Solutions Homework 4 - UC Davis Mathematics

SOLUTIONS HOMEWORK 2
Exercise 9
We will consider the function f (z) = 1 − e2πiz = −2ieiπz sin(πz), and we will
integrate
Log(f (z))dz.
(1)
γ
where γ is as described in the book, with small circular indentations of radius
around 0 and 1.
Notice that Arg(f (z)) = Arg(−2i)+Arg(eiπz )+Arg(sin(πz)), but 0 ≤ Arg(eiπz ≤
π for z ∈ γ, and −π/2 ≤ Arg(sin(πz)) ≤ π/2.
Therefore −π ≤ Arg(f (z)) = −π/2+Arg(sin(πz)) ≤ π, so we can use the principal
branch of the logarithm. In this case,
Log(f (z)) = Log(−2i) + Log(eiπz ) + Log(sin(πz))
Log(f (z)) = log(2) − π/2 + Log(eiπz ) + Log(sin(πz)).
Since Log(f (z)) is analytic inside of γ, the integral (1) is zero. Moreover, the
vertical sections of γ cancel each other due to the periodicity of the sine function:
h
log(−2ie−π(h−y) sin(iπ(h − y)))idy changing variables y˜ = h − y
Lef t = −
h−
log(−2ie−πy˜sin(iπ y˜))yd˜
y
=−
0
h
log(−2ieiπ(1+iy) sin(π(1 + iy)))idy
Right =
h
=
and taking limits as
log(−2ie−πy sin(iπy))idy
→ 0 gives us the result.
Now the top integral,
0
0
log(1 − e2πix e−2πh) )dx
log(1 − e2πi(x+ih) )dx =
1
1
since h → ∞, then e−2πh → 0, from which we infer that log(1 − e2πix e−2πh) ) → 0
as h → ∞, and so the integral also goes to zero.
Finally the indented corners:
0
log(1 − e2πi
Log(f (z))dz =
C
π/2
eiθ
) ieiθ dθ
(2)
but again log( ) → 0 as
→ 0, so the integral goes to zero as well.
Putting all together,
1
1
0
1
Log(eiπx )dx +
Log(f (z))dz = log(2) − π/2 +
0=
0
1
0 = log(2) − π/2 +
log(sin(πx))dx
0
1
πxdx +
0
log(sin(πx))dx
0
1
0 = log(2) − π/2 + π/2 +
log(sin(πx))dx
0
1
log(sin(πx))dx = −log(2).
0
Exercise 12
The function f (z) =
compute their residue:
πcos(πz)
sin(πz)(u+z)2
have poles z = −u and z = k k ∈ N, so lets
For z = −u:
d
z→−u dz
lim
(z + u)2
πcos(πz)
sin(πz)(u + z)2
=
−π 2
sin2 (πu)
Now, for z = k:
lim
z→k
(z − k)πcos(πz)
1
=
sin(πz)(u + z)2
(u + k)2
Therefore the integral over the contour described in the hint is,
γ
−π 2
+
sin2 (πu)
πcos(πz)
dz = 2πi
sin(πz)(u + z)2
N
k=−N
1
(u + k)2
but,
γ
πcos(πz)
dz =
sin(πz)(u + z)2
2π
0
πcot(πRN eiθ )
RN eiθ idθ
(u + RN eiθ )2
1
2,
then |sin(πz)| > δ for some δ > 0, so the integrand goes to
Since RN = N +
zero as RN → ∞, so taking limit as N goes to infinity gives the result.
Exercise 16
a.- We will prove that | g(z)| < |f (z)| for all z ∈ |z| ≤ 1, and the result will
follow from Rouche’s theorem.
Since g(z) is continuous, then there exists M > 0 such that |g(z)| ≤ M, ∀z ∈ D.
On the other hand, since f has only one zero in D, f (0) = 0, then there exists
m > 0, such that m < |f (z)| for all z outside a certain neighborhood of the origin.
|f (z)| > m ∀|z| = r for some r > 0, so choosing
= m/M , then
|g(z)| ≤ m < |f (z)| ∀|z| = r,
then f have the same amount of zeros as f . The result follows from taking r
growing to 1.
Exercise 22
Suppose there exists an holomorphic function F in the disc, such that F (z) =
f (z) ∀|z| = 1. Since F is holomorphic, then
F (z)dz = 0,
∂D
but
∂D
1
dz = 2πi = 0,
z
which is a contradiction.
Problem 1
z
, then f (D) = D1/n (0). Therefore, ∀r > 0, ∃n ∈
n
< r, so D1/n (0) ⊂ Dr (0) strictly, which is what we were looking for.
a.- Consider fn (z) =
N, such that
1
n
Observation: fn (0) =
1
n
→ 0 as n → ∞.
b.- Considering f (z) = (ez/ − 1), f (0) = 0, f (0) = 1. Suppose there is an
r > 0 such that Dr (0) ⊂ f (D) for all > 0. Choose < r, then − ∈ Dr (0) ⇒
− ∈ f (D). Therefore there exists z0 ∈ D such that f (z0 ) = − ⇔ ez0 / − =
− ⇔ ez0 / = 0, which is a contradiction.
c.- Following the hint, we would like to compute the area of h(Dρ − {0})c using
the Stokes theorem. In order to do this we need to know the orientation of the
curve h(∂(Dρ − {0}).
Let ξ ∈
/ h(Dρ (0) − {0}), and lets compute the following integral:
1
2πi
h(ρeiθ
dz
1
=
z−ξ
2πi
1
=
2πi
|z |=ρ
h (z )dz
using the change of variables z = h(z )
h(z ) − ξ
|z |=ρ
(h(z ) − ξ) dz
h(z ) − ξ
= {#of zeros of h(z ) − ξinside C} − {# of poles of h(z ) − ξ inside C}
= −1
Therefore h(ρeiθ ) winds up −1 times around the point ξ.
Now the Area,
Area =
dA = 1/2
h(Dρ −{0})c


