiξz
(τ + z)
2
is holomorphic at every point of Γ
− R,b
except at the point z = −τ which is a
double pole, it follows that
∂Γ
− R,b
e
− 2 πiξz
(τ + z)
2
dz = 2πi Resz=−τ
e
− 2 πiξz
(τ + z)
2
where ∂Γ
− R,b
is the boundary of Γ
− R,b
in the counter-clockwise sense. We have
e
− 2 πiξz
(τ + z)
2
e
2 πξ Im z
|τ + z|
2
which implies that, for ξ > 0,
lim R→∞
z=R−ηi, 0 ≤η≤b
e
− 2 πiξz
(τ + z)
2
dz = 0,
lim R→∞
z=−R−ηi, 0 ≤η≤b
e
− 2 πiξz
(τ + z)
2
dz = 0
and
fˆ (ξ) =
x∈R
f (x)e
− 2 πiξx dx =
z=x−ib, −∞<x<∞
e
− 2 πiξz
(τ + z)
2
dz + 2πi Resz=−τ
e
− 2 πiξz
(τ + z)
2
On the other hand,
∣ ∣ ∣ ∣ ∣
z=x−ib, −∞<x<∞
e
− 2 πiξz
(τ + z)
2 dz
≤ e
− 2 πξ b
x=−∞
dx
(x + Re τ )
2
2
approaches 0 as b → ∞, because ξ > 0. Hence
fˆ (ξ) = 2πi Res z=−τ
e
− 2 πiξz
(τ + z)
2
for ξ > 0. We now compute the residue
Resz=−τ
e
− 2 πiξz
(τ + z)
2
d
dz
e
− 2 πiξz
z=−τ
= 2πi ξ e
2 πiξτ
and conclude that
fˆ (ξ) = − 4 π^2 ξ e^2 πiξτ
for ξ > 0. This means that the constant A = −4.
(b) We expand
cot πτ = i
e
iπτ
−iπτ
eiπτ^ − e−iπτ^
= i
e
2 iπτ
e^2 iπτ^ − 1
= i
1 − e^2 iπτ
as a power series in e
2 iπτ to get
cot πτ = i
∞ ∑
n=
e
2 niπτ
We differentiate both sides with respect to τ to get
()
−π
sin
2 (πτ )
= 4π
∞ ∑
n=
n e
2 niπτ .
with α ∈ C and a ∈ C with |a| < 1. Since
1 − |ζ|
2 = 1 −
z − a
¯az − 1
2
|¯az − 1 |
2 − |z − a|
2
|az − 1 |^2
|a|
2 |z|
2 − ¯az − az¯ + 1 − (|z|
2 − a¯z − ¯az + |a|
2 )
|az − 1 |
2
(1 − |a|
2 )(1 − |z|
2 )
|az − 1 |
2
and
dζ
dz
= e
iα −(1^ − |a|
2 )
(¯az − 1)
2
it follows that
|dζ|
1 − |ζ|
2
|dz|
1 − |z|
2
This means that the expression
|dz|
1 − |z|
2
is invariant under any biholomorphic self-map of D. Take any point a ∈ D.
Besides the two complex variables z and w used in w = f (z), we are going
to introduce two more complex variables ζ and ω. Let
z = S(ζ) =
ζ + a
1 + ¯aζ
and
w = T (ω) =
ω + f (a)
1 + f (a)ω
Let ω = h(ζ) be defined by h = T
− 1 ◦ f ◦ S
− 1 so that f = T ◦ h ◦ S. Then
h : D → D is holomorphic and h(0) = 0. Since the quotient
|dz|
1 − |z|^2
is invariant under any biholomorphic self-map of D, it follows that
|dζ|
1 − |ζ|^2
|dz|
1 − |z|^2
and
|dζ|
1 − |ζ|^2
|dz|
1 − |z|^2
At z = a we have ζ = 0 and w = f (a) and ω = 0 and
|dζ|
ζ=
|dz|
1 − |z|
2
z=a
and
|dω|
ω=
|dw|
1 − |w|
2
w=f (a)
The inequality
|f
′ (z)|
1 − |f (z)|
2
1 − |z|
2
at z = a is equivalent to
|dw|
1 − |w|^2
w=f (a)
≤ |dζ|
ζ=
|dz|
1 − |z|^2
z=a
which in turn is equivalent to
|dζ|
ζ=
≤ |dω|
ω=
which in turn is equivalent to
|h
′ (0)| ≤ 1
which now follows from Schwarz’s lemma. If the inequality
|f
′ (z)|
1 − |f (z)|
2
1 − |z|
2
becomes an equality at z = a, then we have |h
′ (0)| = 1 and by the Schwarz
lemma h(z) = e
iβ z for some real number β and is a biholomorphic self-map
of D, which forces f to be the composite of three biholomorphic self-maps S,
h, T of D and be a biholomorphic self-map of D.
Step Four (6 points of the 32 points of the Problem). For any nonnegative
integer k, let ω̂kωk+1 be the open arc on the unit circle C from ωk to ωk+1 in
the counter-clockwise sense. Then
∣ ∣ ∣ ∣ ∣
ωk ̂ωk+
dζ
(1 − ζn)
2 n
is equal to (^) ∫ π
0
dθ
(sin θ)
2 n
times a positive constant which is independent of the nonnegative integer k.
Final Step (6 points of the 32 points of the Problem). Finish the proof after
the preceding four steps.
Solution. This is the same as Exercise #23 from Stein & Shakarchi, p.254.
