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```Laurent Series Review of Taylor Series Laurent Series Expansion The Coefficients 𝑎𝑘  Example  Another Example The Coefficients 𝑎𝑘 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:```

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# Laurent Series

## Review of Taylor Series

Recall: If 𝑓:𝑈→ℂ is analytic and {|𝑧−𝑧0|<𝑅}⊂𝑈 then 𝑓 has representation `𝑓(𝑧)=∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘, where 𝑎𝑘=𝑓(𝑘)(𝑧0)𝑘!, 𝑘≥0.`

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

## Laurent Series Expansion

```Theorem (Laurent Series Expansion)If 𝑓:𝑈→ℂ is analytic and {𝑟<|𝑧−𝑧0|<𝑅}⊂𝑈 then 𝑓 has a Laurent series expansion 𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘 =⋯+𝑎−2(𝑧−𝑧0)2+𝑎−1(𝑧−𝑧0)2+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯, that converges at each point of the annulus and converges absolutely and uniformly in each sub annulus {𝑠≤|𝑧−𝑧0|≤𝑡}, where 𝑟<𝑠<𝑡<𝑅.```

## The Coefficients 𝑎𝑘

Note: The coefficients 𝑎𝑘 are uniquely determined by 𝑓. How do we find them?

### Example

𝑓(𝑧)=1(𝑧−1)(𝑧−2) is analytic in ℂ\{1,2}.

Let's find the Laurent series in the annulus {1<|𝑧|<2}. Trick: ```1(𝑧−1)(𝑧−2)=(𝑧−1)−(𝑧−2)(𝑧−1)(𝑧−2)=1(𝑧−2)−1(𝑧−1)  =−1211−𝑧2−1𝑧(1−1𝑧)  =−12∞∑𝑘=0𝑧2𝑘−1𝑧∞∑𝑘=01𝑧𝑘  =∞∑𝑘=0−12𝑘+1𝑧𝑘+∞∑𝑘=1−1𝑧𝑘  =∞∑𝑘=0−12𝑘+1𝑧𝑘+−1∑𝑘=−∞(−1)𝑧𝑘. ``` Therefore `1(𝑧−1)(𝑧−2)=∞∑𝑘=0−12𝑘+1𝑧𝑘+−1∑𝑘=−∞(−1)𝑧𝑘 in {1<|𝑧|<2}`

What if choosing a different annulus? 𝑓 is also analytic in {2<|𝑧|<∞}. ```1(𝑧−1)(𝑧−2)=(𝑧−1)−(𝑧−2)(𝑧−1)(𝑧−2)=1(𝑧−2)−1(𝑧−1)  =1𝑧(1−2𝑧)−1𝑧(1−1𝑧)  =1𝑧∞∑𝑘=02𝑧𝑘−1𝑧∞∑𝑘=01𝑧𝑘  =∞∑𝑘=12𝑘−1𝑧𝑘−∞∑𝑘=11𝑧𝑘  =−1∑𝑘=−∞(2−𝑘−1−1)𝑧𝑘. ``` Therefore `1(𝑧−1)(𝑧−2)=∞∑𝑘=12𝑘−1𝑧𝑘−∞∑𝑘=11𝑧𝑘=−1∑𝑘=−∞(2−𝑘−1−1)𝑧𝑘 in {2<|𝑧|<∞}`

What if choosing yet another annulus? 𝑓 is also analytic in {0<|𝑧−1|<1}.

Since `1(𝑧−2)=1(𝑧−1)−1=−11−(𝑧−1)=−∞∑𝑘=0(𝑧−1)𝑘 in {0<|𝑧−1|<1}` So ```1(𝑧−1)(𝑧−2)=(𝑧−1)−(𝑧−2)(𝑧−1)(𝑧−2)=1(𝑧−2)−1(𝑧−1)  =−∞∑𝑘=0(𝑧−1)𝑘−1(𝑧−1)  =∞∑𝑘=−1(−1)(𝑧−1)𝑘. ``` Therefore `1(𝑧−1)(𝑧−2)=∞∑𝑘=−1(−1)(𝑧−1)𝑘 in {0<|𝑧−1|<1}`

### Another Example

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

