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Residue Theorem
 Motivation
 The Residue
  More Residue Examples
 The Residue Theorem
  Example
 Recall the Residue Theorem
 Residues at Removable Singularities
 Residues at Simple Poles
 Residues at Double Poles
  Example
 Residues at Poles of Order 𝑛
 More On Residues
 Evaluating Integrals via the Residue Theorem
 More Examples
 Evaluating an Improper Integral via the Residue Theorem
  An Improper Integral

source/reference:
https://www.youtube.com/channel/UCaTLkDn9_1Wy5TRYfVULYUw/playlists

Residue Theorem

Motivation

Recall: 𝑓 has an isolated singularity at 𝑧0 if 𝑓 is analytic in {0<|𝑧−𝑧0|<𝑟} for some 𝑟>0. In that case, 𝑓 has a Laurent series expansion 𝑓(𝑧)=𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟. Observe: If 0<𝜌<𝑟 then  |𝑧−𝑧0|=𝜌𝑓(𝑧)𝑑𝑧=𝑘=−∞𝑎𝑘 |𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧

What is  |𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧? For 𝑘≠−1, the function ℎ(𝑧)=(𝑧−𝑧0)𝑘 has a primitive, namely 𝐻(𝑧)=1𝑘+1(𝑧−𝑧0)𝑘+1. Therefore,  |𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧=0 for 𝑘≠−1 For 𝑘=−1, the integral is  |𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧. We can use the Cauchy Integral Formula (or compute this directly) and find  𝛾𝑓(𝑧)𝑑𝑧=𝑏𝑎𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡  |𝑧−𝑧0|=𝜌1𝑧−𝑧0���𝑧=2𝜋01𝑧0+𝜌ℯ𝑖𝑡−𝑧0⋅𝜌ℯ𝑖𝑡𝑑𝑡  =2𝜋0𝑖𝑑𝑡=2𝜋𝑖  |𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧=2𝜋𝑖 for 𝑘=−1 Hence  |𝑧−𝑧0|=𝜌𝑓(𝑧)𝑑𝑧=𝑘=−∞𝑎𝑘 |𝑧−𝑧0|=𝜌(𝑧−𝑧0)𝑘𝑑𝑧=2𝜋𝑖𝑎−1. Therefore, 𝑎−1 gets special attention!

The Residue

DefinitionIf 𝑓 has an isolated singularity at 𝑧0 with Laurent series expansion 𝑓(𝑧)=𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟, then the residue of 𝑓 at 𝑧0 is Res(𝑓,𝑧0)=𝑎−1.

Examples of finding a residue: to find Laurent series centered at the singular point and read off the term 𝑎−1. 𝑓(𝑧)=1(𝑧−1)(𝑧−2)=−1𝑧−1+𝑘=0(−1)(𝑧−1)𝑘 in 0<|𝑧−1|<1. Therefore, Res(𝑓,1)=−1 𝑓(𝑧)=1(𝑧−1)(𝑧−2)=1𝑧−2𝑛=0(−1)𝑛(𝑧−2)𝑛 in 0<|𝑧−2|<1. Therefore, Res(𝑓,2)=1

More Residue Examples

More examples: 𝑓(𝑧)=sin 𝑧𝑧4=1𝑧313!1𝑧+15!𝑧−17!𝑧3+−⋯ in 0<|𝑧|<∞. Therefore, Res(𝑓,0)=−13!=−16. 𝑓(𝑧)=cos1𝑧=1−12!1𝑧2+14!1𝑧416!1𝑧6+−⋯. Therefore Res(𝑓,0)=0 𝑓(𝑧)=sin1𝑧=1𝑧13!1𝑧3+15!1𝑧5−+⋯. Therefore Res(𝑓,0)=1 𝑓(𝑧)=cos 𝑧−1𝑧2=−12!+𝑧24!−+⋯. Therefore Res(𝑓,0)=0 𝑓(𝑧)=1(𝑧2+1=1(𝑧−𝑖)(𝑧+𝑖)=12!(𝑧+𝑖)−(𝑧−𝑖)(𝑧−𝑖)(𝑧+𝑖)  =12!1(𝑧−𝑖)1(𝑧+𝑖)=12𝑖1(𝑧−𝑖)+(analytic function near 𝑖). Therefore, Res(𝑓,𝑖)=12𝑖=−12𝑖. Similarly, 𝑓(𝑧)=1(𝑧2+1=−12𝑖1(𝑧+𝑖)+(analytic function near −𝑖). Therefore, Res(𝑓,𝑖)=−12𝑖=12𝑖.

