Assignment 2 - Complex Analysis
MATH 440/508 – M.P. Lamoureux
Sketch of solutions
1.
Z
Z
z dz =
Γ
(x − iy)(dx + idy) =
Z
Z
Γ
Γ
Γ
= (1/2)(x2 + y 2 )|endpoints + i
= 0 − 2i
Z Z
(−ydx + xdy)
(xdx + ydy) + i
Z Z
[(∂/∂y) − y − (∂/∂x)x]dxdy
interiorΓ
dxdy = −2i · Area( interior of curve Γ.
interiorΓ
by Green’s theorem for the second term, exact differential on the first term
(evaluated at a common end point, so cancels to zero).
2. a) Does there exist a holomorphic function f on the unit disk with
1
−1
1
f ( ) = f ( ) = 2 , for n = 1, 2, 3, . . .?
n
n
n
Is it unique?
Yes, and Yes. f (z) = z 2 works, and any other function g(z) that works
would agree with f on the sequence zn = 1/n, which has a limit point in
the range of analyticity. Therefore f and g are identical.
b) Does there exist a holomorphic function f on the unit disk with
1
−1
1
f ( ) = f ( ) = 3 , for n = 1, 2, 3, . . .?
n
n
n
Is it unique?
No. On the first sequence zn = 1/n, the function f1 (z) = z 3 works,
while on the other sequence zn = 1/n, the function f2 (z) = −z 3 works. A
function f that satisfies both sequences would have to agree with both f1
and f2 on their domain, by the limit point argument as above. Which is
impossible, since one function is the negative of the other.
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3. a) |z|=1 z n sin(z) dz = 0 for n ≥ −1 since the integrand is holomorphic,
or has at most a removable singularity at z = 0. For other values, we use
the Cauchy integral formula for derivatives, noting that the k-th derivative
of sin(z) at z = 0 is ±1 or zero, depending on the value of k (modulo 4). So
R
n
|z|=1 z sin(z) dz = 0 when n is negative and odd, and
R
Z
z n sin(z) dz =
Z
|z|=1
|z|=1
sin(z)
−2πi
dz =
(−1)n/2 ,
z −n
(−n − 1)!
when n is negative and even.
3. b) |z|=2 z n (1−z)m dz = 0 when n, m ≥ 0, as the integrand is holomorphic.
It is also zero for n, m < 0 because we can replace the circle of radius 2 with
a circle of radius R > 2 (by the fact the integrand is holomorphic on the
region R > 1). Letting R goes to infinity, we see the integral grow no faster
than R/R|n|+|m| , which tends to zero. So the integral on the original curve
is zero.
The only interesting cases are when n ≥ 0, m < 0 or m ≥ 0, n < 0.
Either way, we have a polynomial divided by a power of z or 1 − z, in which
case we apply the Cauchy integral formula to a simple polynomial.
For n ≥ 0, m < 0,
R
Z
z n (1 − z)m dz = (−1)m
Z
|z|=2
|z|=2
zn
2πi(−1)m
dz
=
(d/dz)−m−1 [z n ]|z=1
(z − 1)−m
(−m − 1)!
!
=
n
2πi(−1)m
,
n(n − 1)(n − 2) · · · (n + m + 2) = 2πi(−1)m
−m − 1
(−m − 1)!
where the binomial coefficient is defined to be zero when the bottom term
is negative. (You might want to check that the signs are correct. Try
n = 0, m = −1 to see that you sometimes need negatives!)
Similarly, for n < 0, m ≥ 0,
Z
z n (1 − z)m dz =
|z|=2
Z
|z|=2
(z − 1)m
2πi
dz =
(d/dz)−n−1 [(1 − z)m ]|z=0
−n
z
(−n − 1)!
!
=
2πi(−1)−n−1
m
m(m − 1)(m − 2) · · · (m + n + 2) = 2πi(−1)n+1
,
(−n − 1)!
−n − 1
(You might want to check that the signs are correct. )
3. c) For |z|=ρ |z − z0 |−4 |dz|
|z0 | 6= ρ, note the integrand and |dz|
are all positive, so we expect a positive answer. Use the parameterization
R
2
z = ρeit , dz = iρeit so |dz| = ρdt = ρdz/iz. Then
Z
−4
|z|=ρ
|z − z0 |
|dz| =
=
=
=
1
1
ρdz
2
2
|z|=ρ (z − z0 ) (z − z0 ) iz
Z
z
ρ
1
dz
2
i |z|=ρ (z − z0 ) (zz − zz0 )2
Z
ρ
1
z
dz
i |z|=ρ (z − z0 )2 (ρ2 − zz0 )2
ρ
z
2πif 0 (z0 ), where f (x) = 2
i
(ρ − zz0 )2
Z
by the Cauchy integral formula for derivatives, with |z0 | < ρ. First derivatives of quotients are easy to compute, so
f 0 (z) =
1
2zz0
1
2|z0 |
ρ2 + |z0 |2
0
+
,
f
(z
)
=
+
=
.
