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COMPLEX ANALYSIS: SOLUTIONS 5

1. Find the poles and residues of the following functions

1

z4 + 5z2 + 6,

1

(z2 − 1)2,

π cot(πz)

z2,

1

zm(1− z)n(m,n ∈ Z>0)

Solution: Throughout we use the following formula for calculating residues: If f(z)has a pole of order k at z = z0 then

res(f, z0) =1

(k − 1)!

dk−1

dzk−1

((z − z0)kf(z)

)∣∣∣z=z0

.

In particular, if f(z) has a simple pole at z0 then the residue is given by simplyevaluating the non-polar part: (z−z0)f(z), at z = z0 (or by taking a limit if we havean indeterminate form).

Let

f(z) :=1

z4 + 5z2 + 6=

1

(z2 + 2)(z2 + 3)=

1

(z + i√

2)(z − i√

2)(z + i√

3)(z − i√

3).

This has simple poles at z = ±i√

2,±i√

3 with residues

res(f, i√

2) = (z − i√

2)1

z4 + 5z2 + 6

∣∣∣z=i√

2=

1

(z + i√

2)(z2 + 3)

∣∣∣z=i√

2=

1

2i√

2,

res(f,−i√

2) = (z + i√

2)1

z4 + 5z2 + 6

∣∣∣z=−i

√2

= − 1

2i√

2,

res(f, i√

3) = − 1

2i√

3,

res(f,−i√

3) =1

2i√

3.

1

COMPLEX ANALYSIS: SOLUTIONS 5 2

For the second one let

f(z) =1

(z2 − 1)2=

1

(z + 1)2(z − 1)2.

This has double poles at ±1. From the formula we get

res(f, 1) =d

dz

1

(z + 1)2

∣∣∣z=1

= −1/4,

res(f,−1) =d

dz

1

(z − 1)2

∣∣∣z=−1

= 1/4.

For the third let

f(z) =π cot(πz)

z2.

Now, cot(πz) has poles wherever sin(πz) = 0, so at z = n ∈ Z. About these pointswe have

sin(πz) = sin(πn) + π cos(πn)(z − n)− π2 sin(πn)(z − n)2

2!− π3 cos(πn)

(z − n)3

3!+ · · ·

=(−1)nπ(z − n)[1− π2 (z − n)2

3!+O((z − n)4)

]and

cos(πz) = cos(πn)− π sin(πn)(z − n)− π2 cos(πn)(z − n)2

2!+ π3 sin(πn)

(z − n)3

3!+ · · ·

=(−1)n[1− π2 (z − n)2

2!+O((z − n)4)

].

Hence, for z close to n ∈ Z, we have

cot(πz) =1− π2(z − n)2/2 +O((z − n)4)

π(z − n)[1− π2(z − n)2/6 +O((z − n)4)

]=

1− π2(z − n)2/2 +O((z − n)4)

π(z − n)

[1 + π2(z − n)2/6 +O((z − n)4)

]=

1

π(z − n)− π

3(z − n) +O((z − n)3)

Therefore, f(z) = π cot(πz)/z2 has simple poles at z = n 6= 0 and a triple pole atz = 0. For the simple poles we have

res(f, n) = (z − n)π cot(πz)

z2

∣∣∣∣z=n

=1

n2.


For the triple pole at at z = 0 we have

f(z) =1

z3− π2

3

1

z+O(z)

so the residue is −π2/3.

Finally, the function

f(z) =1

zm(1− z)n

has a pole of order m at z = 0 and a pole of order n at z = 1. From the formula forresidues we have

res(f, 0) =1

(m− 1)!

dm−1

dzm−1

(1

(1− z)n

)∣∣∣∣z=0

=n(n+ 1) · · · (n+m− 2)

(m− 1)!

=(n+m− 2)!

(n− 1)!(m− 1)!

and

res(f, 1) =1

(n− 1)!

dn−1

dzn−1

((−1)n

zm

)∣∣∣∣z=1

=−m(m+ 1) · · · (m+ n− 2)

(n− 1)!

=− (m+ n− 2)!

(m− 1)!(n− 1)!

=− res(f, 0).

2. Use the substitution eiθ = z along with the residue theorem to show that∫ 2π

0

dθ

2 + cos θ=

2π√3.