The Residue Theorem

Theorem (Residue Theorem)Let 𝐷 be a simply connected domain, and let 𝑓 be analytic in 𝐷, except for isolated singularities. Let 𝐶 be a simple closed curve in 𝐷 (oriented counterclockwise), and let 𝑧1,⋯,𝑧𝑛 be those isolated singularities of 𝑓 that lie inside of 𝐶. Then  𝐶𝑓(𝑧)𝑑𝑧=2𝜋𝑖𝑛𝑘=1Res(𝑓,𝑧𝑘).

Example

𝑓(𝑧)=1(𝑧2+1 is analytic in 𝐷=𝐶, except for isolated singularities at 𝑧=±𝑖.  𝐶1𝑓(𝑧)𝑑𝑧=2𝜋𝑖Res(𝑓,𝑖)=2𝜋𝑖(−12𝑖)=𝜋  𝐶2𝑓(𝑧)𝑑𝑧=2𝜋𝑖Res(𝑓,−𝑖)=2𝜋𝑖(12𝑖)=−𝜋  𝐶3𝑓(𝑧)𝑑𝑧=2𝜋𝑖(Res(𝑓,𝑖)+Res(𝑓,−𝑖))=2𝜋𝑖(−12𝑖+12𝑖)=−𝜋  𝐶4𝑓(𝑧)𝑑𝑧=0 In order to be able to fully take advantage of this powerful theorem, strategies and techniques that can help calculating residues are needed

Recall the Residue Theorem

Recall: 𝑓 has an isolated singularity at 𝑧0 if 𝑓 is analytic in the punctured disk {0<|𝑧−𝑧0|<𝑟}. In that case, 𝑓 has a Laurent series representation 𝑓(𝑧)=𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘 in this punctured disk. The representation is unique. The residue of 𝑓 at 𝑧0 is Res(𝑓,𝑧0)=𝑎−1, the coefficient of term 1𝑧−𝑧0.

Residues at Removable Singularities

𝑓(𝑧)=𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟.

Recall: 𝑧0 is a removable singularity if 𝑎𝑘=0 for all 𝑘<0. In particular: 𝑎−1=0 in that case, so that Res(𝑓,𝑧0)=0. Example: 𝑓(𝑧)=sin 𝑧𝑧=1𝑧𝑛=0(−1)𝑛(2𝑛+1)!𝑧2𝑛+1  =1𝑧𝑧−13!𝑧3+15!𝑧517!𝑧7+−⋯  =1−13!𝑧2+15!𝑧417!𝑧6+−⋯ Thus Res(𝑓,0)=0

Residues at Simple Poles

Example: Res(𝑓,𝑧0)= Lim𝑧→𝑧0(𝑧−𝑧0)𝑓(𝑧).

Example: 𝑓(𝑧)=1𝑧2+1 has a simple pole at 𝑧0=𝑖 (and another one at −𝑖). Res1𝑧2+1,𝑖= Lim𝑧→𝑖(𝑧−𝑖)1𝑧2+1  = Lim𝑧→𝑖(𝑧−𝑖)1(𝑧−𝑖)(𝑧+𝑖)  = Lim𝑧→𝑖(𝑧−𝑖)1(𝑧+𝑖)=12𝑖=−𝑖2

Residues at Double Poles

𝑓(𝑧)=𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟.

Recall: 𝑧0 is a double pole if 𝑎−2≠0 and 𝑎𝑘=0 for all 𝑘≤−3. So 𝑓(𝑧)=𝑎−2(𝑧−𝑧0)2+𝑎𝑧−𝑧0+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ How do we isolate 𝑎? Idea: (𝑧−𝑧0)2𝑓(𝑧)=𝑎−2+𝑎(𝑧−𝑧0)+𝑎0(𝑧−𝑧0)2+⋯, so that 𝑑𝑑𝑧(𝑧−𝑧0)2𝑓(𝑧)=𝑎+2𝑎0(𝑧−𝑧0)+⋯, Hence Res(𝑓,𝑧0)=𝑎= Lim𝑧→𝑧0𝑑𝑑𝑧(𝑧−𝑧0)2𝑓(𝑧)

Example

𝑓(𝑧)=1(𝑧−1)2(𝑧−3) has a double pole at 𝑧0=1 (and a simple one at 3). Res1(𝑧−1)2(𝑧−3),1= Lim𝑧→1𝑑𝑑𝑧(𝑧−1)21(𝑧−1)2(𝑧−3)  = Lim𝑧→1𝑑𝑑𝑧1(𝑧−3)  = Lim𝑧→1−1(𝑧−3)2=−14.

