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Laurent Series
 Review of Taylor Series
 Laurent Series Expansion
 The Coefficients π‘Žπ‘˜
  Example
  Another Example
 The Coefficients π‘Žπ‘˜ Continued
 Isolated singularities
 Laurent Series
 Three Types of Isolated Singularities
 Classification of Isolated Singularities
 Types of Singularities
 Removable Singularities
 Poles
 Essential Singularities
 Casorati-Weierstra𝛽
 Picard's Theorem
  Example:

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

Laurent Series

Review of Taylor Series

Recall: If 𝑓:π‘ˆβ†’β„‚ is analytic and {|π‘§βˆ’π‘§0|<𝑅}βŠ‚π‘ˆ then 𝑓 has representation 𝑓(𝑧)=βˆžβˆ‘π‘˜=0π‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜, where π‘Žπ‘˜=𝑓(π‘˜)(𝑧0)π‘˜!, π‘˜β‰₯0.

  • What if 𝑓 is not differentiable at some point?
  • Example: 𝑓(𝑧)=𝑧𝑧2+4 is not differentiable at 𝑧=Β±2𝑖 (undefined there).
  • 𝑓(𝑧)=Log 𝑧 not continuous on (βˆ’βˆž,0], so not differentiable there.

Laurent Series Expansion

Theorem (Laurent Series Expansion)If 𝑓:π‘ˆβ†’β„‚ is analytic and {π‘Ÿ<|π‘§βˆ’π‘§0|<𝑅}βŠ‚π‘ˆ then 𝑓 has a Laurent series expansion 𝑓(𝑧)=βˆžβˆ‘π‘˜=βˆ’βˆžπ‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜ =β‹―+π‘Žβˆ’2(π‘§βˆ’π‘§0)2+π‘Žβˆ’1(π‘§βˆ’π‘§0)2+π‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹―, that converges at each point of the annulus and converges absolutely and uniformly in each sub annulus {𝑠≀|π‘§βˆ’π‘§0|≀𝑑}, where π‘Ÿ<𝑠<𝑑<𝑅.

The Coefficients π‘Žπ‘˜

Note: The coefficients π‘Žπ‘˜ are uniquely determined by 𝑓. How do we find them?

Example

𝑓(𝑧)=1(π‘§βˆ’1)(π‘§βˆ’2) is analytic in β„‚\{1,2}.

Let's find the Laurent series in the annulus {1<|𝑧|<2}. Trick: 1(π‘§βˆ’1)(π‘§βˆ’2)=(π‘§βˆ’1)βˆ’(π‘§βˆ’2)(π‘§βˆ’1)(π‘§βˆ’2)=1(π‘§βˆ’2)βˆ’1(π‘§βˆ’1)  =βˆ’12
1
1βˆ’π‘§2
βˆ’
1
𝑧(1βˆ’1𝑧)
 =βˆ’12βˆžβˆ‘π‘˜=0𝑧2π‘˜βˆ’1π‘§βˆžβˆ‘π‘˜=01π‘§π‘˜  =βˆžβˆ‘π‘˜=0βˆ’12π‘˜+1π‘§π‘˜+βˆžβˆ‘π‘˜=1βˆ’1π‘§π‘˜  =βˆžβˆ‘π‘˜=0βˆ’12π‘˜+1π‘§π‘˜+βˆ’1βˆ‘π‘˜=βˆ’βˆž(βˆ’1)π‘§π‘˜.
Therefore 1(π‘§βˆ’1)(π‘§βˆ’2)=βˆžβˆ‘π‘˜=0βˆ’12π‘˜+1π‘§π‘˜+βˆ’1βˆ‘π‘˜=βˆ’βˆž(βˆ’1)π‘§π‘˜ in {1<|𝑧|<2}

What if choosing a different annulus? 𝑓 is also analytic in {2<|𝑧|<∞}. 1(π‘§βˆ’1)(π‘§βˆ’2)=(π‘§βˆ’1)βˆ’(π‘§βˆ’2)(π‘§βˆ’1)(π‘§βˆ’2)=1(π‘§βˆ’2)βˆ’1(π‘§βˆ’1)  =
1
𝑧(1βˆ’2𝑧)
βˆ’
1
𝑧(1βˆ’1𝑧)
 =1π‘§βˆžβˆ‘π‘˜=02π‘§π‘˜βˆ’1π‘§βˆžβˆ‘π‘˜=01π‘§π‘˜  =βˆžβˆ‘π‘˜=12π‘˜βˆ’1π‘§π‘˜βˆ’βˆžβˆ‘π‘˜=11π‘§π‘˜  =βˆ’1βˆ‘π‘˜=βˆ’βˆž(2βˆ’π‘˜βˆ’1βˆ’1)π‘§π‘˜.
Therefore 1(π‘§βˆ’1)(π‘§βˆ’2)=βˆžβˆ‘π‘˜=12π‘˜βˆ’1π‘§π‘˜βˆ’βˆžβˆ‘π‘˜=11π‘§π‘˜=βˆ’1βˆ‘π‘˜=βˆ’βˆž(2βˆ’π‘˜βˆ’1βˆ’1)π‘§π‘˜ in {2<|𝑧|<∞}

What if choosing yet another annulus? 𝑓 is also analytic in {0<|π‘§βˆ’1|<1}.

