## Slide 2 Restate Residue Theorem

• f has isolated singularity at $z_0$ and is analytic on the punctured disc 0 < | z- $z_0$| < r
• f has a unique Laurent series representation …
• The residue of f at $z_0$ is represented by Res(f,$z_0$) = $a_1$ the coefficient of $\frac{1}{z - z_0}$

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

## Slide 3 Residue at Removable singularity

$z_0$ is a removable singularity of a Laurent series if there are no negative powers of the terms $(z -z_0)^k$

so $a_1$ = 0

so Res(f,$z_0$) = 0

## Slide 4 Residue at simple pole

$f(z) = \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 + ...$ (*)

How to isolate $a_1$? Cross-multiply then take limit as z tends to $z_0$

$(z-z_0)f(z) ={a_{-1}} + a_0(z-z_0)^1 + a_1(z-z_0)^2 + a_2(z-z_0)^3 + ...$

$Res(f, z_0) = \underset{{z \to z_0}}{\lim}(z-z_0)f(z)$

We need to take limit because of the division by $z - z_0$ in the original Laurent series (*)

## Slide 5 Residue at simple pole example

(From previous slide) $Res(f, z_0) = \underset{{z \to z_0}}{\lim}(z-z_0)f(z)$

Consider $f(z) = \frac{1}{z^2 + 1}$ has simple poles at $z_0$= +i, -i

$Res(\frac{1}{1+z^2}, i) = \underset{{z \to i}}{\lim}(z-i)\frac{1}{z^2 + 1}$

$= \underset{{z \to i}}{\lim}(z-i)\frac{1}{(z+i) (z-i)} = \underset{{z \to i}}{\lim}\frac{1}{(z+i)} = \frac{1}{2i}$

## Slide 6 Residue at double pole example

Laurent series: $f(z) = ...+ \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 + ...$ (*)

$z_0$ is double pole when $a_{-2} \neq 0 \textrm{ and } a_{-k} = 0$ for k > 2.

To isolate $a_{-1}$. (i) Multiply by (z –  z_0)^2 and (ii) differentiate (iii) take limit.

$f(z) = \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 + ...$

$(z-z_0)^2f(z) = {a_{-2}} + {a_{-1}}{(z-z_0)}^1 + a_0(z-z_0)^2 + a_1(z-z_0)^3 + a_2(z-z_0)^4 + ...$

$\frac{d}{dz}(z-z_0)^2f(z) = {a_{-1}} + 2a_0(z-z_0)+ 3a_1(z-z_0)^2 + ...$

$Res(f, z_0) = a_{-1}= \underset{{z \to z_0}}{\lim}\frac{d}{dz} (z-z_0)^2f(z)$

## Slide 7 Residue at double pole example

$f(z) = \frac{1}{(z-1)^2(z-3)}$ double pole at z = 1, simple pole at z = 3.

$Res(\frac{1}{(z-1)^2(z-3)}, 1) = \underset{{z \to 1}}{\lim}\frac{d}{dz} (z-1)^2\frac{1}{(z-1)^2(z-3)}$

$= \underset{{z \to 1}}{\lim}\frac{d}{dz} \frac{1}{(z-3)}$

$= \underset{{z \to 1}}{\lim} \frac{-1}{(z-3)^2} =-\frac{1}{4}$

## 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) = $\frac{1}{(n-1)!}\underset{{z \to z_0}}{\lim}\frac{d^{n-1}}{dz^{n-1}}[(z - z_0)^n f(z)]$

## Slide 9 Residue of “quotient” functions

f(z) = $\frac{g(z)}{h(z)}$

Res(f,$z_0$) = $\frac{g(z_0)}{h\; '(z_0)}$

## 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 $z_0$ e.g. when z = $z_0$, we have division by zero somewhere.

Then if function is analytic everywhere in a (Punctured) disc around $z_0$ (but excluding only the point $z_0$) since a punctured disc is an annulus, the function has a Laurent series representation:

f(z) = $\Sigma_{k=-\infty}^{+\infty}a_k(z - z_0)^k, \; 0 < |z - z_0| < r$  (inequality excludes $z_0$ the centre of disc).