Residues at Poles of Order 𝑛

𝑓(𝑧)=𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟.

Recall: 𝑧0 is a pole of order 𝑛 if 𝑎−𝑛≠0 and 𝑎𝑘=0 for all 𝑘≤−(𝑛+1). 𝑓(𝑧)=𝑎−𝑛(𝑧−𝑧0)𝑛+⋯+𝑎𝑧−𝑧0+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ Then Res(𝑓,𝑧0)=𝑎=1(𝑛−<)!Lim𝑧→𝑧0𝑑𝑛−𝑑𝑧𝑛−(𝑧−𝑧0)𝑛𝑓(𝑧)

More On Residues

RemarkIf 𝑓(𝑧)=𝑔(𝑧)ℎ(𝑧), where 𝑔 and ℎ are analytic near 𝑧0, and ℎ has a simple zero at 𝑧0, then Res(𝑓(𝑧),𝑧0)=𝑔(𝑧0)ℎ′(𝑧0).

Example: 𝑓(𝑧)=1(𝑧−1)2(𝑧−3), choose 𝑔(𝑧)=1(𝑧−1)2 and ℎ(𝑧)=(𝑧−3). Then 𝑔 and ℎ are analytic near 𝑧0=3, and ℎ has a simple zero at 𝑧0=3. Thus Res(𝑓,𝑧0)=𝑔(3)ℎ′(3)=1(3−1)21
 
=14

Evaluating Integrals via the Residue Theorem

Recall: The Residue Theorem  𝐶𝑓(𝑧)𝑑𝑧=2𝜋𝑖𝑛𝑘=1Res(𝑓,𝑧𝑘)

Examples:  |𝑧|=13𝑧𝑑𝑧=2𝜋𝑖Res(𝑓,0), where 𝑓(𝑧)=ℯ3𝑧=𝑘=11𝑘!3𝑧𝑘. Thus Res(𝑓,0)=3, so that  |𝑧|=13𝑧𝑑𝑧=6𝜋𝑖  |𝑧|=2tan 𝑧𝑑𝑧=2𝜋𝑖Res(𝑓,𝜋2)+Res(𝑓,−𝜋2), where 𝑓(𝑧)=tan 𝑧=sin 𝑧cos 𝑧. To find Res(𝑓,𝜋2): Note that 𝑓(𝑧)=𝑔(𝑧)ℎ(𝑧), where 𝑔(𝑧)=sin(𝑧) and ℎ(𝑧)=cos(𝑧) are analytic near 𝜋2 and ℎ(𝜋2)=0. Thus Res(𝑓,𝜋2)=𝑔(𝜋2)ℎ′(𝜋2)=sin(𝜋2)sin(𝜋2)=−1 Similarly, Res(𝑓,−𝜋2)=𝑔(−𝜋2)ℎ′(−𝜋2)=sin(−𝜋2)sin(−𝜋2)=−1−(−1)=−1 Thus  |𝑧|=2tan 𝑧𝑑𝑧=2𝜋𝑖(−1−1)=−4𝜋𝑖.  𝐶11(𝑧−1)2(𝑧−3)𝑑𝑧=2𝜋𝑖Res(𝑓,1). Res(𝑓,1)= Lim𝑧→1𝑑𝑑𝑧(𝑧−1)2(𝑧−1)2(𝑧−3)  = Lim𝑧→1𝑑𝑑𝑧1𝑧−3= Lim𝑧→1−1(𝑧−3)2=−14 Thus  𝐶11(𝑧−1)2(𝑧−3)𝑑𝑧=−142𝜋𝑖=−𝜋𝑖2.  𝐶21(𝑧−1)2(𝑧−3)𝑑𝑧=2𝜋𝑖(Res(𝑓,1)+Res(𝑓,3)) Res(𝑓,3)= Lim𝑧→3(𝑧−3)(𝑧−1)2(𝑧−3)= Lim𝑧→31(𝑧−1)2=14

More Examples

The Residue Theorem can also be used to evaluate real integrals, for example of the following forms: 2𝜋0𝑅(cos 𝑡,sin 𝑡)𝑑𝑡, where 𝑅(𝑥,𝑦) is a rational function of the real variables 𝑥 and 𝑦. −∞𝑓(𝑥)𝑑𝑥, where 𝑓 is a rational function of 𝑥. −∞𝑓(𝑥)cos(𝛼𝑥)𝑑𝑥, where 𝑓 is a rational function of 𝑥. −∞𝑓(𝑥)sin(𝛼𝑥)𝑑𝑥, where 𝑓 is a rational function of 𝑥.