Recall: `sin 𝑧=∞∑𝑘=0(−1)𝑘(2𝑘+1)!𝑧2𝑘+1=𝑧−𝑧33!+𝑧55!−𝑧77!+−⋯` so `sin 𝑧𝑧4=1𝑧3−13!1𝑧+15!𝑧−17!𝑧3+−⋯` Thus `𝑎−3=1, 𝑎−2=0, 𝑎−1=−13!, 𝑎0=0, 𝑎1=15!, 𝑎2=0, 𝑎3=−17!, ⋯`

## The Coefficients 𝑎𝑘 Continued

Recall: For a Taylor series, `𝑓(𝑧)=∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘, |𝑧−𝑧0|<𝑅,` the 𝑎𝑘 can be calculated via 𝑎𝑘=𝑓(𝑘)(𝑧0)𝑘!. How about for a Laurent series `𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 𝑟<|𝑧−𝑧0|<𝑅?` 𝑓 may not be defined at 𝑧0, so needed a new approach! Back to Taylor series: `𝑎𝑘=𝑓(𝑘)(𝑧0)𝑘!Cauchy=12𝜋𝑖 ∫|𝑧−𝑧0|=𝑠𝑓(𝑧)(𝑧−𝑧0)𝑘+1𝑑𝑧` for any 𝑠 between 0 and 𝑅.

One can show a similar fact for Laurent series: ```TheoremIf 𝑓 is analytic in {𝑟<|𝑧−𝑧0|<𝑅}, then 𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, where 𝑎𝑘=12𝜋𝑖 ∫|𝑧−𝑧0|=𝑠𝑓(𝑧)(𝑧−𝑧0)𝑘+1𝑑𝑧 for any 𝑠 between 𝑟 and 𝑅, and all 𝑘∈ℤ. ```

Note: This does not seem all that useful for finding actual values of 𝑎𝑘, but it is useful to estimate 𝑎𝑘. 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<|𝑧−𝑧0|<𝑟} centered at 𝑧0`
• 𝑓(𝑧)=1𝑧 has an isolated singularity at 𝑧0=0.
• 𝑓(𝑧)=1sin 𝑧 has isolated singularities at 𝑧0=0, ±𝜋, ±2𝜋, ⋯.
• 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)=1𝑧−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<|𝑧−𝑧0|<𝑟} for some 𝑟>0) then 𝑓 has a laurent series expansion there: ```𝑓(𝑧)=∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘  =⋯+𝑎−2(𝑧−𝑧0)2+𝑎−1(𝑧−𝑧0)2 principal part +𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ analytic ``` Three fundamentally different things can happen that influence how 𝑓 behaves near 𝑧0.

## Three Types of Isolated Singularities

`𝑓(𝑧)=⋯+𝑎−2(𝑧−𝑧0)2+𝑎−1(𝑧−𝑧0)2+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯, 0<|𝑧−𝑧0|<𝑟`

Examples

• 𝑓(𝑧)=cos 𝑧−1𝑧2=1𝑧2𝑧22!+𝑧44!−+⋯=−12!+𝑧24!−+⋯ No negative powers of 𝑧!
• 𝑓(𝑧)=cos 𝑧𝑧4=1𝑧41−𝑧22!+𝑧44!−+⋯=1𝑧412!1𝑧2+14!𝑧26!+−⋯ Finitely many negative powers of 𝑧!
