COMPLEX ANALYSIS–Spring 2014
Final Exam.
1
Fun with the Bernoulli numbers.
The Bernoulli numbers are a famous sequence of numbers {Bn }, the n-th
Bernoulli number being defined as being n! times the coefficient of z n of the
power series expansion at 0 of z/(ez − 1). That is, they are defined by
∞
X
Bn n
z
=
z .
ez − 1 n=0 n!
Here is a bunch of exercises about these numbers. I hope you’ll like them.
∞
X
Bn n
1. Determine the radius of convergence of the series
z .
n!
n=0
2. Show that B0 = 1, B1 = −1/2, and Bn = 0 for all odd integers n > 1.
3. Show that the Bernoulli numbers satisfy the following recursive equation:
Bn = −
n−1 1 X n+1
Bj .
j
n + 1 j=0
Using this formula calculate B2 , B4 .
4. Show that the Laurent expansion of cot z at 0 is given by
∞
cot z =
1 X (−1)k 4k B2k 2k−1
+
z
.
z
(2k)!
k=1
For which z’s is this equality valid?
5. Recall that we proved (and our textbook proved) that
∞
π cot πz =
1 X 2z
+
,
z n=1 z 2 − n2
the series converging uniformly on all compact subsets of C\Z. Expand
z/(z 2 − n2 ) into a Maclaurin series, use that to get another expression of
the Laurent series of cot z at 0. Equating coefficients prove:
∞
X
1
(−1)k−1 B2k 4k π 2k
=
n2k
2(2k)!
n=1
for k = 1, 2, 3, . . .. These formulas were first derived by Euler.
1
2
Holomorphic functions as mappings
Definition 1 Let U be an open subset of C. A holomorphic function f : U → C
is said to be a conformal mapping iff f 0 (z) 6= 0 for all z ∈ U . One can show
(it isn’t hard) that such a function preserves angles; that is, if γ1 , γ2 are two
curves in U intersecting at a point z at an angle of α, then f (γ1 ), f (γ2 ) intersect
at f 9z) with an angle of α. Two open sets U, V are said to be conformally
equivalent iff there exists a one-to-one holomorphic function f on U such that
f (U ) = V . As you are required to prove below, such a function is automatically
conformal; i.e., its derivative never vanishes.
1. Let f be holomorphic in the open subset Ω of C. Assume f has a zero of
order m ≥ 1 at z0 ∈ Ω. Prove: There exists an open neighborhood U of
z0 in Ω and a holomorphic function g : U → C such that g is an invertible
mapping of U onto some disc Dr (0) and such that f (z) = g(z)m for all
z ∈ U.
2. Show that if f is analytic and one-to-one in the open set Ω, then f 0 (z) 6= 0
for all z ∈ Ω. (Possible hint: If f 0 (z0 ) = 0, then h = f − f (z0 ) has a zero
of order m ≥ 2 at z0 .)
3. Let ϕ be an injective conformal mapping of the unit disc D onto itself.
That is, assume that ϕ : D → C is holomorphic, one-to-one and ϕ(D) = D.
Prove: If ϕ(0) = 0, then ϕ(z) = αz for all z ∈ D, some α ∈ C, |α| = 1.
Conclude: If Ω is open in C, and z0 ∈ Ω, up to multiplication by a complex
number of absolute value 1, there exists at most one conformal equivalence
ϕ : Ω → D such that ϕ(z0 ) = 0. (Riemann’s Mapping Theorem proves that
every simply connected open set other than C is conformally equivalent
to D. It is easy to see if Ω is not simply connected, then it cannot be
conformally equivalent to D.)
4. Let w ∈ D. Show that all conformal mappings ϕ from D onto itself are
given by
z−w
ϕ(z) = eiθ
1 − w̄z
for some θ ∈ R.
5. Show that C is not conformally equivalent to D.
6. Let Ω be an open subset of C and for each n ∈ N let fn be a one-to-one
holomorphic function in Ω. Assume {fn } converges uniformly on compact
subsets of Ω to a function f . Prove: f is one-to-one or constant.
Hint: Assume f is not constant and let z1 6= z2 ∈ Ω. Let γ be a circle
centered at z2 such that z1 is outside of the circle and such that f (z) 6=
f (z1 ) for all z ∈ γ ∗ (why is there such a circle?) For n sufficiently large one
should have |f (z) − fn (z)| < |fn (z) − fn (z1 )| for all z ∈ γ ∗ and Rouché’s
Theorem can now be used to show that f (z2 ) 6= f (z1 ).