Solution: As suggested we let eiθ = z so that dθ = dz/(iz) and the integral becomes

1

i

∫|z|=1

dz

z(2 + (z + z−1)/2)=

2

i

∫|z|=1

dz

z2 + 4z + 1.

Now z2 + 4z + 1 has zeros of order 1 at z = z± = −2±√

3 and so the integrand hassimple poles at z+ and z−. Only z+ lies in the unit disk and therefore by the residue


theorem

2

i

∫|z|=1

dz

z2 + 4z + 1=

2

i× 2πi× res

( 1

z2 + 4z + 1, z+

)=4π(z − z+)

1

z2 + 4z + 1

∣∣∣∣z=z+

=4π(z − z+)1

(z − z+)(z − z−)

∣∣∣∣z=z+

=4π1

z+ − z−

=4π1

2√

3

=2π√

3.

3. Evaluate the following integrals via residues. Show all estimates.(i) ∫ ∞

0

x2

x4 + 5x2 + 6dx

(ii) ∫ ∞0

x sinx

x2 + a2dx, a real

(iii) ∫ ∞0

log x

1 + x2dx

Solution: (i) Since the integrand is an even function the integral in question is equalto I/2 where

I =

∫ ∞−∞

x2

x4 + 5x2 + 6dx.

As a function of a complex variable, the integrand has simple poles at ±i√

2, ±i√

3.We will be considering a semicircular contour in the upper half plane so we only needcalculate the residues at z = i

√2, i√

3. A slight modification of the first calculationin question 1 gives

res

(z2

z4 + 5z2 + 6, i√

2

)=

(i√

2)2

2i√

2=i√

2

2


and

res

(z2

z4 + 5z2 + 6, i√

3

)= −(i

√3)2

2i√

3= −i

√3

2.

Now, Consider the semicircular contour ΓR, which starts at R, traces a semicirclein the upper half plane to −R and then travels back to R along the real axis. Then,on taking R large enough, by the residue theorem∫

ΓR

z2

z4 + 5z2 + 6dz = 2πi×

∑residues inside ΓR = 2πi(

i√

2

2− i√

3

2) = π(

√3−√

2).

On the other hand

limR→∞

∫ΓR

z2

z4 + 5z2 + 6dz = I + lim

R→∞

∫γR

z2

z4 + 5z2 + 6dz

where γR is the semicircle in the upper half plane. But by the Estimation Lemma∣∣∣ ∫γR

z2

z4 + 5z2 + 6dz∣∣∣ 6 πRmax

z∈γR

∣∣∣ z2

z4 + 5z2 + 6

∣∣∣� 1

R→ 0

as R→∞. Hence, I = π(√

3−√

2) and so∫ ∞0

x2

x4 + 5x2 + 6dx =

π

2(√

3−√

2).

The last estimate in the inequality for the integral over γR should be clear since thedominant term is the z4 term in the denominator, but for completeness:

πRmaxz∈γR

∣∣∣ z2

z4 + 5z2 + 6

∣∣∣ =π

Rmaxz∈γR

∣∣∣ 1

1 + 5z−2 + 6z−4

∣∣∣6π

Rmaxz∈γR

1

1− 5|z|−2 − 6|z|−4

=π

R

1

1− 5R−2 − 6R−4

where we have used |z+w| > |z|− |w| in the second line. On taking R > 5, say, thisis 6 2π/R and the constant 2π is absorbed into the � symbol.

(ii). We have∫ ∞0

x sinx

x2 + a2dx =

1

2

∫ ∞−∞

x sinx

x2 + a2dx =

1

2=(∫ ∞

−∞

xeix

x2 + a2dx

).

Denote this last integral by J . Again, we will consider J as the horizontal section ofthe contour ΓR from part (i).


In the upper half plane the integrand has a simple pole at z = ia with residue

res( zeiz

z2 + a2, ia)

= (z − ia)zeiz

z2 + a2

∣∣∣∣z=ia

=iae−a

2ia=e−a

2.

Hence, by the residue theorem

πie−a = limR→∞

∫ΓR

zeiz

z2 + a2dz = J + lim

R→∞

∫γR

zeiz

z2 + a2dz.

