Complex Analysis Revision Lecture

MATH20101 Complex Analysis Revision Lecture

Charles Walkden

Dec 11th 2014

The exam

I The exam paper contains 8 questions:I 4 on Real Analysis (Section A)I 4 on Complex Analysis (Section B).

I You have to answer 5 questions, with at least 2 from eachsection.

I Answer more than 5 then only your best 5 count (subject toat least 2 from each section).

Main topics in the course

I 2: The Cauchy-Riemann TheoremI 3: Power seriesI 4: Contour integration and Cauchys TheoremI 5: Cauchys integral formula and its applicationsI 6: Laurent seriesI 7: Poles, residues, the Cauchy Residue Theorem and its uses.

(6 topics, 4 questions...)

2 The Cauchy-Riemann Theorem

Let f : D C.

Theorem (Cauchy Riemann Theorem)

IF f is differentible at z0 THEN the partial derivatives exist andthe Cauchy-Riemann equations hold at z0:

u

x(z0) =

v

y(z0),

u

y(z0) = v

x(z0)

Theorem (Converse to the Cauchy-Riemann Theorem)

IF the partial derivatives exist AND ARE CONTINUOUS at z0and the Cauchy-Riemann equations hold at z0 THEN f isdifferentiable at z0.

2 The Cauchy-Riemann Theorem

Let f : D C.Theorem (Cauchy Riemann Theorem)

IF f is differentible at z0 THEN the partial derivatives exist andthe Cauchy-Riemann equations hold at z0:

u

x(z0) =

v

y(z0),

u

y(z0) = v

x(z0)

Theorem (Converse to the Cauchy-Riemann Theorem)

IF the partial derivatives exist AND ARE CONTINUOUS at z0and the Cauchy-Riemann equations hold at z0 THEN f isdifferentiable at z0.

3 Power series

A power series at z0 has the form

n=0

an(z z0)n.

an are the coefficients. Think of this as a function of z .

By changing co-ordinates we look at power series at 0:

n=0

anzn ()

Q: When does (*) converge?A: (*) converges for z such that |z | < R, R= radius ofconvergence.(R = means (*) converges for ALL z C.)

3 Power series


n=0

an(z z0)n.

an are the coefficients. Think of this as a function of z .By changing co-ordinates we look at power series at 0:

n=0

anzn ()


3 Power series


n=0

an(z z0)n.


n=0

anzn ()

Q: When does (*) converge?

A: (*) converges for z such that |z | < R, R= radius ofconvergence.(R = means (*) converges for ALL z C.)

3 Power series


n=0

an(z z0)n.


n=0

anzn ()

Q: When does (*) converge?A: (*) converges for z such that |z | < R, R= radius ofconvergence.

(R = means (*) converges for ALL z C.)

3 Power series


n=0

an(z z0)n.


n=0

anzn ()


3 Radius of convergenceLet

n=0 anz

n be a power series.

Proposition

SUPPOSE the following limit exists

limn

an+1an = 1R .

THEN the radius of convergence is R. (1/0 =, 1/ = 0.)

Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.an+1an = n!(n + 1)!

= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.


n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! .

Then an = 1/n!.an+1an = n!(n + 1)!



n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.

an+1an = n!(n + 1)!



n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.an+1an = n!(n + 1)!

=

1

n + 1 0 = 1

Ras n.

So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.


n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.an+1an = n!(n + 1)!

= 1n + 1

0 = 1R

as n.



n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.an+1an = n!(n + 1)!

= 1n + 1 0 = 1R as n.



n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.an+1an = n!(n + 1)!

= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.

RemarkIf you get a z in your formula for R then youve gone wrong.


n=0 anz


Proposition


limn

an+1an = 1R .


Example

Define exp(z) =

n=0zn

n! . Then an = 1/n!.an+1an = n!(n + 1)!


3 Properties of power seriesTheoremLet f (z) =

n=0 anz

n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =

n=1 nanz

n1.

Example

exp(z) =n=0

zn

n!

= 1 + z +z2

2!+

z3

3!+

z4

4!+ .

exp(z) = 1 + z +z2

2!+

z3

3!+ = exp(z).


n=0 anz


n=1 nanz

n1.

Example

exp(z) =n=0

zn

n!

