## 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.

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 7 Removable singularities

$z_0$ is a removable singularity if its Laurent series centred at $z_0$ satisfies $a_k$ = 0 for all k < 0 (i.e. negative powers).

$f(z) = \frac{sin z }{ z} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - ..$ 0 < |z| < $\infty$

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 >>

$z_0$ is an isolated singularity of f. $z_0$ is a removable singularity if and only if f is bounded near $z_0$.

## Slide 8 Poles

$\frac{sin z}{ z^5}$ has order 4. $z_0$ is a pole if and only if function approaches infinity as z approaches $z_0$.

Note if f(z) has a pole at $z_0$ then 1/f(z) has a removable singularity at $z_0$.

## Slide 9 Essential singularities

e.g. $e^{(\frac{1}{z})} = 1 + \frac{1}{z} + \frac{1}{2!} \frac{1}{z^2} + \frac{1}{3!} \frac{1}{z^3} + ...$

has an essential singularity at $z_0$ = 0.

Consider z is a real number.

$e^{(\frac{1}{z})}$ approaches infinity as z approaches 0 from the right (the positive side) because 1/z is getting large and positive.

Whereas

$e^{(\frac{1}{z})}$ 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 $z_0$ (zero in this case)

<< casorati weierstrass theorem >>