output.to from Sideway
Complex Analysis

Draft for Information Only

Content

```Laurent SeriesβReview of Taylor SeriesβLaurent Series ExpansionβThe Coefficients ππββExampleββAnother ExampleβThe Coefficients ππ ContinuedβIsolated singularitiesβLaurent SeriesβThree Types of Isolated SingularitiesβClassification of Isolated SingularitiesβTypes of SingularitiesβRemovable SingularitiesβPolesβEssential SingularitiesβCasorati-Weierstraπ½βPicard's TheoremββExample:```

source/reference:

Laurent Series

Review of Taylor Series

Recall: If π:πββ is analytic and {|π§βπ§0|<π}βπ then π has representation `π(π§)=ββπ=0ππ(π§βπ§0)π, where ππ=π(π)(π§0)π!, πβ₯0.`

• What if π is not differentiable at some point?
• Example: π(π§)=π§π§2+4 is not differentiable at π§=Β±2π (undefined there).
• π(π§)=Log π§ not continuous on (ββ,0], so not differentiable there.

Laurent Series Expansion

```Theorem (Laurent Series Expansion)If π:πββ is analytic and {π<|π§βπ§0|<π}βπ then π has a Laurent series expansion π(π§)=ββπ=ββππ(π§βπ§0)π =β―+πβ2(π§βπ§0)2+πβ1(π§βπ§0)2+π0+π1(π§βπ§0)+π2(π§βπ§0)2+β―, that converges at each point of the annulus and converges absolutely and uniformly in each sub annulus {π β€|π§βπ§0|β€π‘}, where π<π <π‘<π.```

The Coefficients ππ

Note: The coefficients ππ are uniquely determined by π. How do we find them?

Example

π(π§)=1(π§β1)(π§β2) is analytic in β\{1,2}.

Let's find the Laurent series in the annulus {1<|π§|<2}. Trick: ```1(π§β1)(π§β2)=(π§β1)β(π§β2)(π§β1)(π§β2)=1(π§β2)β1(π§β1)  =β1211βπ§2β1π§(1β1π§)  =β12ββπ=0π§2πβ1π§ββπ=01π§π  =ββπ=0β12π+1π§π+ββπ=1β1π§π  =ββπ=0β12π+1π§π+β1βπ=ββ(β1)π§π. ``` Therefore `1(π§β1)(π§β2)=ββπ=0β12π+1π§π+β1βπ=ββ(β1)π§π in {1<|π§|<2}`

What if choosing a different annulus? π is also analytic in {2<|π§|<β}. ```1(π§β1)(π§β2)=(π§β1)β(π§β2)(π§β1)(π§β2)=1(π§β2)β1(π§β1)  =1π§(1β2π§)β1π§(1β1π§)  =1π§ββπ=02π§πβ1π§ββπ=01π§π  =ββπ=12πβ1π§πβββπ=11π§π  =β1βπ=ββ(2βπβ1β1)π§π. ``` Therefore `1(π§β1)(π§β2)=ββπ=12πβ1π§πβββπ=11π§π=β1βπ=ββ(2βπβ1β1)π§π in {2<|π§|<β}`

What if choosing yet another annulus? π is also analytic in {0<|π§β1|<1}.

Since `1(π§β2)=1(π§β1)β1=β11β(π§β1)=βββπ=0(π§β1)π in {0<|π§β1|<1}` So ```1(π§β1)(π§β2)=(π§β1)β(π§β2)(π§β1)(π§β2)=1(π§β2)β1(π§β1)  =βββπ=0(π§β1)πβ1(π§β1)  =ββπ=β1(β1)(π§β1)π. ``` Therefore `1(π§β1)(π§β2)=ββπ=β1(β1)(π§β1)π in {0<|π§β1|<1}`

Another Example

sin π§π§4 is analytic in β\{0}. What is its Laurent series, centered at 0?

Recall: `sin π§=ββπ=0(β1)π(2π+1)!π§2π+1=π§βπ§33!+π§55!βπ§77!+ββ―` so `sin π§π§4=1π§3β13!1π§+15!π§β17!π§3+ββ―` Thus `πβ3=1, πβ2=0, πβ1=β13!, π0=0, π1=15!, π2=0, π3=β17!, β―`

The Coefficients ππ Continued

Recall: For a Taylor series, `π(π§)=ββπ=0ππ(π§βπ§0)π, |π§βπ§0|<π,` the ππ can be calculated via ππ=π(π)(π§0)π!. How about for a Laurent series `π(π§)=ββπ=ββππ(π§βπ§0)π, π<|π§βπ§0|<π?` π may not be defined at π§0, so needed a new approach! Back to Taylor series: `ππ=π(π)(π§0)π!Cauchy=12ππ β«|π§βπ§0|=π π(π§)(π§βπ§0)π+1ππ§` for any π  between 0 and π.