= 1 + z +z2

2!+

z3

3!+

z4

4!+ .

exp(z) =

1 + z +z2

2!+

z3

3!+ = exp(z).


n=0 anz


n=1 nanz

n1.

Example

exp(z) =n=0

zn

n!

= 1 + z +z2

2!+

z3

3!+

z4

4!+ .

exp(z) = 1 + z +z2

2!+

z3

3!+ =

exp(z).


n=0 anz


n=1 nanz

n1.

Example

exp(z) =n=0

zn

n!

= 1 + z +z2

2!+

z3

3!+

z4

4!+ .

exp(z) = 1 + z +z2

2!+

z3

3!+ = exp(z).

3 Properties of power series

TheoremLet f (z) =

n=0 anz


n=1 nanz

n1.

How did we prove this?

I Step 1: Show that the power series g(z) =

n=1 nanzn1 has

radius of convergence at least R.

I Step 2: Show that f (z) is differentiable and f (z) = g(z).Step 2 is hard (so not examinable).Step 1 is fun!


TheoremLet f (z) =

n=0 anz


n=1 nanz

n1.How did we prove this?


n=1 nanzn1 has




TheoremLet f (z) =

n=0 anz


n=1 nanz



n=1 nanzn1 has


I Step 2: Show that f (z) is differentiable and f (z) = g(z).

Step 2 is hard (so not examinable).Step 1 is fun!


TheoremLet f (z) =

n=0 anz


n=1 nanz



n=1 nanzn1 has



4 Contour integrationSuppose f : D C, is a path in D.

Let : [a, b] D be a parametrisation of .Definition

f =

ba

f ((t))(t) dt.

Properties of the contour integral:1+2

f =

1

f +

2

f .

f =

f .

(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)

4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .

Definition

f =

ba

f ((t))(t) dt.


f =

1

f +

2

f .

f =

f .


4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .Definition

f =

ba

f ((t))(t) dt.


f =

1

f +

2

f .

f =

f .



f =

ba

f ((t))(t) dt.

Properties of the contour integral:

1+2

f =

1

f +

2

f .

f =

f .



f =

ba

f ((t))(t) dt.


f =

1

f +

2

f .

f =

f .


4 The Fundamental Theorem of Contour Integration

DefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)

Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then

f = F (z1) F (z0).

Example

f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then

f = F (i) F (0) = i

3

3= i

3.

RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .

4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .

Theorem (Fundamental Theorem of Contour Integration)


f = F (z1) F (z0).

Example


f = F (i) F (0) = i

3

3= i

3.


4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)


f = F (z1) F (z0).

Example


f = F (i) F (0) = i

3

3= i

3.




f = F (z1) F (z0).

Example

f (z) = z2.

f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then

f = F (i) F (0) = i

3

3= i

3.




f = F (z1) F (z0).

Example

f (z) = z2. f has antiderivative F (z) = z3/3.

Let be any path from 0 to i .Then

f = F (i) F (0) = i

3

3= i

3.




f = F (z1) F (z0).

Example

f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .

Then

f = F (i) F (0) = i3

3= i

3.




f = F (z1) F (z0).

Example


f =

F (i) F (0) = i3

3= i

3.




f = F (z1) F (z0).

Example


f = F (i) F (0) = i

3

3= i

3.

RemarkMost functions DO NOT have an anti-derivative.

Examplef : C \ {0} C, f (z) = 1/z .



f = F (z1) F (z0).

Example


f = F (i) F (0) = i

3

3= i

3.


4 Cauchys TheoremIf is a closed contour (= starts and ends at the same point, sayz0) AND f has an anti-derivative then by the Fund Thm ofContour Integ

f = F (z0) F (z0) = 0.

Q: What if f does NOT have an anti-derivative?

Theorem (Cauchys theorem)

Suppose:

I f : D C is holomorphicI is a closed contour in D s.t. w(, z) = 0 for all z / D.

Then

f = 0.

4 The Generalised Cauchy Theorem

Theorem (The Generalised Cauchy Theorem)

Suppose

I f : D C is holomorphicI 1, . . . , n are closed contours in D s.t.

w(1, z) + + w(n, z) = 0 for all z / D.

Then 1

f + +n

f = 0.

Example

1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate

1

f ,2

f .

1(t) = eit , 0 t 2pi.