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

Then integrate the function around the disc,

$\int_{|z - z_0|=\rho}f(z) \;dz =\Sigma_{k=-\infty}^{+\infty}a_k \int_{|z - z_0|=\rho}(z - z_0)^k \;dz\;, \; 0 < |z - z_0| < r$ (*) (not explained, that because convergence is uniform over disc then sigma and integral can be interchanged). Note we have taken $a_k$ 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, $\int_{|z - z_0|=\rho}(z - z_0)^k \;dz$ (which are contained within the sigma)?

## Slide 3

For all values of k EXCEPT -1, the function $h(z) = (z - z_0)^k \textrm{ has a primitive, } H(z) = \frac{1}{(k+1)} (z - z_0)^{k+1}$

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

$\int_{|z - z_0|=\rho}(z - z_0)^k \;dz\;=\;0$, see,

So almost all the terms that are integrated have disappeared.

For the remaining integral $\int_{|z - z_0|=\rho}(z - z_0)^{-1} \;dz\; = \int_{|z - z_0|=\rho}\frac{1}{(z - z_0)} \;dz\;$ we can parameterise the curve , or use the Cauchy Integral theorem, to give a value of $2i\pi$

And because this integral value is for k=-1 (all other integral being zero), looking back at (*) the final value for the sigma is $a_{-1}2i\pi$

## Slide 4

From (*) above, at an isolated singularity, $z_0$, the Residue is the $a_{-1}$ coefficient of the Laurent expansion of f(z) around the isolated singularity $z_0$, all other coefficients being zero.

If annulus around isolated singularity $z_0 \textrm{ is } 0<|z-z_0| the the residue of f at $z_0$ is represented as $Res(f,z_0) = a_{-1}$ (of the Laurent expansion of f around z_0).

examples

f(z) = $\frac{1}{(z-1)(z-2)}$

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) = $-1(z-1)^{-1}$+ sum rest of terms

Res(f,1) = -1

or

f(z) = $1(z-2)^{-1}$+ sum rest of terms

Res(f,2) = +1

## Slide 5 Residue examples

$f(z) = \frac{sin z}{z^4} = \frac{1}{z^3} -\frac{1}{3!} \; \frac{1}{z} + \frac{1}{5!} z - ...$

Isolated singularity z = 0

Laurent series over annulus 0 < | z – 0 | < $\infty$

So Res(f,0) = $- \frac{1}{3!}$

• f(z) = cos(1/z) = ….

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

$cos (\frac{1}{z}) = 1 - \frac{1}{z^2}\frac{1}{2!} + \frac{1}{z^4}\frac{1}{4!} - \frac{1}{z^6}\frac{1}{6!} + ...$

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

$sin (\frac{1}{z}) = \frac{1}{z}\frac{1}{1!} - \frac{1}{z^3}\frac{1}{3!} + \frac{1}{z^5}\frac{1}{5!} + ...$

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

Res(f,0) = 1

• f(z) = $\frac{-1 + cos z}{z^2} = ....$

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

$\frac{-1 + cos z}{z^2} = -\frac{1}{2!} + \frac{z^2}{4!} - ...$

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 $z_0$, the centre is called an isolated singularity. The disk is called a punctured disc and is represented 0 < | z – $z_0$ | < r.

### Examples

f(z) = 1/z has an isolated singularity at $z_0$ = 0

f(z) = 1/sin z has (multiple) isolated singularities at $z_0$ = 0 , +/- \pi,   +/- 2\pi etc.

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

Counter examples

f(z)  = $\sqrt{z}$ 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. $a_k = 0$ 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 $a_{-N} \neq 0$ but $a_k = 0$ for all k < -N. N is the order of the pole. if N = 1 this is a simple pole.

## 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:

$f(z) = \Sigma_{k=-\infty}^{+\infty}a_k(z-z_0)^k = ... + \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{(z-z_0)^1} + a_{0} + {a_{1}}{(z-z_0)^1} + {a_{2}}{(z-z_0)^2} + ...$

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 $\Sigma_{k=0}^{\infty}a_k(z - z_0)^k$there is a value of R such that if distance of any z from the “centre” $z_0$ is less than R, i.e. | z – $z_0$| < R then the power series converges whereas if |z – $z_0$| > R the series diverges.

How do we find R.?

Theorem: if $|\frac{a_k}{a_{k+1}}|$ reaches a limit R as n tends to infinity, then this limit is the radius of convergence.

