## 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 >>