Since 1(π‘§βˆ’2)=1(π‘§βˆ’1)βˆ’1=βˆ’11βˆ’(π‘§βˆ’1)=βˆ’βˆžβˆ‘π‘˜=0(π‘§βˆ’1)π‘˜ in {0<|π‘§βˆ’1|<1} So 1(π‘§βˆ’1)(π‘§βˆ’2)=(π‘§βˆ’1)βˆ’(π‘§βˆ’2)(π‘§βˆ’1)(π‘§βˆ’2)=1(π‘§βˆ’2)βˆ’1(π‘§βˆ’1)  =βˆ’βˆžβˆ‘π‘˜=0(π‘§βˆ’1)π‘˜βˆ’1(π‘§βˆ’1)  =βˆžβˆ‘π‘˜=βˆ’1(βˆ’1)(π‘§βˆ’1)π‘˜. Therefore 1(π‘§βˆ’1)(π‘§βˆ’2)=βˆžβˆ‘π‘˜=βˆ’1(βˆ’1)(π‘§βˆ’1)π‘˜ in {0<|π‘§βˆ’1|<1}

Another Example

sin 𝑧𝑧4 is analytic in β„‚\{0}. What is its Laurent series, centered at 0?

Recall: sin 𝑧=βˆžβˆ‘π‘˜=0(βˆ’1)π‘˜(2π‘˜+1)!𝑧2π‘˜+1=π‘§βˆ’π‘§33!+𝑧55!βˆ’π‘§77!+βˆ’β‹― so sin 𝑧𝑧4=1𝑧3βˆ’13!1𝑧+15!π‘§βˆ’17!𝑧3+βˆ’β‹― Thus π‘Žβˆ’3=1, π‘Žβˆ’2=0, π‘Žβˆ’1=βˆ’13!, π‘Ž0=0, π‘Ž1=15!, π‘Ž2=0, π‘Ž3=βˆ’17!, β‹―

The Coefficients π‘Žπ‘˜ Continued

Recall: For a Taylor series, 𝑓(𝑧)=βˆžβˆ‘π‘˜=0π‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜, |π‘§βˆ’π‘§0|<𝑅, the π‘Žπ‘˜ can be calculated via π‘Žπ‘˜=𝑓(π‘˜)(𝑧0)π‘˜!. How about for a Laurent series 𝑓(𝑧)=βˆžβˆ‘π‘˜=βˆ’βˆžπ‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜, π‘Ÿ<|π‘§βˆ’π‘§0|<𝑅? 𝑓 may not be defined at 𝑧0, so needed a new approach! Back to Taylor series: π‘Žπ‘˜=𝑓(π‘˜)(𝑧0)π‘˜!Cauchy=12πœ‹π‘– βˆ«|π‘§βˆ’π‘§0|=𝑠𝑓(𝑧)(π‘§βˆ’π‘§0)π‘˜+1𝑑𝑧 for any 𝑠 between 0 and 𝑅.

One can show a similar fact for Laurent series: TheoremIf 𝑓 is analytic in {π‘Ÿ<|π‘§βˆ’π‘§0|<𝑅}, then 𝑓(𝑧)=βˆžβˆ‘π‘˜=βˆ’βˆžπ‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜, where π‘Žπ‘˜=12πœ‹π‘– βˆ«|π‘§βˆ’π‘§0|=𝑠𝑓(𝑧)(π‘§βˆ’π‘§0)π‘˜+1𝑑𝑧 for any 𝑠 between π‘Ÿ and 𝑅, and all π‘˜βˆˆβ„€.

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

Isolated singularities

DefinitionA point 𝑧0 is an isolated singularity of 𝑓 if 𝑓 is analytic in a punctured disk {0<|π‘§βˆ’π‘§0|<π‘Ÿ} centered at 𝑧0
  • 𝑓(𝑧)=1𝑧 has an isolated singularity at 𝑧0=0.
  • 𝑓(𝑧)=1sin 𝑧 has isolated singularities at 𝑧0=0, Β±πœ‹, Β±2πœ‹, β‹―.
  • 𝑓(𝑧)=𝑧 and 𝑓(𝑧)=Log 𝑧 do not have isolatd singularities at 𝑧0=0 since these functions cannot be defined to be analytic in any punctured disk around 0.
  • 𝑓(𝑧)=1π‘§βˆ’2 has an isolated singularity at 𝑧0=2.