One can show a similar fact for Laurent series: ```TheoremIf π is analytic in {π<|π§βπ§0|<π}, then π(π§)=ββπ=ββππ(π§βπ§0)π, where ππ=12ππ β«|π§βπ§0|=π π(π§)(π§βπ§0)π+1ππ§ for any π  between π and π, and all πββ€. ```

Note: This does not seem all that useful for finding actual values of ππ, but it is useful to estimate ππ. Will using this when calculating integrals later

Isolated singularities

`DefinitionA point π§0 is an isolated singularity of π if π is analytic in a punctured disk {0<|π§βπ§0|<π} centered at π§0`
• π(π§)=1π§ has an isolated singularity at π§0=0.
• π(π§)=1sin π§ has isolated singularities at π§0=0, Β±π, Β±2π, β―.
• π(π§)=π§ and π(π§)=Log π§ do not have isolatd singularities at π§0=0 since these functions cannot be defined to be analytic in any punctured disk around 0.
• π(π§)=1π§β2 has an isolated singularity at π§0=2.

Laurent Series

By Laurent's Theorem, if π has an isolated singularity at π§0 (so π is analytic in the annulus {0<|π§βπ§0|<π} for some π>0) then π has a laurent series expansion there: ```π(π§)=ββπ=ββππ(π§βπ§0)π  =β―+πβ2(π§βπ§0)2+πβ1(π§βπ§0)2 principal part +π0+π1(π§βπ§0)+π2(π§βπ§0)2+β― analytic ``` Three fundamentally different things can happen that influence how π behaves near π§0.

Three Types of Isolated Singularities

`π(π§)=β―+πβ2(π§βπ§0)2+πβ1(π§βπ§0)2+π0+π1(π§βπ§0)+π2(π§βπ§0)2+β―, 0<|π§βπ§0|<π`

Examples

• π(π§)=cos π§β1π§2=1π§2βπ§22!+π§44!β+β―=β12!+π§24!β+β― No negative powers of π§!
• π(π§)=cos π§π§4=1π§41βπ§22!+π§44!β+β―=1π§4β12!1π§2+14!βπ§26!+ββ― Finitely many negative powers of π§!
• π(π§)=cos1π§=1β12!1π§2+14!1π§4β16!1π§6+ββ― Infinitely many negative powers of π§!

Classification of Isolated Singularities

We classify singularities based upon these differences: ```DefinitionSuppose π§0 is an isolated singularity of an analytic function π with Laurent series ββπ=ββππ(π§βπ§0)π, 0<|π§βπ§0|<π. Then the singularity π§0 is removable if ππ=0 for all π<0. a pole if there exist π>0 so that πβπβ 0 but ππ=0 for all π<βπ. The index π is the order of the pole. essential if ππβ 0 for infinitely many π<0. ```

Types of Singularities

The following table illustrates this definition:

π§0 is a β―Laurent series in 0<|π§βπ§0|<π Removable singularityπ0+π1(π§βπ§0)+π2(π§βπ§0)2+β― Pole of order ππβπ(π§βπ§0)π+β―+πβ1(π§βπ§0)2+π0+π1(π§βπ§0)+π2(π§βπ§0)2+β― Simple poleπβ1(π§βπ§0)2+π0+π1(π§βπ§0)+π2(π§βπ§0)2+β― Essential singularityβ―+πβ2(π§βπ§0)2+πβ1(π§βπ§0)2+π0+π1(π§βπ§0)+π2(π§βπ§0)2+β―

Removable Singularities

Recall: π§0 is a removable singularity of π if its Laurent series, centered at π§0 satisfies that ππ=0 for all π<0. `Example: π(π§)=sin π§π§=1βπ§22!+π§44!β+β―, 0<|π§|<β`

The Laurent series looks like a Taylor series! Taylor series are analytic within their region of convergence. Thus, if we define π(π§) to have the value 1 at π§0=0, then π becomes analytic in β: `π(π§)={sin π§π§,π§β 01 π§=0 is analytic in β.` The singularity have then been removed. `Theorem (Riemann's Theorem)Let π§0 be an isolated singularity of π. Then π§0 is a removable singularity if and only if π is bounded near π§0`