## Slide 3 Examples

$\Sigma_{k=0}^{\infty}(z)^k$, so $z_0$ = 0, a_k = 1, so $|\frac{a_k}{a_{k+1}}| = \frac{1}{1} \to 1$ then R = 1 (about the point z = 0).

$\Sigma_{k=0}^{\infty}k(z )^k$, so $z_0$ = 0, a_k = k, $|\frac{a_k}{a_{k+1}}| = \frac{k}{k+1} \to 1$, so R = 1

$\Sigma_{k=0}^{\infty}\frac{(z)^k}{k!}$, so $z_0$ = 0, a_k = 1/k!, $|\frac{a_k}{a_{k+1}}| = \frac{(k+1)!}{k\;!} \to k + 1$ therefore R = $\infty$

## Slide 7 Relationship Analytic Functions and Power series

If f: U -> C is analytic

and ${|z - z_0| < r } \subset U$ i.e. considering the z values in a disk of radius r surrounding the point $z_0$ << diagram >>, then in this disk f HAS a power series representation and that representation is $\Sigma_{k=0}^{\infty}a_k(z - z_0)^k \textrm{ where } a_k=\frac{f^{(k)}z_0}{k!}$ and its radius of convergence is at least the radius of the disk surrounding $z_0$, i.e. $|z \;-z_0|$.

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

Summary of Taylor series expansion

$\Sigma_{k=0}^{\infty}a_k(z - z_0)^k, |z \;-z_0|. Note carefully the expression for the coefficient.

consider $f(z) = e^z$, then all derivatives $f^{(k)}(z) = e^z$. I fin the Taylor expansion we set z_0 = 0, then  $f^{(k)}(0) = e^0 = 1$ for every derivative, thus $a_k = \frac{1}{k!}$ for all k so

$e^z = \Sigma_{k=0}^{\infty}\frac{z ^k}{k!}\; ,z \in C$

Similarly, about $z_0 = 1$

$e^z = \Sigma_{k=0}^{\infty}\frac{e}{k!}z ^{k-1}\; ,z \in C$

## 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 $sin (z) = 0 + \frac{1}{1!}(z) + \frac{0}{2!}(z)^2 + \frac{-1}{3!}(z)^3 + \frac{0}{4!}z^4 + \frac{1}{5!}(z)^5 + ...$

$z - \frac{1}{3!}(z)^3 + \frac{1}{5!}(z)^5 - \frac{1}{7!}(z)^7 + \frac{1}{9!}(z)^9 + ...$

## Slide 10 Series cos z about 0

Differentiating term by term

Cos z = $1 - \frac{1}{2!}(z)^2 + \frac{1}{4!}(z)^4 - \frac{1}{6!}(z)^6 + \frac{1}{8!}(z)^8 + ...$

## Slide 11 Analytic function

An analytic function is determined by all its derivatives $f^{(k)}(z_0)$ at the centre $z_0$ of the disc.

## Slide 2 Definition Taylor series

Series of form, Centred at z_0 \in C.

$\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$

Examples

$\Sigma_{k=0}^{\infty} (z )^k \textrm{ where } a_k = 1 \textrm{ and } z_0 = 0$, converges for |z| < 1

$\Sigma_{k=0}^{\infty} \frac{(-1)^k}{2^k}z^{2k} \;= \;1 - \; \frac{z^2}{2}\; + \; \frac{z^4}{4}\; - \; \frac{z^6}{8}= \Sigma_{k=0}^{\infty} (\frac{-z^2}{2})^k = \Sigma_{k=0}^{\infty} (w)^k$

where w = $\frac{-z^2}{2}$

this converges when $|w| < 1$

converges when $|z| < \sqrt{2}$ and diverges when $|z| \geq \sqrt{2}$

## Slide 3 Radius Convergence Theorem *important *

Let $\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$ be a power series.