Proof of Step One. Since z must be of the form e
(θ+^
2 `π n )
i for some 0 < θ <
2 π n
we can rewrite
arg
z
(z − ωk)
2
= arg
e
(θ+^
2 `π n )i
(
e
(θ+^
2 `π n )i^ − e
2 kπi n
= arg
e
(θ+^
2 `π n )i
e
(θ+^
2(`+k)π n )i
e
1 2 (θ+^
2(`−k)π n )i^ − e
− 1 2 (θ+^
2(`−k)π n )i
= arg
e
−
2 kπ n i (
e
1 2 (θ+^
2(`−k)π n )i^ − e
− 1 2 (θ+^
2(`−k)π n )i
= arg
e−^
2 kπ n i^
4 sin
2
1 2
θ +
2(`−k)π n
(^) = arg
−e
− 2 kπi n
which is independent of θ, because 0 < θ <
2 π n
and
sin
2
θ +
2(` − k)π
n
Proof of Step Two. The statement in Step Two is the same as the function
arg
(1 − zn)
2
dz
dθ
)n)
= arg
n∏− 1
j=
dz dθ
(z − ωj )
2
being constant for z in the open arc ω̂kωk+1 of the unit circle C from ωk to
ωk+1 in the counter-clockwise sense. Using z = e
iθ , we have
dz
dθ
= i e
iθ = i z
and the statement to be verified is the same as the function
arg
(1 − z
n )
2
dz
dθ
)n)
= arg
n− 1 ∏
j=
i z
(z − ωj )
2
n− 1 ∑
j=
arg
i z
(z − ωj )
2
being constant for z in the open arc ω̂kωk+1 of the unit circle C from from ωk
to ωk+1 in the counter-clockwise sense. Since we have already verified earlier
in Step One that each summand
arg
i z
(z − ωj )
2
is constant for z in the open arc ω̂kωk+1 of the unit circle C from from ωk to
ωk+1 in the counter-clockwise sense, it follows that for z = e
iθ the function
arg
(1 − z
n )
2 n
dz
dθ
is constant for z in the open arc ω̂kωk+1 of the unit circle C from ωk to ωk+
in the counter-clockwise sense.
Proof of Step Three. Let k be a nonnegative integer. Let P be a point on
the unit circle C which is very close to ωk such that to go from P to ωk
requires a very small counter-clockwise movement. Let Q be a point on the
unit circle C which is very close to ωk such that to go from ωk to Q requires
a very small counter-clockwise movement. Then the direction of the vector −→
ωk P (joining ωk to P ) rotates to the direction of the vector
−→
O ωk (joining
the origin O to ωk) by an angle
π 2
- εP for some very small positive number
which after a change of variables ζ 7 → ζ e
− 2 kπ ı becomes
ζ=eiθ , 0 <θ< (^2) nπ
dζ
(1 − ζ
n )
2 n
because
ζ e
− 2 kπ ı
)n
= ζ
n
. Thus the length of the image line-segment Lk is
given by
∫ 2 π n
θ=
dθ
(1 − e
inθ )
2 n
∫ 2 π n
θ=
dθ
(
e
nθ 2 i^ − e−^
nθ 2 i
n
∫ 2 π n
θ=
dθ
2 n (^) sin
2 n
nθ 2
which becomes 2 n
2
2 n
∫ (^) π
θ=
dθ
sin
2 n (^) θ
after we replace θ by
n 2
θ. Thus the total perimeter (which is equal to n times
the length of the image line-segment Lk) is equal to
1 − 2 n
∫ (^) π
θ=
dθ
sin
2 n (^) θ
and the number A is equal to 2
1 − (^2) n .
Final Step of the Solution of Problem 5. In the definition
F (z) =
∫ (^) z
1
dζ
(1 − ζ
n )
2 n
for the map F (z), for some real number α we can rewrite the integrand as
(]) F (z) = e
iα
∫ (^) z
1
dζ
∏n− 1
k=0 (ζ^ −^ ωk)^
2 n
Step Two tells us that as z goes along the open arc ω̂kωk+1 of the unit circle
C from ωk to ωk+1 in the counter-clockwise sense, the value F (z) stays in
a straight line segment Lk. From (]) Step Three tells us that as z along C
passes ωk in the counter-clockwise sense, its image F (z) turns an angle equal
to −
2 n
(−π) which is equal to
2 π n
. By Step Four the length of the line-segments
Lk is independent of k and the sum of such lengths over 0 ≤ k ≤ n − 1 is
equal to
1 − (^) n^2
∫ (^) π
θ=
dθ
sin
2 n (^) θ
Hence the image of the closed unit disk under z 7 → F (z) is a regular polygon
of n sides whose total parameter is equal to
1 − (^) n^2
∫ (^) π
θ=
dθ
sin
2 n (^) θ
Problem 7. (a) (10 points). Let z = sin w with z = x + iy and w = u + iv
and x, y, u, v ∈ R. Verify that
x
2
sin
2 u
y
2
cos^2 u
by using the addition formula for the sine function of a complex variable,
expressed in terms of the sine function, the cosine function, the hyperbolic
sine function, and the hyperbolic cosine function. The hyperbolic sine and
cosine functions are defined respectively by
sinh ξ =
e
ξ − e
−ξ
, cosh ξ =
e
ξ
−ξ
Hence show that
(x + 1)
2
2 −
(x − 1)
2
2 = 2 sin u
for 0 ≤ u ≤
π 2
by using the characterization of a hyperbola as the locus of a
point whose distances from two fixed points have a constant difference.
(b) (8 points). Let w = sin z. Use the mapping behavior of the map
w =
ζ +
ζ
from the polar coordinates for ζ ∈ C and the Cartesian coordinates for w ∈ C
and use ζ = e
iZ and z =
π 2
− Z to show that w = sin z maps the semi-infinite
strip
π
≤ Re z ≤
π
, Im z ≥ 0
bo the upper half-plane { Im w ≥ 0 }.
