Assignment 12

Partial Solution 12
June 11, 2011
3)h.f (z) = sinz z . f has isolated singularities in C, which are at zk = πk.
Since zk have accumulation point at innity, z = ∞ is not an isolated singularity
for f . The singularity at z = 0 is removable (since limz→0 f (z) = 1). At zk ,
πk
z−πk
+(−1)k sin(z−πk)
. The rst summand is analytic near z =
f (z) = (−1)k sin(z−πk)
πk
πk
z−πk
= (−1)k z−πk
πk and does not aect the residue. Also, (−1)k sin(z−πk)
sin(z−πk)
k
, the last ratio is again analytic with value 1 at zk , so res(f ; πk) = (−1) πk.
j. f (z) = sin12 z has isolated singularities in C, which are at zk = πk. Again
z = ∞ is not an isolated singularity for f . The resizdue at zk is the same for
all k since f is π -periodic. Finally, f is even, so the residue is 0.
5) If the function is merorphic at 0, there is a representation f (z) = z −k g(z)
with g holomorphic and non-vanishing at 0, and k ∈ Z. Then
|f (1/n)| = nk |g(1/n)| ≤
1
n!
implying g(1/n) → 0, a contradiction. p
8)d. Dene the function f (z) = z φ(z) where φ(z) = 1−z
is a Mobius
z
1
transformation mapping
CP
\
[0,
1]
to
C
\
[0,
∞)
,
and
the
square
root
is given
√
√
by the branch of z in C \ [0, ∞) satisfying −1 = i. f (z) is holomorphic and
non-vanishing in CP1 \ [0, 1]. Let γ denote the dog-bone path. Then
ˆ
I=
γ
1
dz =
f (z)
ˆ
|z|=R
1
dz
f (z)
1
for large R (by Cauchy's theorem, since f (z)
is holomorphic in the region G
bounded by the two curves γ and |z| = R, where care should be taken of
orientation of curves: The internal curve is γ which violates the left foot rule,
so the outside circle |z| = R should also be oriented anti-left-foot-wise, that is clockwise ). We need to compute the residue at ∞, and the easiest way to do
this is by changing coordinates immediately inside the integral: we let w = 1/z .
Then dw = −dz/z 2 (also note that the curve reverses its orientation after the
transformation)
ˆ
I=
|z|=R
dz
p
=
z φ(z)
ˆ
|w|=1/R
−dw/w2
p
=−
1/w φ(1/w)
1
ˆ
|w|=1/R
dw
w
p
φ(1/w)
1
The function 1/w has residue 1 at 0. the function √φ(1/w)
is analytic and
nonzero near w = 0, so
Res(0;
1
w
p
φ(1/w)
)= p
1
(φ(1/0)
= 1/
p
√
φ(∞) = 1/ −1 = −i
And we conclude by the residue theorem that I = −2πi(−i) = −2π .
Now we do a second computation
by breaking the curve into 4 parts. The two
√
circular parts are of order (trivial bound) and give nothing in the limit.
The two p
linear segments give equal contributions
of opposite sign (essentially
p
because (1 + i) → 1 as → 0+ while (1 − i) → −1 as → 0− ). Being
´
dx
careful with signs, you should get in the limit , δ → 0 that I → −2 01 √x(1−x)
,
´
dx
and we conclude 01 √x(1−x)
= π . Which is amazingly correct, as wolfram alpha
has just testied.
k. We will use a slightly dierent curve: the bump on the lower curve goes
downward, not upward (it seems to make the solution a bit simpler). Dene
f+ (z) =
e−iaz
eiaz
, f− (z) =
sinh z
sinh z
sinh z has zeros when ez = e−z , i.e. z = πik . Inside our domain, it happens
only at z = πi and z = 0. The poles for f there are simple (since e±iaz 6= 0,
and (sinh z)0 = cosh z does not vanish at z = 0, πi ). So the residues there are
(z − πi)e±iaz
e±iaz ± (z − πi)ae±iaz
e∓πa
= lim
=
= −e∓πa
z→πi
z→πi
sinh z
cosh z
cosh πi
res(f± , πi) = lim
and
ze±iaz
e±iaz ± zae±iaz
= lim
=1
z→0 sinh z
z→1
cosh z
´
so γ f±, (z)dz = 2πi(res(f± , 0) + res(f± , πi)) = 2πi(1 − e∓πa ). Next, we
res(f± , 0) = lim
calculate the integral by breaking the curve into parts.
Denote the parts of the curve
L, R,´ T, B (left, right, top, bottom). Then
´
the trivial bound shows that L f± (z), B f± (z) → 0 as A → ∞. Hence
ˆ
ˆ
f± (z)dz → 2πi(1 − e∓πa )
f± (z)dz +
B
T
Also note that f± (z + 2πi) = e∓2πa f± (z), so
ˆ
ˆ
f± (z)dz = −e∓2πa
T
f± (z)dz
B
The minus is because the top and bottom parts point in opposite directions.
This is the equation for which we modied the curve (the alternative is to study
the integrals on the little circles). So
ˆ
(1 − e
∓2πa
f± (z)dz = 2πi(1 − e∓πa )
)
B
2
The desired integral is of course
ˆ
0
∞
sin(ax)
1
dx =
sinh(x)
2
ˆ
∞
−∞
sin(ax)
1
dx = lim
A→∞ 2
sinh(x)
ˆ
B
1
(f+ (z) − f− (z))dz
2i
there is no mistake, one half is because we need half the real line. The other
it
−it
half is because sin t = e −e
. Note that f+ (z) − f− (z) is holomorphic near
2i
0 (since sin az has a zero that cancels that pole of sinh z ´) so the bump of the
curve B doesn't prevent the integral from converging to R (f+ (z) − f− (z))dz .
We conclude that
ˆ
0
∞
sin(ax)
1 − eπa π 1
1 2πi 1 − e−πa
−
=
−
dx =
sinh(x)
4i 1 − e−2πa
1 − e2πa
2 1 + e−πa
1 + eπa
(and this is actually true, at least if you believe wolfram alpha).
It is a good idea to try substitute a = 0 and see if we get a correct answer.
This is how I discovered one minus somehow got replaced by a plus.
3

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