Thus it remains to show that this last integral vanishes in the limit. This is similarto question 7 (ii) of Problems 3; a trivial estimate of the integrand is� 1/R which isnot enough for the Estimation Lemma. Instead we apply integration by parts whichis probably the quickest way (see Problems 3). Integrating eiz and differentiating therest gives∫

γR

zeiz

z2 + a2dz =

zeiz

i(z2 + a2)

∣∣∣∣−RR

− 1

i

∫γR

(1

z2 + a2− 2z2

(z2 + a2)2

)eizdz.

The first term on the right is −2R cosR/i(R2 + a2)� 1/R. For the integrals we usethe Estimation Lemma to give∣∣∣∣ ∫

γR

eiz

z2 + a2dz

∣∣∣∣ 6 πRmaxz∈γR

∣∣∣∣ eiz

z2 + a2

∣∣∣∣ 6 πR1

R2 − a2� 1

R,

∣∣∣∣ ∫γR

z2eiz

(z2 + a2)2dz

∣∣∣∣ 6 πRmaxz∈γR

∣∣∣∣ z2eiz

(z2 + a2)2

∣∣∣∣ 6 πR3 1

(R2 − a2)2� 1

R,

as R→∞. Hence, J = πie−a and so∫ ∞0

x sinx

x2 + a2dx =

1

2=(J) =

π

2e−a.

(iii) This is quite hard and, as I discovered recently, the solution is in Conway’sbook anyway (pg. 117–118). My bad. It’s still instructive to attempt this beforereading Conway though.

We see that, as a function of a complex variable, the integrand has a branch cutand simple poles at z = i,−i. Taking the branch of the log with −π < arg(z) < π,we would like to choose a contour which lies just above and below the cut and thatalso picks up the residues at i,−i. A natural choice is the contour we used in thelectures. This consisted of a small circle about z = 0, horizontal lines just above andbelow the negative real axis, and a large circle completing the contour.


Unfortunately, with this choice the integrals over the horizontal lines L1, L2 aregiven by ∫

L1

log z

1 + z2dz +

∫L2

log z

1 + z2dz

=

∫ 0

−∞

log(x+ iδ)

1 + (x+ iδ)2dx+

∫ −∞0

log(x− iδ)1 + (x− iδ)2

dx

→∫ 0

−∞

log |x|+ πi

1 + x2dx+

∫ −∞0

log |x| − πi1 + x2

dx

=2πi

∫ 0

−∞

1

1 + x2dx

and the integral we’re after has disappeared.This motivates the choice of a new contour: we want something with a horizontal

line over the whole real axis, since then the integral over this line is given by∫ ∞−∞

log |x|1 + x2

dx = 2

∫ ∞0

log x

1 + x2dx

and we avoid the problem of cancellation. Consequently, we choose a branch of log zwith a branch cut along the negative imaginary axis. To avoid z = 0 and the branchcut we indent our contour with a small circle in the upper half plane. We completethe contour with a large circle.

Thus, let Γr,R be the contour consisting of a line from r to R, a semicircle in theupper half plane traced from R to −R, a line from −R to −r, and a semicircle inthe upper half plane traced from −r to r. Follwing the details in Conway we find∫ ∞

0

log x

1 + x2dx = 0.

4. Let ΓN be the square which crosses the real axis at ±(N + 1/2) with N ∈ N.(i) Show that cot(πz) is bounded on ΓN and hence show that

limN→∞

∫ΓN

π cot(πz)

z2dz = 0.

(ii) For a given N compute the above integral via residues. Conclude somethinginteresting.

Solution: (i) We have

cot(πz) = ieπiz + e−πiz

eπiz − e−πiz= i

e2πiz + 1

e2πiz − 1= i

e2πixe−2πy + 1

e2πixe−2πy − 1.


Thus, if y > 1

| cot(πz)| 6 1 + e−2πy

1− e−2πy6

1 + e−2π

1− e−2π.

Since cot(πz) = cot(πz) we obtain the same bound for y 6 −1. It remains to showthat cot(πz) is bounded for z = ±(N + 1/2) + iy with |y| < 1. But in this case

| cot(πz)| =∣∣∣∣e±2πi(N+1/2)e−2πy + 1

e±2πi(N+1/2)e−2πy − 1

∣∣∣∣ =| − e−2πy + 1|| − e−2πy − 1|

<e2π − 1

e−2π + 1=: C.