Laurent Series

By Laurent's Theorem, if 𝑓 has an isolated singularity at 𝑧0 (so 𝑓 is analytic in the annulus {0<|π‘§βˆ’π‘§0|<π‘Ÿ} for some π‘Ÿ>0) then 𝑓 has a laurent series expansion there: 𝑓(𝑧)=βˆžβˆ‘π‘˜=βˆ’βˆžπ‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜  =β‹―+π‘Žβˆ’2(π‘§βˆ’π‘§0)2+π‘Žβˆ’1(π‘§βˆ’π‘§0)2 principal part +π‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹― analytic Three fundamentally different things can happen that influence how 𝑓 behaves near 𝑧0.

Three Types of Isolated Singularities

𝑓(𝑧)=β‹―+π‘Žβˆ’2(π‘§βˆ’π‘§0)2+π‘Žβˆ’1(π‘§βˆ’π‘§0)2+π‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹―, 0<|π‘§βˆ’π‘§0|<π‘Ÿ

Examples

  • 𝑓(𝑧)=cos π‘§βˆ’1𝑧2=1𝑧2βˆ’π‘§22!+𝑧44!βˆ’+β‹―=βˆ’12!+𝑧24!βˆ’+β‹― No negative powers of 𝑧!
  • 𝑓(𝑧)=cos 𝑧𝑧4=1𝑧41βˆ’π‘§22!+𝑧44!βˆ’+β‹―=1𝑧4βˆ’12!1𝑧2+14!βˆ’π‘§26!+βˆ’β‹― Finitely many negative powers of 𝑧!
  • 𝑓(𝑧)=cos1𝑧=1βˆ’12!1𝑧2+14!1𝑧4βˆ’16!1𝑧6+βˆ’β‹― Infinitely many negative powers of 𝑧!

Classification of Isolated Singularities

We classify singularities based upon these differences: DefinitionSuppose 𝑧0 is an isolated singularity of an analytic function 𝑓 with Laurent series βˆžβˆ‘π‘˜=βˆ’βˆžπ‘Žπ‘˜(π‘§βˆ’π‘§0)π‘˜, 0<|π‘§βˆ’π‘§0|<π‘Ÿ. Then the singularity 𝑧0 is removable if π‘Žπ‘˜=0 for all π‘˜<0. a pole if there exist 𝑁>0 so that π‘Žβˆ’π‘β‰ 0 but π‘Žπ‘˜=0 for all π‘˜<βˆ’π‘. The index 𝑁 is the order of the pole. essential if π‘Žπ‘˜β‰ 0 for infinitely many π‘˜<0.

Types of Singularities

The following table illustrates this definition:

𝑧0 is a β‹―Laurent series in 0<|π‘§βˆ’π‘§0|<π‘Ÿ Removable singularityπ‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹― Pole of order π‘π‘Žβˆ’π‘(π‘§βˆ’π‘§0)𝑁+β‹―+π‘Žβˆ’1(π‘§βˆ’π‘§0)2+π‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹― Simple poleπ‘Žβˆ’1(π‘§βˆ’π‘§0)2+π‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹― Essential singularityβ‹―+π‘Žβˆ’2(π‘§βˆ’π‘§0)2+π‘Žβˆ’1(π‘§βˆ’π‘§0)2+π‘Ž0+π‘Ž1(π‘§βˆ’π‘§0)+π‘Ž2(π‘§βˆ’π‘§0)2+β‹―

Removable Singularities

Recall: 𝑧0 is a removable singularity of 𝑓 if its Laurent series, centered at 𝑧0 satisfies that π‘Žπ‘˜=0 for all π‘˜<0. Example: 𝑓(𝑧)=sin 𝑧𝑧=1βˆ’π‘§22!+𝑧44!βˆ’+β‹―, 0<|𝑧|<∞

The Laurent series looks like a Taylor series! Taylor series are analytic within their region of convergence. Thus, if we define 𝑓(𝑧) to have the value 1 at 𝑧0=0, then 𝑓 becomes analytic in β„‚: 𝑓(𝑧)={sin 𝑧𝑧,𝑧≠01 𝑧=0 is analytic in β„‚. The singularity have then been removed. Theorem (Riemann's Theorem)Let 𝑧0 be an isolated singularity of 𝑓. Then 𝑧0 is a removable singularity if and only if 𝑓 is bounded near 𝑧0

Poles

Recall: 𝑧0 is a pole of order 𝑁 of 𝑓 if its Laurent series, centered at 𝑧0 satisfies that π‘Žβˆ’π‘β‰ 0 and π‘Žπ‘˜=0 for all π‘˜<βˆ’π‘. Example: 𝑓(𝑧)=sin 𝑧𝑧5=1𝑧4βˆ’13!1𝑧2+15!βˆ’17!𝑧2+βˆ’β‹― has a pole of order 4 at 0 TheoremLet 𝑧0 be an isolated singularity of 𝑓. Then 𝑧0 is a pole if and only if |𝑓(𝑧)|β†’βˆž as 𝑧→𝑧0. Note: If 𝑓(𝑧) has a pole at 𝑧0 then 1𝑓(𝑧) has a removable singularity at 𝑧0 (and vice versa).

