## IMPORTANT

## Slide 2 Restate Residue Theorem

- f has isolated singularity at and is analytic on the punctured disc 0 < | z- | < r
- f has a unique Laurent series representation …
- The residue of f at is represented by Res(f,) = the coefficient of

Following slides are to calculate residues at removable singularities and poles (not essential singularities?)

## Slide 3 Residue at Removable singularity

is a removable singularity of a Laurent series if there are no negative powers of the terms

so = 0

so Res(f,) = 0

## Slide 4 Residue at simple pole

(*)

How to isolate ? Cross-multiply then take limit as z tends to

We need to take limit because of the division by in the original Laurent series (*)

## Slide 5 Residue at simple pole example

(From previous slide)

Consider has simple poles at = +i, -i

## Slide 6 Residue at double pole example

Laurent series: (*)

is double pole when for k > 2.

To isolate . (i) Multiply by (z – z_0)^2 and (ii) differentiate (iii) take limit.

## Slide 7 Residue at double pole example

double pole at z = 1, simple pole at z = 3.

## Slide 8 Residue at poles of order n

Use method above: (i) Multiply by (z – z_0)^n and (ii) differentiate multiple times (iii) take limit.

Giving

Res(f,z_0) =

## Slide 9 Residue of “quotient” functions

f(z) =

Res(f,) =

## Coursera Complex Analysis Lesson 7.3 Residue Theorem

### Aug 16, 2016

God this stuff goes on and on

## Slide 2

Recap: f has an isolated singularity at e.g. when z = , we have division by zero somewhere.

Then if function is **analytic** everywhere in a (Punctured) disc around (but excluding **only** the point ) since a punctured disc *is* an annulus, the function has a Laurent series representation:

f(z) = (inequality excludes the centre of disc).

Now take some closed circular path somewhere in the punctured disc, 0 < < r.

Then integrate the function around the disc,

(*) (not explained, that because convergence is uniform over disc then sigma and integral can be interchanged). Note we have taken outside the integral as it is a constant coefficient for each integral. And in theory we have to perform an infinite number of integrations as k ranges between – and + infinity.

How do we calculate these integrals, (which are contained within the sigma)?

## Slide 3

For all values of k EXCEPT -1, the function

and since we are integrating an (analytic?) function over a closed curve then

, see,

So almost all the terms that are integrated have disappeared.

For the remaining integral we can parameterise the curve , or use the Cauchy Integral theorem, to give a value of

And because this integral value is for k=-1 (all other integral being zero), looking back at (*) the final value for the sigma is

## Slide 4

From (*) above, at an **isolated singularity**, , the Residue **is** the coefficient of the Laurent expansion of f(z) around the isolated singularity , all other coefficients being zero.

If annulus around isolated singularity the the residue of f at is represented as (of the Laurent expansion of f around z_0).

examples

f(z) =

This has isolated singularities at z = 1,2

Therefore annuli around the singularities

0 < |z-1| < 1 and 0 < |z-2| < 1

Laurent series

f(z) = + sum rest of terms

Res(f,1) = -1

or

f(z) = + sum rest of terms

Res(f,2) = +1

## Slide 5 Residue examples

Isolated singularity z = 0

Laurent series over annulus 0 < | z – 0 | <

So Res(f,**0**) =

- f(z) = cos(1/z) = ….

this is an essential singularity (see) because infinitely many terms with negative powers

so z = 0 is an essential singularity and since coefficient of z^-1 is **zero**

Res(f,0) = **0**

- f(z) = sin(1/z) = ….

this has an essential singularity (see) because infinitely many terms with negative powers

so z = 0 is an essential singularity and since coefficient of z^-1 is **1**

Res(f,0) = **1**

- f(z) =

this has a removable singularity (see) because no terms with negative powers

so z = 0 is an removable singularity and since coefficient of z^-1 is **0**

Res(f,0) = **0**

<< to follow examples of residues and use in calculating integrals >>

## Slide 2

If a function is analytic **everywhere** on a disc **apart** from the point at its centre , the centre is called an **isolated singularity**. The disk is called a **punctured disc** and is represented 0 < | z – | < r.

### Examples

f(z) = 1/z has an isolated singularity at = 0

f(z) = 1/sin z has (multiple) isolated singularities at = 0 , +/- \pi, +/- 2\pi etc.

f(z) = 1/ (z-2) has isolated singularity at z = 2

Counter examples

f(z) = and Log(z) do not have isolated singularities at 0 since there is no punctured disc around 0 where the functions are analytic. The functions are not analytic on the negative real axis.