Since C is bigger than the bound for |y| > 1, we may say that | cot(πz)| < C on allof ΓN .

(ii) By the estimation lemma∣∣∣∣ ∫ΓN

π cot(πz)

z2dz

∣∣∣∣ 6 8π(N+1/2) maxz∈ΓN

∣∣∣∣cot(πz)

z2

∣∣∣∣ < 8π(N+1/2)C

(N + 1/2)2=

8πC

N + 1/2

and this tends to zero as N →∞.On the other hand, for a fixed N this integral is given by 2πi× the sum of residues

inside ΓN . Using the results from question 1 then gives∫ΓN

π cot(πz)

z2dz = 2πi

( ∑−N6n6N

n6=0

1

n2− π2

3

).

Therefore, on letting N →∞ we have

0 =∑

−∞6n6∞n6=0

1

n2− π2

3= 2

∞∑n=1

1

n2− π2

3

and so

ζ(2) =∞∑n=1

1

n2=π2

6,

which is interesting.

5. (i) Let f(z) = z6 + cos z. Find the change in argument of f(z) as z travels oncearound the circle of radius 2, center zero, in the positive direction.

(ii) How many solutions does 3ez − z = 0 have in the disk |z| 6 1?

(iii) Use Rouche’s Theorem to prove that a polynomial of degree n has n roots in C.


Solution: (i) By the argument principle, the change in argument of f(z) as z travelsaround the circle is equal to (2π×) the number of zeros minus poles of f(z) insidethe circle, so just the number of zeros then.

To find the number of zeros of f(z) inside the circle we compare f(z) with itsdominant term z6 and apply Rouche’s Theorem. So let g(z) = z6. On the circle wehave

|g(z)| = |z|6 = 26 = 64

and

|f(z)− g(z)| = | cos(z)| 6 |eiz|+ |e−iz|

26 e|z| = e2 < 64.

Thus, Rouche’s Theorem applies and f(z) has the same number of zeros inside thecircle as z6, which is 6. Hence, the change in argument is 2π × 6 = 12π.

(ii). Rephrasing the question we ask for the number of zeros of f(z) = 3ez − z inthe closed unit disk. We apply Rouche’s Theorem again. This time the dominantterm is g(z) = 3ez. On the unit circle we have

|g(z)| = 3e<(z) > 3e−1 > 1

and

|f(z)− g(z)| = |z| = 1.

By Rouche’s Theorem 3ez − z has the same number of zeros in the unit disk as 3ez,which is none. So the answer is no solutions.

(iii) In the lectures we showed that the polynomial z6 + z + 2 has 6 zeros in thedisk |z| 6 2 by comparing it with z6 and applying Rouche’s Theorem. We generalisethis by comparing a general polynomial

f(z) = anzn + an−1z

n−1 + · · ·+ a0, an 6= 0

with its leading term

g(z) = anzn

on arbitrarily large circles. On the circle |z| = R > 1 we have

|g(z)| = |an|Rn


and

|f(z)− g(z)| =|an−1zn−1 + an−2z

n−2 + · · ·+ a0|6|an−1|Rn−1 + |an−2|Rn−2 + · · ·+ |a0|<n max

06j6n−1|aj|Rn−1

=(nmaxn−1

j=0 |aj||an|R

)|an|Rn

Therefore, on a circle of radius R > nmaxn−1j=0 |aj|/|an| we have |f(z)−g(z)| < |g(z)|.

By Rouche’s Theorem f(z) has the same number of zeros inside the circle as g(z),which is n.

6. Prove that the function

f(z) =∑n∈Z

1

(n+ z)2

is meromorphic on C.