1f =

2pi0 f (1(t))

1(t) dt

= 2pi

01e it

ie it dt = 2pii .

Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence

2

f =2

f = 2pii .

Example


1

f ,2

f .

1(t) = eit , 0 t 2pi.

1

f = 2pi

0 f (1(t))1(t) dt

= 2pi

01e it

ie it dt = 2pii .


2

f =2

f = 2pii .

Example


1

f ,2

f .

1(t) = eit , 0 t 2pi.

1f =

2pi0 f (1(t))

1(t) dt

= 2pi

01e it

ie it dt = 2pii .


2

f =2

f = 2pii .

Example


1

f ,2

f .

1(t) = eit , 0 t 2pi.

1f =

2pi0 f (1(t))

1(t) dt

= 2pi

01e it

ie it dt = 2pii .

Apply GCT to 1, 2 and the domain D =C \ {0}.

The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence

2

f =2

f = 2pii .

Example


1

f ,2

f .

1(t) = eit , 0 t 2pi.

1f =

2pi0 f (1(t))

1(t) dt

= 2pi

01e it

ie it dt = 2pii .

Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.

w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence

2

f =2

f = 2pii .

Example


1

f ,2

f .

1(t) = eit , 0 t 2pi.

1f =

2pi0 f (1(t))

1(t) dt

= 2pi

01e it

ie it dt = 2pii .

Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.

GCT 1 f + 2 f = 0.Hence

2

f =2

f = 2pii .

Example


1

f ,2

f .

1(t) = eit , 0 t 2pi.

1f =

2pi0 f (1(t))

1(t) dt

= 2pi

01e it

ie it dt = 2pii .


2

f =2

f = 2pii .

5 Cauchys Integral formula and its applications

Theorem (Cauchys Integral Formula)

Suppose

I f is holomorphic on the disc {z C | |z z0| < R}I Cr is the circular path with centre z0 radius r , r < R.

Then if w is a point s.t. |w z0| < r we have

f (w) =1

2pii

Cr

f (z)

z w dz

Proof: hard (so not examinable).

5 Consequences of Cauchys Integral Formula: TaylorsTheorem

Theorem (Taylors Theorem)

Suppose

I f is holomorphic on D.

Then

I all the higher derivatives of f exist in DI on any disc {z C | |z z0| < R} D, f is equal to its

Taylor series

f (z) =n=0

an(z z0)n

I

an =f (n)(z0)

n!=

1

2pii

Cr

f (z)

(z z0)n+1 dz

where Cr is a circular path centre z0 radius r < R.




Suppose


Then

I all the higher derivatives of f exist in D

I on any disc {z C | |z z0| < R} D, f is equal to itsTaylor series

f (z) =n=0

an(z z0)n

I

an =f (n)(z0)

n!=

1

2pii

Cr

f (z)

(z z0)n+1 dz





Suppose


Then

I all the higher derivatives of f exist in DI on any disc {z C | |z z0| < R} D, f is equal to its

Taylor series

f (z) =n=0

an(z z0)n

I

an =f (n)(z0)

n!=

1

2pii

Cr

f (z)

(z z0)n+1 dz



5 Consequences of Cauchys Integral Formula: CauchysEstimate

Proposition (Cauchys Estimate)

Suppose

I f is holomorphic on {z C | |z z0| < R}.I |f (z)| M for all z s.t. |z z0| = r .

Then

I for all n 0 we have

|f (n)(z0)| Mn!rn

.

Proof: easy consequence of Taylors Theorem and the EstimationLemma. (See this weeks support class sheet.)

5 Consequences of Cauchys Integral Formula: LiouvillesTheorem

Theorem (Liouvilles Theorem)

Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.

Proof: easy consequence of Cauchys Estimate when n = 1.

Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.





Choose M > 0 s.t. |f (z)| M for all z C.

Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.





Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.

f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.





Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.

By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.





Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .

Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.





Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.

Hence f is a constant.





Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.

5 Consequences of Cauchys Integral Formula: Fund Thmof Algebra

Theorem (The Fundamental Theorem of Algebra)

Let p(z) = zn + an1zn1 + + a1z + a0 be a polynomial ofdegree n 1 with coefficients aj C.Then p has a zero: there exists w C such that p(w) = 0.

