IMPORTANT

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)}

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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|<r 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.

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

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 >>

Slide 10 casorati weierstrass example I

Slide 11 casorati weierstrass example II

Slide 12 Picards Theorem

Slide 2 Review of Taylor Series

Assumes function is analytic over the whole disc.

What if function not differentiable at some point?

.e.g. f(z) = \frac{z}{z^2 + 4} 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:

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)^kthere 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 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 {|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|<r, \textrm{ where } a_k=\frac{f^{(k)}z_0}{k!}. 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

Slide 9 Example of absolute convergence inequality

<< to do >>