1

−ydx + xdy = − Im 
2

zdz 
h(ρeiθ )
h(ρeiθ )
2π
h(ρeiθ )h (ρeiθ )ρieiθ dθ
zdz =
h(ρeiθ )
0
∞
∞
2π
cn ρn e−inθ
=i
0
cm mρm−1 ei(m−1)θ ρeiθ idθ
m=−1
n=−1
∞
∞
2π
cn cm ρn+m m
=i
ei(m−n)θ dθ = 2πi
0
n,m≥−1
|cn |2 ρ2n n
n=−1
∞
−1
+
|cn |2 ρ2n n
ρ2
n=0
= 2πi
Now, 0 ≤ Area = −π
−1
ρ2
∞
+
|cn |2 ρ2n n , therefore
n=0
∞
|cn |2 ρ2n n ≤
n=0
1
ρ2
Passing to the limit as ρ → 1 gives the result.
d.- First note that f (z)/z is nowhere vanishing since f is injective, and vanishes only at zero, but f (z)/z evaluated at zero gives 1. Now we define ψ(z) =
f (z)
e1/2log( z ) , and g(z) = zψ(z 2 ).
It’s easy to check that g(0) = 0 and g (0) = 1, now let’s prove it is injective:
Suppose that g(z1 ) = g(z2 ), then,
z1 ψ(z12 ) = z2 ψ(z22 ) taking squares
⇔ z12
f (z12 )
f (z 2 )
= z22 22
2
z1
z2
⇔ f (z12 ) = f (z22 )
⇔ z12 = z22
⇔ z1 = z2 or z1 = −z2
If z1 = −z2 then z1 ψ(z12 ) = −z2 ψ(z22 ) ⇔ 2z1 ψ(z12 ) = 0 ⇔ z1 = 0 or ψ(z12 ) = 0
which can’t be since ψ is nowhere vanishing.
e.- Lets write the power series expasion of
f (z 2 )
= 1 + a2 z 2 + a3 z 3 + · · ·
z2
1
1
f (z 2 )
− log( 2 ) = −
2
z
2
⇒ g(z)−1
n≥1
(−1)n
(a2 z 2 + a3 z 3 + · · · )n
n
1
1
= − (a2 z 2 + a3 z 3 + · · · ) + (a2 z 2 + a3 z 3 + · · · )2 + · · ·
2
4
a22
4 −a3
2 −a2
)+z (
+ ) + ···
=z (
2
2
4
f (z 2 )
a2
1 2 −a2
−a3
(z (
) + z4(
+ 2 ) + · · · )n
= z −1 e−1/2log( z2 ) = z −1
n!
2
2
4
n≥0
1
−a2
−a3
a2
= + z(
) + z3(
+ 2) + ···
z
2
2
4
∞
By part (c),
n|cn |2 ≤ 1, in particular |c1 |2 ≤ 1 ⇔ |a2 |2 ≤ 4 ⇔ |a2 | ≤ 2.
n=1
If |a2 | = 2 ⇒ a2 = 2eiθ ⇒ cn = 0 ∀n ≥ 2,
1
g(z)−1 = − eiθ z
z
z2
⇒ g(z)2 =
= f (z 2 )
(1 − z 2 eiθ )2
z
⇒ f (z) =
.
(1 − zeiθ )2
f.- We consider the function g(z) =
1
h(z)−zj ,
and we look at the series expansion,
1
1
+ (c0 − zj ) + c1 z + · · · =
1 + (c0 − zj )z + c1 z 2 + · · ·
z
z
1
⇒ g(z) =
= z (1 − (c0 − zj )z + · · · )
h(z) − zj
h(z) − zj =
Therefore g(0) = 0, g (0) = 1 and it is injective since h is injective (and avoids zj ),
so by the previous part,
| − c0 + zj | ≤ 2 ⇒ |z1 − z2 | ≤ |z1 − c0 | + |z2 − c0 | ≤ 4.
1
, and from part (f)
g.- Let w ∈
/ f (D), then w1 and 0 are avoided by h(z) = f (z)
1
1
c
c
we know then that | w − 0| ≤ 4 ⇒ |w| ≥ 4 ⇒ f (D) ⊂ D 41 (0) ⇔ D 41 (0) ⊂ f (D).
Solution Problem 5 ??

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