Proof: straightforward consequence of Liouvilles Theorem

Idea: Suppose not, i.e. p(z) 6= 0 for all z C. Then 1/p(z) isdefined and is holomorphic on C. Show thart 1/p(z) is bounded:there exists M > 0 such that |1/p(z)| < M for all z C. ThenLiouville implies 1/p(z) is a constant, hence p(z) is constant - acontradiction.

6 Laurent SeriesA power series at z0 has the form

n=0 an(z z0)n.

This converges for {z C | |z z0| < R}.

A Laurent series has the form

n=an(z z0)n.

Often we write this instead as

n=1

bn(z z0)n +

n=0

an(z z0)n

= + bn(z z0)n + +

b1z z0 +a0+a1(zz0)+ +an(zz0)

n+ .

A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)


n=0 an(z z0)n.



n=an(z z0)n.


n=1

bn(z z0)n +

n=0

an(z z0)n

= + bn(z z0)n + +

b1z z0 +a0+a1(zz0)+ +an(zz0)

n+ .

A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.

(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)


n=0 an(z z0)n.



n=an(z z0)n.


n=1

bn(z z0)n +

n=0

an(z z0)n

= + bn(z z0)n + +

b1z z0 +a0+a1(zz0)+ +an(zz0)

n+ .

A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)

6 Laurents Theorem

Theorem (Laurents Theorem)

Suppose f is holomorphic on the annulus{z C | R1 < |z z0| < R2}, 0 R1 < R2 .

Then we can write f as a Laurent series

f (z) =n=1

an(z z0)n +n=0

an(z z0)n

on {z C | R1 < |z z0| < R2}.Moreover

an =1

2pii

Cr

f (z)

(z z0)n+1 dz

where Cr is the circular path centre z0 radius r (R1 < r < R2)described once anticlockwise.

6 Laurents Theorem


Suppose f is holomorphic on the annulus{z C | R1 < |z z0| < R2}, 0 R1 < R2 .Then we can write f as a Laurent series

f (z) =n=1

an(z z0)n +n=0

an(z z0)n

on {z C | R1 < |z z0| < R2}.

Moreover

an =1

2pii

Cr

f (z)

(z z0)n+1 dz


6 Laurents Theorem


Suppose f is holomorphic on the annulus{z C | R1 < |z z0| < R2}, 0 R1 < R2 .Then we can write f as a Laurent series

f (z) =n=1

an(z z0)n +n=0

an(z z0)n

on {z C | R1 < |z z0| < R2}.Moreover

an =1

2pii

Cr

f (z)

(z z0)n+1 dz


6 Laurents Theorem: an example

Let f (z) = 1z(1z) .

This is holomorphic on 0 < |z | < 1.Note

1

1 z = 1 + z + z2 + z3 + if |z | < 1.

So

f (z) =1

z(1 + z + z2 + z3 + ) = 1

z+ 1 + z + z2 + z3 +

if 0 < |z | < 1. This is the Laurent series on the annulus0 < |z | < 1.


Let f (z) = 1z(1z) .

This is holomorphic on 0 < |z | < 1.

Note1

1 z = 1 + z + z2 + z3 + if |z | < 1.

So

f (z) =1

z(1 + z + z2 + z3 + ) = 1

z+ 1 + z + z2 + z3 +



Let f (z) = 1z(1z) .


1

1 z = 1 + z + z2 + z3 + if |z | < 1.

So

f (z) =1

z(1 + z + z2 + z3 + ) = 1

z+ 1 + z + z2 + z3 +



Let f (z) = 1z(1z) .


1

1 z = 1 + z + z2 + z3 + if |z | < 1.

So

f (z) =1

z(1 + z + z2 + z3 + ) = 1

z+ 1 + z + z2 + z3 +

if 0 < |z | < 1.

This is the Laurent series on the annulus0 < |z | < 1.


Let f (z) = 1z(1z) .


1

1 z = 1 + z + z2 + z3 + if |z | < 1.

So

f (z) =1

z(1 + z + z2 + z3 + ) = 1

z+ 1 + z + z2 + z3 +



Let f (z) = 1z(1z) .

f is also holomorphic on |z | > 1.Note if |1/z | < 1 (i.e. |z | > 1)

1

1 z =1

z

(1

1 1z

)=

1

z

(1 +

1

z+

1

z2+

1

z3+

).