0
(ρ2 − zz0 )2 (ρ2 − zz0 )3
(ρ2 − |z0 |)2 (ρ2 − |z0 |)3
(ρ2 − |z0 |)3
Thus
Z
|z|=ρ
|z − z0 |−4 |dz| = 2πρ
ρ2 + |z0 |2
(ρ2 − |z0 |)3
when |z0 | < ρ. A similar calculation shows that
Z
|z|=ρ
|z − z0 |−4 |dz| = 2πρ
ρ2 + |z0 |2
(|z0 | − ρ2 )3
when |z0 | > ρ. Notice both answers are positive, as expected.
4. a) Use the residue theorem to compute
Z ∞
−∞
1
dx.
1 + x4
1
The function f (z) = 1+z
4 has simple poles at the four points zk =
√
(±1 ± i)/ 2 and the residues are computed using L’Hopital’s rule:
Reszk = lim
z→zk
1
1
zk
z − zk
= lim
=
=
= −zk /4,
4
3
3
z→z
k 4z
1+z
4(zk )
4(zk )4
since zk4 = −1. Using a semicircular arc on the upper half plane, and letting
the radius go to infinite (standard argument), we are left with picking up
the residues on the upper half plane, so
Z ∞
−∞
√
√
1
dx
=
2πi(Res
2
=
π/
2.
+
Res
)
=
2πi(1
+
i
+
−1
+
i)/4
z
z
0
1
1 + x4
3
b) For any even integer 2N , how would you compute
Z ∞
−∞
1
dx.
1 + x2N
A similar calculation of residues for f (z) =
Reszk = lim
z→zk
1
1+z 2N
shows
z − zk
1
1
zk
= lim
=
=
= −zk /2N,
2N
2N
−1
2N
−1
z→zk 2N z
1+z
2N (zk )
2N (zk )2N
since (zk )2N = −1. There are N poles in the upper half plane, zk = eπi(2k+1)/2N , k =
0, 1, . . . , N − 1, so we add those residues to get
Z ∞
−∞
N
−1
−1
X
1
−2πi NX
dx
=
2πi
Res
eπi(2k+1)/2N .
=
z
k
1 + x2N
2N
k=0
k=0
This last sum we recognize as a geometric series, summing to
zN − z0
−2eπi/2N
2
=
= πi/2N
= −i csc(π/2N ).
πi/N
πi/N
1−e
1−e
e
− e−πi/2N
Thus
Z ∞
−∞
π
1
dx =
csc(π/2N ).
1 + x2N
N
5. Prove, that if f is a non-constant, entire function satisfying
|f (z)| < |z|N
for some fixed N and all z sufficiently large, then f has at most N zeros in
the complex plane.
Proof: use the Cauchy integral formula for derivatives to observe that the
N-th derivative f (N ) is bounded on the plane, and therefore a constant by
Liouville’s theorem. Computing N antiderivatives, we find f is a polynomial
of at most degree N , hence it has no more than N zeros (or roots).
6. a) Show that the real part u(x, y) = Real{f (x + iy)} of an analytic function f in the open unit disk D is harmonic. That is, u is twice continuously
differentiable (in the real variables sense) and satisfies the PDE
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
Proof: With f = u + i · v, the fact that f is holomorphic tells us it is
infinitely differentiable in the complex sense, and tis by computing difference
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quotients along horizontal and vertical directions, we conclude both u and v
are infinitely differentiable in the real sense (i.e. with respect to variables x
and y). The Cauchy-Riemann equations tell us ∂x u = ∂y v and ∂y u = −∂x v,
so
∂x ∂x u + ∂y ∂y u = ∂x ∂y v − ∂y ∂x v = 0,
since the order of differentiation does not matter for infinitely differentiable
function v. Thus u is harmonic (as is v).
b) Show that given a (real-valued) harmonic function u(x, y) on the open
unit disk, there exists another harmonic function v(x, y) such that the function
f (z) = f (x + iy) = u(x, y) + iv(x, y)
is analytic in the unit disk.