2
7. Let U be an open convex subset of C and assume that f : U → C is
holomorphic. Assume Ref 0 (z) ≥ 0 for all z ∈ U . Prove: If f is not
constant, then f is one-to-one in U . Show that this result fails to hold if
one replaces the assumption of convexity by simple connectedness, even if
one replaces ≥ by >. That is, give an example of a simply connected set
and a function f holomorphic on the set such that Ref 0 (z) > 0 for al z
but f is not one to one.
Hint: For the direct part, assume first Ref 0 (z) > 0 for all z ∈ Ω. Then
show that every function with Ref 0 ≥ 0 can be obtained as uniform limit
over compact subsets of functions satisfying Ref 0 > 0.
3
Integrals
Evaluate the following integrals.
Z ∞
1
dx, for n − 2, 3, 4, . . ..
1.
1 + xn
0
Note: If n is even, the integral is half the integral over the interval
(−∞, ∞); withZ γR the halfcircle on the upper half plane from R to −R
of the function f (z) = 1/(1 + z n ) by residues
one computes
L[−R,R]+γR
and one sees that as R → ∞ the integral over γR goes to 0. This does not
work if n is odd, and has some disadvantages even if n is even; too many
poles for large n. A better path is suggested by the picture. Let R → ∞.
∞
xα
dx, where α ∈ C, 0 ≤ Reα < 1 by integrating the
1 + x2
0
analytic function f (z) = z α (1 + z 2 )−1 over the contour ΓR, , 0 < << R,
pictured below. Use an analytic determination of the logarithm valid in
C\{x : x ∈ R, x ≥ 0}; for example, log z = log |z|+i arg z, 0 < arg z < 2π.
Z
2. Compute
3
The contour is the sum of four paths: ΓR, = γ1 + γ2 + γ3 + γ4 , where γ1
is the line segment from i to Reiθ , where θ = arcsin(/R);, γ2 is the big
circular arc from Reiθ to Re−iθ , γ3 is the line segment from Re−iθ to
−i, and γ4 is the arc in the second and third quadrants from −i to i.
For your convenience, here are possible parameterizations:
γ1 (t)
=
L[i, Reiθ ](t) = t + i,
γ2 (t)
=
Reit ,
γ3 (t)
=
−L[−i, Re−iθ ](t),
0 ≤ t ≤ Rcos(θ ),
θ ≤ t ≤ 2π − θ ,
L[−i, Re−iθ ](t) = t − i,
0 ≤ t ≤ Rcos(θ ),
(− cos t + i sin t), −π/2 ≤ t ≤ π/2.
Z ∞
Z ∞
2
3. The Fresnel integrals
cos(x ) dx,
sin(x2 ) dx are important in opγ4 (t)
=
0
0
tics. Obviously, the integrand in these integrals is not integrable over
[0, ∞) so the interpretation is
Z ∞
Z R
Z ∞
Z R
2
2
2
cos(x ) dx = lim
cos(x ) dx,
sin(x ) dx = lim
sin(x2 ) dx.
R→∞
0
Prove
Z
0
R→∞
0
∞
2
Z
cos(x ) dx =
0
0
−z 2
∞
1
sin(x ) dx =
2
2
r
0
π
2
by integrating z 7→ e
over a contour like that of Exercise 1; with the
angle at the vertex equal to π/4. That is the contour should be the line
segment from 0 to R, followed by the arc from R to Reiπ/4 , followed by
the line segment from Reiπ/4 to 0.
Z ∞
cos bx
dx, a, b ∈ R, a > 0. The trick here is to use cos bx = Reeibx .
4.
2 + a2
x
0
4
Z
∞
5.
0
cos bx
dx, a, b ∈ R, a > 0.
(x2 + a2 )2
Z
e−1/z dz.
6.
|z|=1
Z
7.
0
π
dθ
.
(5 + cos θ)2
8. Let Ω be an open subset of C, let f be holomorphic in Ω, and let γ
be a closed, piecewise differentiable curve in Ω. Assume that Wγ (z) is
either 0 or 1 for all z ∈ C\γ ∗ , that Wγ (z) = 0 for all z ∈
/ Ω so that
Int(γ) = {z ∈ C\γ ∗ : Wγ (z) = 1} ⊂ Ω. Assume f (z) 6= 0 for all z ∈ γ ∗ .
For n = 0, 1, 2, . . ., evaluate
Z 0
f (z) n
1
z dz.
2πi γ f (z)
5