## Slide 3

As we have removed the isolated singularity from the disc, we have created an annulus around this point and hence the (analytic) function has a Laurent series.

## Slide 4 Behaviour of Laurent series near the isolated singularity

f(z) = (cos z – 1 )/ z^2 = -1/2! +z^2 /4! – …

**No** negative powers

f(z) = cos z / z^4 = 1/z^4 – (1/2!)(1/z^2) + (1/4!) –

**Finitely** many negative powers

f(z) = 1 / cos z = 1 – (1/2!)(1/z^2) + (1/4!)(1/z^4) – (1/6!)(1/z^6) + …

**Infinitely** many negative powers

## Slide 5 Definition according to number of negative powers of Laurent series

In the Laurent series surrounding an isolated singularity

If coefficients of negative powers are all zero (there are **no negative power** terms, i.e. for k < 0) in Laurent series then the singularity is **removable**.

If there are **finitely** many negative terms in Laurent series around singularity then these singularities also called **poles**. i.e there exists an N > 0, such that but for all k < -N. N is the order of the pole. if N = 1 this is a simple pole.

If there are **infinitely** many negative powers in Laurent series around singularity then this is an **essential** singularity. i.e. a_{k} \neq 0$ for infinitely many k < 0.

## Slide 6 Summary table of type of singularity based on Laurent expansion and number of negative power indices

## Slide 7 Removable singularities

is a removable singularity if its Laurent series centred at satisfies = 0 for all k < 0 (i.e. negative powers).

0 < |z| <

which looks like a Taylor series

and here if we **define** f(0) =1

f has **become analytic** in C and the singularity has been **removed.**

<< Statement of Riemann Theorem >>

is an **isolated** singularity of f. is a **removable** singularity if and only if **f is bounded** near .

## Slide 8 Poles

has order 4. is a pole if and only if function approaches infinity as z approaches .

Note if f(z) has a pole at then 1/f(z) has a removable singularity at .

## Slide 9 Essential singularities

e.g.

has an essential singularity at = 0.

Consider z is a real number.

approaches infinity as z approaches 0 from the right (the positive side) because 1/z is getting large and positive.

Whereas

approaches 0 as z approaches 0 from the left (the negative side) because 1/z is getting large and negative.

So f does not have a limit as z approaches the isolated singularity (zero in this case)

<< casorati weierstrass theorem >>

## Slide 10 casorati weierstrass example I

## Slide 11 casorati weierstrass example II

## Slide 12 Picards Theorem

## Coursera Complex Analysis 7.1 Laurent Series

### Aug 14, 2016

## Slide 2 Review of Taylor Series

Assumes function is analytic over the whole disc.

What if function not differentiable at some point?

.e.g. is not differentiable (or even defined) at +/- 2i.

e.g. f(z) = Log(z) not continuous, so not differentiable, on(along) $late (-\infty,0]$

## Slide 3 Laurent Series Expansion

f is analytic on some complex region U (note: U may contain some points or small regions where f is not defined or *would not be analytic*, so these have been *excluded from U*} and there is some annulus in U, {r <| z – z_0|<R } around those regions which have *already been excluded from U*,

<< picture here>>

then f has a **Laurent series expansion:**

Notice similar to Taylor series but limits of the sigma run between **minus and plus infinity**, therefore series runs infinitely in “both directions”. There are **negative powers** of z-z_0 as well as positive powers.

This series converges at each point in the annulus …

## Slide 4 & 5 Partial series example

<<forum query posted >>

## Slide 2 Determining Radius of Convergence for a power series. The Ratio Test

In earlier lectures we were told for a power series there is a value of R such that if distance of any z from the “centre” is less than R, i.e. | z – | < R then the power series converges whereas if |z – | > R the series diverges.

How do we find R.?

Theorem: **if** **reaches a limit** R as n tends to infinity, then this limit **is** the radius of convergence.

## Slide 3 Examples

, so = 0, a_k = 1, so then R = 1 (about the point z = 0).