=1

z+

1

z2+

1

z3+

1

z4+

Hence

f (z) =1

z2+

1

z3+

1

z4+

1

z5+

if 1 < |z |


Let f (z) = 1z(1z) .

f is also holomorphic on |z | > 1.

Note if |1/z | < 1 (i.e. |z | > 1)

1

1 z =1

z

(1

1 1z

)=

1

z

(1 +

1

z+

1

z2+

1

z3+

).

=1

z+

1

z2+

1

z3+

1

z4+

Hence

f (z) =1

z2+

1

z3+

1

z4+

1

z5+

if 1 < |z |


Let f (z) = 1z(1z) .

f is also holomorphic on |z | > 1.Note if |1/z | < 1 (i.e. |z | > 1)

1

1 z =1

z

(1

1 1z

)=

1

z

(1 +

1

z+

1

z2+

1

z3+

).

=1

z+

1

z2+

1

z3+

1

z4+

Hence

f (z) =1

z2+

1

z3+

1

z4+

1

z5+

if 1 < |z |

7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.

RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).

z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:

f (z) =n=1

bn(z z0)n +

n=0

an(z z0)n

= principal part + a power series

If the principal part has finitely many terms in it then z0 is a pole.



z0 is an isolated singularity if there are no other singularitiesnearby:

i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:

f (z) =n=1

bn(z z0)n +

n=0

an(z z0)n





z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.

Apply Laurents Theorem to this punctured disc:

f (z) =n=1

bn(z z0)n +

n=0

an(z z0)n





z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:

f (z) =n=1

bn(z z0)n +

n=0

an(z z0)n



7 Poles and residues

Suppose f has a pole of order m at z0. Then on 0 < |z z0| < Rfor some R > 0 we can write

f (z) =bm

(z z0)m + +b1

z z0 +a0+a1(zz0)+ +an(zz0)n+

where bm 6= 0.

m=order of the pole (most negative power).

b1 = residue of the pole (coefficient of (z z0)1 term) =Res(f , z0).

7 Poles and residues

Suppose f has a pole of order m at z0. Then on 0 < |z z0| < Rfor some R > 0 we can write

f (z) =bm

(z z0)m + +b1

z z0 +a0+a1(zz0)+ +an(zz0)n+

where bm 6= 0.m=order of the pole (most negative power).

b1 = residue of the pole (coefficient of (z z0)1 term) =Res(f , z0).

7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.

Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i . (Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)

LemmaSuppose f (z) = p(z)/q(z) where p, q holomorphic. Suppose

I q has a zero of order m at z0I p(z0) 6= 0.

Then f has a pole of order m at z0.

Example:

f (z) =z + 1

(z2 + 25)2

has poles of order 2 at 5i .

7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i .

(Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)




Example:

f (z) =z + 1

(z2 + 25)2


7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i . (Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)




Example:

f (z) =z + 1

(z2 + 25)2


7 Calculating residues

We had various formulae for calculating residues:

LemmaIf f has a simple pole at z0 then Res(f , z0) = limzz0(z z0)f (z).

If f (z) has a pole of order m at z0 then

Res(f , z0) = limzz0

(1

(m 1)!dm1

dzm1((z z0)mf (z))

).

7 Calculating residuesSuppose f has a pole of order 2 at z0.

Then f has a Laurent series

f (z) =b2

(z z0)2 +b1

z z0 + a0 + a1(z z0) + a2(z z0)2 +

on 0 < |z z0| < R.

(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .

d

dz

((z z0)2f (z)

)= b1 + 2a0(z z0) + 3a1(z z0)2 +

limzz0

d

dz

((z z0)2f (z)

)= b1.

7 Calculating residuesSuppose f has a pole of order 2 at z0. Then f has a Laurent series

f (z) =b2

(z z0)2 +b1

z z0 + a0 + a1(z z0) + a2(z z0)2 +

on 0 < |z z0| < R.

(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .

d

dz

((z z0)2f (z)

)= b1 + 2a0(z z0) + 3a1(z z0)2 +

limzz0

d

dz

((z z0)2f (z)

)= b1.

7 Cauchys Residue Theorem

Theorem (Cauchys Residue Theorem)

Let D be a domain. Suppose

I f : D C is meromorphic (=holomorphic except for finitelymany removable singularities or poles)

I is a simple closed loop in D s.t. all the points inside lie inD

I f has poles at z1, . . . , zn inside .