Solution: Clearly, f(z) has poles only at the integers so we need to show that f(z) isanalytic on C\Z. By the results given in the lectures, this will follow if we can showthat the series is uniformly convergent on compact subsets of said region. Since

f(z) =1

z2+∑n6=0

1

(n+ z)2

it suffices to show the uniform convergence of this last sum.So let |z| 6 R, let z be fixed away from integers: |z + n| > ρ > 0 ∀n ∈ Z, and let−M be the nearest integer to z. Note that the series looks like 2

∑n 6=0 n

−2 which isconvergent. So we first pull out the convergent part by writing∑

6=0

∣∣∣∣ 1

(n+ z)2

∣∣∣∣ =∑n6=0

1

n2|1 + z/n|2.

Then if |n| 6 |M |

|z + n| > |z +m| =⇒ |1 + z/n| > (|M |/|n|)|1 + z/m| > |1 + z/M | > ρ/|M |.

If |n| > |M |+ 1

|1 + z/n| > |1 + <(z)/n| > 1− |<(z)||n|

> 1− |M |+ 1

|M |+ 2=

1

|M |+ 2.


Therefore,∑n6=0

∣∣∣∣ 1

(n+ z)2

∣∣∣∣ 6 |M |2ρ2

∑06=|n|6|M |

1

n2+ (|M |+ 2)2

∑|n|>|M |+1

1

n2+

1

|z ± (M ± 1)|2

62ζ(2)(R + 1)2

ρ2+ 2ζ(2)(R + 3)2 +

4

ρ2

<∞.

Hence, by the Weierstrass M test with Mn = n−2 max((R + 3)2, (R + 1)2/ρ2) theseries is uniformly convergent on compact subsets of C\Z.

7. Let ϕ : R→ R be a smooth function whose derivatives ϕ(k)(t), k > 0, are of rapiddecay at ∞ i.e.

limt→∞

tAϕ(k)(t) = 0

for all A ∈ R and all k > 0. The Mellin transform of ϕ is defined as

ϕ̃(z) =

∫ ∞0

ϕ(t)tz−1dt.

(i) Prove that ϕ̃(z) is analytic in the region <(z) > 0.(ii) Use integration by parts to show that ϕ̃(z) can be analytically continued to

C\Z60, with possible simple poles at z = −n.(iii) Find the residues at these poles and compute the formal sum

∞∑n=0

res(ϕ̃(z)t−z,−n).

What does this look like?(iv) Given part (iii), suggest an argument that would prove

1

2πi

∫<(z)=c

ϕ̃(z)t−zdz = ϕ(t)

where the integral is over the vertical line from c− i∞ to c+ i∞ with c > 0.

Solution: (i). Since ϕ is smooth on R it is bounded in a neighbourhood of t = 0:|ϕ(t)| 6 C for t ∈ [0, t0], say. Then∫ t0

0

|ϕ(t)tz−1|dt 6 C

∫ t0

0

t<(z)−1dt = Ct<(z)0 /<(z) <∞.


Also, since ϕ is of rapid decay, for any z there exists a t1 = t1(<(z)) such that|ϕ(t)tz−1| 6 Dt−2 for all t > t1. Then∫ ∞

t1

|ϕ(t)tz−1|dt 6 D/t1 <∞.

Since the remaining integral∫ t1t0|ϕ(t)tz−1|dt is finite the integral is absolutely con-

vergent in the region <(z) > 0.Let γ be a closed curve in the region <(z) > 0. By absolute convergence and

Fubini’s Theorem∫γ

ϕ̃(z)dz =

∫γ

∫ ∞0

ϕ(t)tz−1dtdz =

∫ ∞0

ϕ(t)t−1

∫γ

tzdzdt.

But since tz is analytic in the region <(z) > 0 for all t ∈ [0,∞) this last inner integralis zero by Cauchy’s Theorem. Hence,∫

γ

ϕ̃(z)dz = 0.

The continuity of ϕ̃(z) follows exactly as it does for the Gamma function, hence itis analytic by Morera’s Theorem.

(ii). Integrating by parts once gives

ϕ̃(z) =ϕ(t)tz

z

∣∣∣∣∞t=0

− 1

z

∫ ∞0

ϕ′(t)tzdt = −1

z

∫ ∞0

ϕ′(t)tzdt.

By the same reasoning used in part (i), the integral∫∞

0ϕ′(t)tzdt is analytic in the

region <(z) > −1. Hence, the above expression provides a meromorphic continuationof ϕ̃(z) to said region. At z = 0 we have a residue of

z · −1

z

∫ ∞0

ϕ′(t)tzdt

∣∣∣∣z=0

= −∫ ∞

0

ϕ′(t)dt = ϕ(0).