Essential Singularities

Recall: 𝑧0 is an essential singularity of 𝑓 if its Laurent series, centered at 𝑧0 satisfies that π‘Žπ‘˜β‰ 0 for infinitely many π‘˜<0. Example: 𝑓(𝑧)=β„―1𝑧=βˆžβˆ‘π‘˜=01π‘˜!1π‘§π‘˜=1+1𝑧+12!1𝑧2+13!1𝑧3+β‹― has an essential singularity at 𝑧0=0. Note that if 𝑧=π‘₯βˆˆβ„, then 𝑓(𝑧)=β„―1/π‘₯β†’βˆž as π‘₯β†’0 from the right and 𝑓(𝑧)=β„―1/π‘₯β†’0 as π‘₯β†’0 from the left.

Also, if 𝑧=𝑖π‘₯βˆˆπ‘–β„ the 𝑓(𝑧)=β„―1/𝑖π‘₯=β„―βˆ’π‘–/π‘₯ lies on the unit circle for all π‘₯ It appears that 𝑓 does not have a limit as 𝑧→𝑧0. Theorem (Casorati-Weierstra𝛽)Suppose that 𝑧0 is an essential singularity of 𝑓. Then for every 𝑀0βˆˆβ„‚ there exists a sequence {𝑧𝑛} with 𝑧𝑛→𝑧0 such that 𝑓(𝑧𝑛)→𝑀0 as π‘›β†’βˆž.

Casorati-Weierstra𝛽

Casorati-Weierstra𝛽: Suppose that 𝑧0 is an essential singularity of 𝑓. Then for every 𝑀0βˆˆβ„‚ there exists a sequence {𝑧𝑛} with 𝑧𝑛→𝑧0 such that 𝑓(𝑧𝑛)→𝑀0. Example: Let 𝑓(𝑧)=β„―1𝑧. Then 𝑓 has an essential singularity at 0. Let's pick a point 𝑀0βˆˆβ„‚, say, 𝑀0βˆˆβ„‚=1+3𝑖. By Casorati-Weierstra𝛽 there must exist π‘§π‘›βˆˆβ„‚\{0} such that β„―1𝑧𝑛→1+3𝑖 as π‘›β†’βˆž. How do we find 𝑧𝑛 Idea: We can find 𝑧𝑛 such that β„―1𝑧𝑛=1+3𝑖, namely 1𝑧𝑛=log(1+3𝑖). Recall: log(𝑧)=ln(|𝑧|)+𝑖arg(𝑧). So log(1+3𝑖)=ln 2+π‘–πœ‹3+2π‘›πœ‹π‘–. Pick 𝑧𝑛=
1
ln 2+π‘–πœ‹3+2π‘›πœ‹π‘–
Then 𝑧𝑛→0 as π‘›β†’βˆž. Furthermore: β„―1𝑧𝑛=β„―ln 2+π‘–πœ‹3+2π‘›πœ‹π‘–  =2β„―π‘–πœ‹3  =212+𝑖32  =1+3𝑖  =𝑀0 for all 𝑛 We thus found 𝑧𝑛 with 𝑧𝑛→0 such that 𝑓(𝑧𝑛)=𝑀0 for all 𝑛

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 𝑧0 is an essential singularity of 𝑓. Then for every 𝑀0βˆˆβ„‚ with at most one excepton there exists a sequence {𝑧𝑛} with 𝑧𝑛→𝑧0 such that 𝑓(𝑧𝑛)=𝑀0.

Example:

𝑓(𝑧)=β„―1/𝑧 has an essential singularity at 𝑧0=0. Also, 𝑓(𝑧)β‰ 0 for all 𝑧, and so by Picard's theorem, for every 𝑀0β‰ 0 there must exist infnitely many 𝑧𝑛 with 𝑧𝑛→0 such that 𝑓(𝑧𝑛)=𝑀0.

Pick 𝑀0=1 for example. Then 𝑓(𝑧)=𝑀0 if β„―1/𝑧=1, that is 1𝑧=2π‘›πœ‹π‘– for some π‘›βˆˆβ„€. Now let 𝑧𝑛=12π‘›πœ‹π‘–. Then 𝑧𝑛→0 as π‘›β†’βˆž, and 𝑓(𝑧𝑛)=1 for all 𝑛.


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