There exists a real number R, $0 \leq R \leq \infty$ , such that the series

converges absolutely in $|z \; - \; z_0|$ < R

and diverges in $|z \; - \; z_0|$ > R

convergence is uniform for each r < R

## Slide 4 Examples Radius Convergence

$\Sigma_{k=0}^{\infty} k^kz^k$ Pick an arbitrary $z \in C \backslash{0}$

Note: $|k^kz^k| = (k|z|)^k \geq 2^k \textrm{ as soon as } k \geq \frac{2}{|z|}$

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: $\Sigma_{k=0}^{\infty} \frac{z^k}{k^k}$ . Similarly to above pick an arbitrary z, then $| \frac{z^k}{k^k}| = \frac{|z^k|}{k^k} \textrm { which will eventually } \leq (\frac{1}{2})^k$ when k reaches the value $k \geq 2|z|$ 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 $\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$ with radius of convergence R > 0, the claim is:

f(z) = $\Sigma_{k=0}^{\infty} a_k (z \; - \; z_0)^k$ is analytic in $|z \; - \; z_0| < R$

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

f ‘ (z) = $\Sigma_{k=1}^{\infty} a_k .k(z \; - \; z_0)^{k-1}$

f ” (z) = $\Sigma_{k=2}^{\infty} a_k .k.(k-1)(z \; - \; z_0)^{k-2}$

eventually

$f^{(k)}(z_0) = a_k .k!$ (since it makes sense to define $0^0= 1$).

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

$a_k = \frac{f^{(k)}(z_0)}{k!} \textrm{ for } k \geq 0$

## Slide 6 Differentiating term-by-term

Because $\Sigma_{k=0}^{\infty} z^k \; = \; \frac{1}{1-z}$ with radius of convergence 1, the sum is analytic when |z| < 1

So we can differentiate term by term and end up with

$\Sigma_{k=0}^{\infty} (k+1)z^k \; = \; \frac{1}{(1-z)^2}$

## 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 $\int_0^w\;\Sigma_{k=0}^{\infty} z^k dz\; = \;-Log(1-z) \; = \;\Sigma_{k=1}^{\infty} \frac{w^k}{k}$ when |z|< 1.

By fiddling with variable we can get: $\Sigma_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(z-1)^k = Log(z)$ for |z – 1| < 1, we have a power series for Log.

## Slide 2 Definition

$\Sigma_{k=0}^\infty {a_k} = a_0 + a_1 +a_2 + ... + a_n + ...$

which converges if sequence of partial sums, Sn, converges

## Slide 3 Example

$\Sigma_{k=0}^\infty {z^k}$

$S_n = 1 + z + z^2+ ... z^n$

Formula in usual way $S_n = \frac{1-z^{n+1}}{1-z}$

Since $z^{n+1} \to 0 \textrm{ as } n \to \infty$ as long as |z| < 1

then $\Sigma_{k=0}^\infty {z^k} = \frac{1}{1-z}$ for |z| < 1.

## Slide 4 Convergence and Divergence

Theorem: If a series converges, then $a_k \to 0 \textrm{ as k } \to \infty$ (*)

The converse (swap propositions), if $a_k \to 0$ then sum converges is not necessarily true – classic example is the harmonic series.

Contrapositive (*) $a_k \to \infty \textrm{ or constant as k } \to \infty$ 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 = $re^{i \theta}$ so  $z^k \; = r^ke^{ik \theta} = r^kcos\theta \; + \; i r^k sin \theta$

So $\Sigma_{k=0}^\infty {z^k} = \Sigma_{k=0}^\infty r^kcos\theta \; + \; i \Sigma_{k=0}^\infty r^k sin \theta$

<< more , separate real and imaginary parts>>

## Slide 6 Another example of convergence. Part I

Does $\Sigma_{k=1}^\infty \frac{i^k}{k}$ converge?

If take modulus of the terms, $|\frac{i^k}{k}|$ 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, $i^k = i^{2n} = (-1)^n$ ( a real number)

Similarly, k is odd, $i^k = i^{2n+1} = i(-1)^n$ (a purely imaginary number)

So $\Sigma_{k=1}^\infty \frac{i^k}{k} = \Sigma_{n=1}^\infty \frac{i^{2n}}{2n} + \Sigma_{n=0}^\infty \frac{i^{2n+1}}{2n+1}$ (note the limits).

Using prelim results and simplifying: $\frac{1}{2}\Sigma_{n=1}^\infty \frac{-1^{n}}{n} + i \Sigma_{n=0}^\infty \frac{-1^{n}}{2n+1}$

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 ? $\Sigma_{k=0}^{\infty}a_k$ converges absolutely if the series $\Sigma_{k=0}^{\infty}|a_k|$ converges

<< examples >>

If $\Sigma_{k=0}^{\infty}a_k$ converges absolutely then it converges and $|\Sigma_{k=0}^{\infty}a_k| \leq \Sigma_{k=0}^{\infty}|a_k|$ converges absolutely

<< to do >>