In particular, if ϕ(0) = 0 then there is no pole at z = 0.More generally, integrating by parts n times gives

(1) ϕ̃(z) =(−1)n

z(z + 1) · · · (z + n− 1)

∫ ∞0

ϕ(n)(t)tz+n−1dt.

Again, this expression is seen to be analytic in the region <(z) > −n, with theexception of possible simple poles at z = 0,−1, . . . ,−n + 1. Since n is arbitrary,we’re done.


(iii). From equation (1) we see that

res(ϕ̃(z), z = −n) =(z + n)(−1)n+1

z(z + 1) · · · (z + n)

∫ ∞0

ϕ(n+1)(t)tz+ndt

∣∣∣∣z=−n

=(−1)n+1

(−n)(−n+ 1) · · · (−1)

∫ ∞0

ϕ(n+1)(t)dt

=− 1

n!

∫ ∞0

ϕ(n+1)(t)dt

=1

n!ϕ(n)(0).

In particular, if ϕ(t) is of rapid decay at zero as well as at infinity, then ϕ̃(z) is entire.Since all the poles are simple, an extra factor of t−z in the above residue calculations

gives a factor of t−z|z=−n = tn and hence

∞∑n=0

res(ϕ̃(z)t−z,−n) =∞∑n=0

1

n!ϕ(n)(0)tn.

This, of course, looks like ϕ(t) in the form of a Taylor expansion about t = 0.

(iv). Assuming that the above series converges and equals ϕ(t) (so ϕ is analyticin the real analysis sense) then we would like a contour integral which captures allthese residues. As suggested, we should look at

1

2πi

∫<(z)=c

ϕ̃(z)t−zdz.

We would like to truncate this integral at heights z = c ± iT and then considerthe truncated integral as part of a rectangular contour which encloses some of theresidues. To estimate these integrals we need bounds on ϕ̃(z).

From equation (1) we see that for fixed <(z), ϕ̃(z) → 0 as |=(z)| → ∞. Moreprecisely, as =(z)→∞ equation (1) gives

|ϕ̃(z)| 6 1

|z(z + 1) · · · (z + n− 1)|

∫ ∞0

|ϕ(n)(t)|t<(z)+n−1dt 6C<(z),n

|=(z)|n,

for some constant C<(z),n. Consequently,

1

2πi

∫<(z)=c

ϕ̃(z)t−zdz =1

2πi

∫ c+iT

c−iTϕ̃(z)t−zdz +

1

2π

[ ∫ ∞T

+

∫ −T−∞

]ϕ̃(c+ iy)t−c−iydy

=1

2πi

∫ c+iT

c−iTϕ̃(z)t−zdz +O(T−n+1)


We now consider this last integral as part of a rectangular contour Γ whose left edgecrosses the real axis at −N − 1/2. Then,

1

2πi

∫ c+iT

c−iTϕ̃(z)t−zdz =

1

2πi

∫Γ

ϕ̃(z)t−zdz − 1

2πi

∫other edges of Γ

ϕ̃(z)t−zdz

=N∑n=0

1

n!ϕ(n)(0)tn − 1

2πi

∫other edges of Γ

ϕ̃(z)t−zdz

by the residue theorem. Using our bound on |ϕ̃(z)| we can estimate the remainingintegrals. The largest contribution comes from the integral over the left edge of Γ,and this is seen to be O(tM+1/2). The other integrals are small in size. On lettingT →∞ we get

1

2πi

∫<(z)=c

ϕ̃(z)t−zdz =N∑n=0

1

n!ϕ(n)(0)tn +O(tN+1/2) = ϕ(t).

COMPLEX ANALYSIS: SOLUTIONS 5 - NTNU · COMPLEX ANALYSIS: SOLUTIONS 5 1. ... dx= 1 2 Z 1 1 xsinx x...

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Transcript of COMPLEX ANALYSIS: SOLUTIONS 5 - NTNU · COMPLEX ANALYSIS: SOLUTIONS 5 1. ... dx= 1 2 Z 1 1 xsinx x...