Poles

Recall: π§0 is a pole of order π of π if its Laurent series, centered at π§0 satisfies that πβπβ 0 and ππ=0 for all π<βπ. `Example: π(π§)=sin π§π§5=1π§4β13!1π§2+15!β17!π§2+ββ― has a pole of order 4 at 0` `TheoremLet π§0 be an isolated singularity of π. Then π§0 is a pole if and only if |π(π§)|ββ as π§βπ§0.` Note: If π(π§) has a pole at π§0 then 1π(π§) has a removable singularity at π§0 (and vice versa).

Essential Singularities

Recall: π§0 is an essential singularity of π if its Laurent series, centered at π§0 satisfies that ππβ 0 for infinitely many π<0. `Example: π(π§)=β―1π§=ββπ=01π!1π§π=1+1π§+12!1π§2+13!1π§3+β― has an essential singularity at π§0=0.` Note that if π§=π₯ββ, then π(π§)=β―1/π₯ββ as π₯β0 from the right and π(π§)=β―1/π₯β0 as π₯β0 from the left.

Also, if π§=ππ₯βπβ the `π(π§)=β―1/ππ₯=β―βπ/π₯ lies on the unit circle for all π₯` It appears that π does not have a limit as π§βπ§0. `Theorem (Casorati-Weierstraπ½)Suppose that π§0 is an essential singularity of π. Then for every π€0ββ there exists a sequence {π§π} with π§πβπ§0 such that π(π§π)βπ€0 as πββ.`

Casorati-Weierstraπ½

Casorati-Weierstraπ½: Suppose that π§0 is an essential singularity of π. Then for every π€0ββ there exists a sequence {π§π} with π§πβπ§0 such that π(π§π)βπ€0. ```Example: Let π(π§)=β―1π§. Then π has an essential singularity at 0. Let's pick a point π€0ββ, say, π€0ββ=1+3π. By Casorati-Weierstraπ½ there must exist π§πββ\{0} such that β―1π§πβ1+3π as πββ. How do we find π§π Idea: We can find π§π such that β―1π§π=1+3π, namely 1π§π=log(1+3π). Recall: log(π§)=ln(|π§|)+πarg(π§). So log(1+3π)=ln 2+ππ3+2πππ. Pick π§π=1ln 2+ππ3+2πππ Then π§πβ0 as πββ. Furthermore: β―1π§π=β―ln 2+ππ3+2πππ  =2β―ππ3  =212+π32  =1+3π  =π€0 for all π We thus found π§π with π§πβ0 such that π(π§π)=π€0 for all π ```

Picard's Theorem

We just observed a much stronger result that is true (but much harder to prove) for essential singularities: `Theorem (Picard)Suppose that π§0 is an essential singularity of π. Then for every π€0ββ with at most one excepton there exists a sequence {π§π} with π§πβπ§0 such that π(π§π)=π€0.`

Example:

π(π§)=β―1/π§ has an essential singularity at π§0=0. Also, π(π§)β 0 for all π§, and so by Picard's theorem, for every π€0β 0 there must exist infnitely many π§π with π§πβ0 such that π(π§π)=π€0.

Pick π€0=1 for example. Then π(π§)=π€0 if β―1/π§=1, that is 1π§=2πππ for some πββ€. Now let π§π=12πππ. Then π§πβ0 as πββ, and π(π§π)=1 for all π.

ID: 190500008 Last Updated: 2019/5/8 Revision:

Home (5)

Management

HBR (3)

Information

Recreation

Hobbies (7)

Culture

Chinese (1097)

English (336)

Reference (66)

Computer

Hardware (149)

Software

Application (187)

Digitization (24)

Numeric (19)

Programming

Web (618)

CSS (SC)

ASP.NET (SC)

HTML

Knowledge Base

Common Color (SC)

Html 401 Special (SC)

OS (388)

MS Windows

Windows10 (SC)

.NET Framework (SC)

DeskTop (7)

Knowledge

Mathematics

Formulas (8)

Number Theory (206)

Algebra (20)

Trigonometry (18)

Geometry (18)

Calculus (67)

Complex Analysis (21)

Engineering

Tables (8)

Mechanical

Mechanics (1)

Rigid Bodies

Statics (92)

Dynamics (37)

Fluid (5)

Control

Acoustics (19)

Biology (1)

Geography (1)