Evaluating an Improper Integral via the Residue Theorem

GoalEvaluate 0cos 𝑥1+𝑥2𝑑𝑥.

An Improper Integral

Note: 0⋯𝑑𝑥 means Lim𝑅→∞𝑅0⋯𝑑𝑥, so we need to consider 𝑅0cos 𝑥1+𝑥2𝑑𝑥 and then let 𝑅→∞. Idea: 𝑅0cos 𝑥1+𝑥2𝑑𝑥=12𝑅−𝑅cos 𝑥1+𝑥2𝑑𝑥  =12𝑅−𝑅cos 𝑥+𝑖sin 𝑥1+𝑥2𝑑𝑥  =12𝑅−𝑅𝑖𝑥1+𝑥2𝑑𝑥. Idea: 12𝑅−𝑅𝑖𝑥1+𝑥2𝑑𝑥= 12 [−𝑅,𝑅]𝑖𝑧1+𝑧2𝑑𝑧  =12 𝐶𝑅𝑖𝑧1+𝑧2𝑑𝑧−12 Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧  =122𝜋𝑖Res𝑖𝑧1+𝑧2,𝑖12 Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧 we thus need to find the residue of 𝑓(𝑧)=𝑖𝑧1+𝑧2 at 𝑧0=𝑖 and estimate the integral over Γ𝑅. Finding the residue of 𝑓(𝑧)=𝑖𝑧1+𝑧2 at 𝑧0=𝑖 𝑓 has a simple pole at 𝑧=𝑖. Thus Res(𝑓,𝑖)= Lim𝑧→𝑖(𝑧−𝑖)𝑓(𝑧)= Lim𝑧→𝑖(𝑧−𝑖)ℯ𝑖𝑧1+𝑧2= Lim𝑧→𝑖𝑖𝑧𝑧+𝑖=-12𝑖. Hence 12 𝐶𝑅𝑖𝑧1+𝑧2𝑑𝑧=122𝜋𝑖12𝑖ℯ=𝜋2ℯ. Estimating 12 Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧 We are only interested in what happens as 𝑅→∞ Want to show 12 Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧→0 as 𝑅→∞ Therefore, it suffices to show that  Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧const(𝑅), where the constant, const(𝑅) goes to zero as 𝑅→∞ Recall:  Γ𝑅𝑓(𝑧)𝑑𝑧length𝑅)⋅ max𝑧∈Γ𝑅|𝑓(𝑧)|. Thus:  Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧length𝑅)⋅ max𝑧∈Γ𝑅𝑖𝑧1+𝑧2. 𝑖𝑧1+𝑧2=Re(𝑖𝑧)|1+𝑧2|=−𝑦|1+𝑧2|−𝑦𝑅2−11𝑅2−1 for 𝑧∈Γ𝑅, since |𝑧|=𝑅 and 𝑦≥0 on Γ𝑅. So  Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧≤𝜋𝑅1𝑅2−1→0 as 𝑅→∞ Thus  Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧→0 as 𝑅→∞ To find: 0cos 𝑥1+𝑥2𝑑𝑥. 0cos 𝑥1+𝑥2𝑑𝑥=Lim𝑅→∞𝑅0cos 𝑥1+𝑥2𝑑𝑥 𝑅0cos 𝑥1+𝑥2𝑑𝑥=12 𝐶𝑅𝑖𝑧1+𝑧2𝑑𝑧−12 Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧 12 𝐶𝑅𝑖𝑧1+𝑧2𝑑𝑧=12⋅2𝜋𝑖Res𝑖𝑧1+𝑧2,𝑖=𝜋2ℯ  Γ𝑅𝑖𝑧1+𝑧2𝑑𝑧→0 as 𝑅→∞ Hence 0cos 𝑥1+𝑥2𝑑𝑥=𝜋2ℯ.


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