Proof: our motivation is that if such an f exists, its derivative, computed
along the x direction would be f 0 (z) = ∂x u + i∂x v = ∂x u − i∂y u, by the
Cauchy-Riemann equation. So, although we don’t know (yet) what f is, or
its derivative, we can define a function
g(z) = ∂x u(x, y) − i∂y u(x, y)
and verify that the real and imaginary parts of g satisfy the Cauchy-Riemann
equation. To check this:
∂x Real(g) = ∂x ∂x u, ∂y Imag(g) = −∂y ∂y u,
and since uxx + uyy = 0, these two guys are equal. i.e.
∂x Real(g) = ∂y Imag(g).
A similar calculation, noting that uxy = uyx since u is twice continuously
differentiable, shows.
∂x Imag(g) = −∂y Real(g).
Thus g satisfies the C-R equation, hence is holomorphic, hence has an
antiderivative f in the open disk, and we can add a constant to ensure that
f (0) = u(0, 0). With f = U + iV , define the difference function D(x, y) =
U (x, y) − u(x, u). Since f 0 = g we compute that
∂x D = ∂y D = 0, with D(0, 0) = 0, (the difference of f (0, 0) and u(0, 0).)
The only solution to this linear PDE is D ≡ 0, so U ≡ u, confirming that
we have found an analytic function f on the open disk whose real part is
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u. Of course, V is also harmonic, by part a), since it is the real part of the
analytic function −if (z).
c) Assume u(x, y) is harmonic in the open unit disk, and continuous on its
closure. Show we can compute its values inside the disk by the real integral
1
u(re ) =
2π
iθ
Z 2π
0
1 − r2
u(eiφ ) dφ.
1 − 2r cos(θ − φ) + r2
That is, we are integrating the values of u on the circle with the Poisson
kernel.
(Hint. Use the Cauchy integral formula on f = u + iv, applied to a disk of
radius slightly less than 1. Then take a limit.)
Proof: Rather than using “magic” to get the right form for the answer, let’s
try to figure this out step by step.
We know the solution has something to do with two points, one inside
the unit disk z = reiθ and one on the unit circle, w = eiφ . The distance
(squared) between them is
|w − z|2 = (eiφ − reiθ )(e−iφ − re−iθ ) = 1 − 2r cos(θ − φ) + r2 ,
so we can write the Poisson kernel as
1 − r2
z
1 − r2
w
=
=
+
,
2
1 − 2r cos(θ − φ) + r
(w − z)(w − z)
w−z w−z
where we get these last two terms by computing the partial fractions. Parametrizing the circle by w = eiφ , dw = ieiφ dφ = iw · dφ, we can write
w
dw
1
1
z
z
1 − r2
dφ =
+
=
+
dw.
2
1 − 2r cos(θ − φ) + r
w − z w − z iw
i w − z 1 − wz
Now we are in good shape, because the first term has the 1/(w − z) which
we need for the Cauchy integral formula, and the second term has a pole
only at w = (1/r)eiθ , which is outside the unit circle, so it is holomorphic
inside the disk.
Thus if f is an analytic function on, and inside, the closed unit disk, we
have
1
2π
Z 2π
0
1 − r2
1
f (eiφ )dφ =
2
1 − 2r cos(θ − φ) + r
2πi
6
Z
|w|=1
1
z
+
f (w)dw = f (z),
w − z 1 − wz
by the Cauchy integral formula (first term) and analyticity (second term).
Taking real parts of this equation, we have
1
2π
Z 2π
0
1 − r2
Real(f (eiφ ))dφ = Real(f (z)).
1 − 2r cos(θ − φ) + r2
(∗∗)
Now, we are not quite done, because the original u we started with
was only harmonic inside the unit disk, not on its boundary. So when we
construct analytic f from u, it is only analytic inside the unit disk, not on
its boundary. So, for a fixed parameter 0 < ρ < 1, define
fρ (z) = f (ρz) = u(ρz) + iv(ρz).
This function is analytic on the open disk of radius 1/ρ > 1 and in particular,
on an open set containing the unit disk and its boundary. Thus we can apply
the above formula (**) to fρ to obtain
1
2π
Z 2π
0
1 − r2
u(ρeiφ )dφ = u(ρz).
1 − 2r cos(θ − φ) + r2
Holding z fixed, and let ρ → 1, by uniform continuity of u on the closed unit
disk (which is compact), we can exchange limits with integrals to find
1
2π
Z 2π
0
1 − r2
u(eiφ )dφ = u(z) = u(reiθ ),
1 − 2r cos(θ − φ) + r2
as desired. (Notice for fixed z, the Poisson kernel is a nice continuous,
bounded function of φ, so it is only the uniform convergence of u(ρeiφ ) →
u(eiφ ) that needs to concern us.)
That was a bit longer than I would expect the students to do for this
question, but I decided to fill in some details as to why the proof should
proceed this way.
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