Then

f (z) dz = 2piin

j=1

Res(f , zj).

7 Cauchys Residue Theorem: an easy exampleTake f (z) = e

iz

(z3)2 .

Then f has a pole of order 2 at z = 3.

Res(f , 3) = limz3

d

dz(z 3)2f (z) = lim

z3d

dze iz = lim

z3ie iz = ie3i .

Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT

C5

f = 2pii Res(f , 3) = 2pie3i .

Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT

C2

f = 0.


iz

(z3)2 .


Res(f , 3) = limz3

d

dz(z 3)2f (z)

= limz3

d

dze iz = lim

z3ie iz = ie3i .


C5



C2

f = 0.


iz

(z3)2 .


Res(f , 3) = limz3

d

dz(z 3)2f (z) = lim

z3d

dze iz = lim

z3ie iz = ie3i .


C5



C2

f = 0.


iz

(z3)2 .


Res(f , 3) = limz3

d

dz(z 3)2f (z) = lim

z3d

dze iz = lim

z3ie iz = ie3i .

Let C5 be the circle centre 0 radius 5, described onceanticlockwise.

Then the pole at z = 3 lies inside . So by the CRTC5



C2

f = 0.


iz

(z3)2 .


Res(f , 3) = limz3

d

dz(z 3)2f (z) = lim

z3d

dze iz = lim

z3ie iz = ie3i .


C5



C2

f = 0.


iz

(z3)2 .


Res(f , 3) = limz3

d

dz(z 3)2f (z) = lim

z3d

dze iz = lim

z3ie iz = ie3i .


C5


Let C2 be the circle centre 0 radius 2, described once anticlockwise.

Then there are no poles of f inside C2. So by the CRTC2

f = 0.


iz

(z3)2 .


Res(f , 3) = limz3

d

dz(z 3)2f (z) = lim

z3d

dze iz = lim

z3ie iz = ie3i .


C5



C2

f = 0.

7 Applications of the CRT: calculating infinite integralsRecall

f (x) dx = lim

A,B

BA

f (x) dx (1)

where the limit is taken in either order.

The principal value of this integral is defined to be

limR

RR

f (x) dx (2)

(2) may exist even if (1) doesnt (eg: f (x) = x).

Lemma (Technical Lemma)

Suppose f : R C is such that: there exist K > 0, C > 0, r > 1s.t. for all |x | > K we have |f (x)| < C/|x |r . Then (1) exists and isequal to its principal value.

7 Applications of the CRT: calculating infinite integralsTo calculate

f (x) dx :

(i) Check that f satisfies the technical lemma.

(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.

(iv) Use the CRT to calculate

Rf .

(v) Write R

f =

[R,R]

f +

SR

f .

(vi) Show (using the Estimation lemma) thatSR

f 0 asR .

(vii) Hence calculate

f (x) dx = limR

[R,R]

f .

See Exercises 7.4, 7.5, 7.6, 7.7.


f (x) dx :


(ii) Construct a D-shaped contour R = [R,R] + SR .

(iii) Find the poles and residues of f inside R for large R.


Rf .

(v) Write R

f =

[R,R]

f +

SR

f .


f 0 asR .


f (x) dx = limR

[R,R]

f .

See Exercises 7.4, 7.5, 7.6, 7.7.


f (x) dx :


(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.


Rf .

(v) Write R

f =

[R,R]

f +

SR

f .


f 0 asR .


f (x) dx = limR

[R,R]

f .

See Exercises 7.4, 7.5, 7.6, 7.7.

Which proofs are examinable?

Proofs of all theorems, propositions, lemmas, corollaries, etc, that Idid in the lectures are examinable EXCEPT FOR:

I Proof of Theorem 3.3.2I Proof of Lemma 4.4.1I Proof of Theorem 5.1.1I Proof of Theorem 5.2.1

ALL OTHER PROOFS THAT I DID IN THE LECTURESARE EXAMINABLE. ANYTHING I DID IN THELECTURES IS EXAMINABLE. YOU SHOULD ALSOMAKE SURE YOUVE DONE THE EXERCISES. (SODONT COMPLAIN I HAVENT WARNED YOU WHAT ISAND WHAT ISNT EXAMINABLE!)

Complex Analysis Revision Lecture

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