, so = 0, a_k = k, , so R = 1

, so = 0, a_k = 1/k!, therefore R =

## Slide 4 Root Test << to be done >>

## Slide 5 Root Test examples << to be done >>

## Slide 6 Cauchy Hadamard

## Slide 7 Relationship Analytic Functions and Power series

If f: U -> C is analytic

and i.e. considering the z values in a disk of radius r surrounding the point << diagram >>, then in this disk f **HAS** a power series representation and that representation **is** and its radius of convergence is at least the radius of the disk surrounding , i.e. .

## Slide 8 Examples of Taylor series of exp(z) about different points in complex plane

Summary of Taylor series expansion

. Note carefully the expression for the coefficient.

consider , then all derivatives . I fin the Taylor expansion we set z_0 = 0, then for every derivative, thus for all k so

Similarly, about

## Slide 9 Series sin z about 0

Sin (z) is analytic in C. Then about 0

f(z) | sin(z) | f(0) | sin(0)=0 |

f’(z) | cos(z) | f’(0) | cos(0)=1 |

f’’(z) | -sin(z) | f’’(0) | -sin(0)=0 |

f’’’(z) | -cos(z) | f’’’(0) | -cos(0)= -1 |

f(4)(z) | sin(z) | f(4)(0) | sin(0)=0 |

so

=

## Slide 10 Series cos z about 0

Differentiating term by term

Cos z =

## Slide 11 Analytic function

An analytic function is determined by all its derivatives at the centre of the disc.

## Coursera Complex Analysis 6.2 Taylor Series

### Aug 9, 2016

## Slide 2 Definition Taylor series

Series of form, Centred at z_0 \in C.

Examples

, converges for |z| < 1

where w =

this converges when

converges when and diverges when

## Slide 3 Radius Convergence Theorem *important *

Let be a power series.

There exists a real number R, , such that the series

converges absolutely in **<** R

and diverges in **>** R

convergence is uniform for each r < R

## Slide 4 Examples Radius Convergence

Pick an arbitrary

Note:

But eventually this will happen, no matter what the value of |z| is, k will cause the terms of the series to be increasing in size and the series does not converge.

In contrast: . Similarly to above pick an arbitrary z, then when k reaches the value and beyond this value of k all subsequent terms are less than powers of 1/2 so the series will converge. Since this is always true the series has infinite radius of convergence.

## Slide 5 Analycity of Power Series

Theorem. For the power series with radius of convergence R > 0, the claim is:

f(z) = is analytic in

and since it is analytic it can be differentiated (infinitely) as follows (note change of limits):

f ‘ (z) =

f ” (z) =

eventually

(since it makes sense to define ).

Then by rearranging we get an expression for every coefficient in the original power series

## Slide 6 Differentiating term-by-term

Because with radius of convergence 1, the sum is analytic when |z| < 1

So we can differentiate term by term and end up with

## Slide 7 Integrating term-by-term

Similarly , with certain certain conditions (to be added), can integrate infinite series term by term.

## Slide 8 Integration example to get series for Log(z)

Justification that when |z|< 1.

By fiddling with variable we can get: for |z – 1| < 1, we have a power series for Log.

## Coursera Complex Analysis 6.1 Infinite Series

### Aug 8, 2016

## Slide 2 Definition

which converges if *sequence of partial sums*, Sn, converges

## Slide 3 Example

Formula in usual way

Since as long as |z| < 1

then for |z| < 1.

## Slide 4 Convergence and Divergence

Theorem: If a series converges, then (*)

The converse (swap propositions), if then sum converges is **not** necessarily true – classic example is the harmonic series.

Contrapositive (*) then series **diverges**, of course is true.

For 1/(1 – z) , this diverges when |z| >= 1

## Slide 5 Real and Imaginary parts of series

z = so

So

<< more , separate real and imaginary parts>>

## Slide 6 Another example of convergence. Part I

Does converge?

If take modulus of the terms, then get harmonic series which does not converge.

What about original series on this slide? Split into real and imaginary parts.

## Slide 7 Another example of convergence. Part II

Preliminaries: k is even, ( a real number)

Similarly, k is odd, (a *purely* imaginary number)

So (note the limits).

Using prelim results and simplifying:

The first sum is the alternating harmonic series which converges (can justify by looking at intervals on real line)

and (probably) because denominators of terms second series getting smaller and series is alternating this also converges.

## Slide 8 Absolute Convergence

Definition ? converges absolutely if the series converges

<< examples >>

If converges absolutely then it converges and converges absolutely

## Slide 9 Example of absolute convergence inequality

<< to do >>