• 𝑓(𝑧)=cos1𝑧=1−12!1𝑧2+14!1𝑧416!1𝑧6+−⋯ 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 ∞∑𝑘=−∞𝑎𝑘(𝑧−𝑧0)𝑘, 0<|𝑧−𝑧0|<𝑟. Then the singularity 𝑧0 is removable if 𝑎𝑘=0 for all 𝑘<0. a pole if there exist 𝑁>0 so that 𝑎−𝑁≠0 but 𝑎𝑘=0 for all 𝑘<−𝑁. The index 𝑁 is the order of the pole. essential if 𝑎𝑘≠0 for infinitely many 𝑘<0. ```

## Types of Singularities

The following table illustrates this definition:

𝑧0 is a ⋯Laurent series in 0<|𝑧−𝑧0|<𝑟 Removable singularity𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ Pole of order 𝑁𝑎−𝑁(𝑧−𝑧0)𝑁+⋯+𝑎−1(𝑧−𝑧0)2+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ Simple pole𝑎−1(𝑧−𝑧0)2+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯ Essential singularity⋯+𝑎−2(𝑧−𝑧0)2+𝑎−1(𝑧−𝑧0)2+𝑎0+𝑎1(𝑧−𝑧0)+𝑎2(𝑧−𝑧0)2+⋯

## Removable Singularities

Recall: 𝑧0 is a removable singularity of 𝑓 if its Laurent series, centered at 𝑧0 satisfies that 𝑎𝑘=0 for all 𝑘<0. `Example: 𝑓(𝑧)=sin 𝑧𝑧=1−𝑧22!+𝑧44!−+⋯, 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 ℂ: `𝑓(𝑧)={sin 𝑧𝑧,𝑧≠01 𝑧=0 is analytic in ℂ.` 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 𝑎−𝑁≠0 and 𝑎𝑘=0 for all 𝑘<−𝑁. `Example: 𝑓(𝑧)=sin 𝑧𝑧5=1𝑧4−13!1𝑧2+15!−17!𝑧2+−⋯ 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)|→∞ 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 𝑎𝑘≠0 for infinitely many 𝑘<0. `Example: 𝑓(𝑧)=ℯ1𝑧=∞∑𝑘=01𝑘!1𝑧𝑘=1+1𝑧+12!1𝑧2+13!1𝑧3+⋯ has an essential singularity at 𝑧0=0.` Note that if 𝑧=𝑥∈ℝ, then 𝑓(𝑧)=ℯ1/𝑥→∞ as 𝑥→0 from the right and 𝑓(𝑧)=ℯ1/𝑥→0 as 𝑥→0 from the left.

Also, if 𝑧=𝑖𝑥∈𝑖ℝ the `𝑓(𝑧)=ℯ1/𝑖𝑥=ℯ−𝑖/𝑥 lies on the unit circle for all 𝑥` 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 𝑤0∈ℂ there exists a sequence {𝑧𝑛} with 𝑧𝑛→𝑧0 such that 𝑓(𝑧𝑛)→𝑤0 as 𝑛→∞.`

## Casorati-Weierstra𝛽

Casorati-Weierstra𝛽: Suppose that 𝑧0 is an essential singularity of 𝑓. Then for every 𝑤0∈ℂ there exists a sequence {𝑧𝑛} with 𝑧𝑛→𝑧0 such that 𝑓(𝑧𝑛)→𝑤0. ```Example: Let 𝑓(𝑧)=ℯ1𝑧. Then 𝑓 has an essential singularity at 0. Let's pick a point 𝑤0∈ℂ, say, 𝑤0∈ℂ=1+3𝑖. By Casorati-Weierstra𝛽 there must exist 𝑧𝑛∈ℂ\{0} such that ℯ1𝑧𝑛→1+3𝑖 as 𝑛→∞. How do we find 𝑧𝑛 Idea: We can find 𝑧𝑛 such that ℯ1𝑧𝑛=1+3𝑖, namely 1𝑧𝑛=log(1+3𝑖). Recall: log(𝑧)=ln(|𝑧|)+𝑖arg(𝑧). So log(1+3𝑖)=ln 2+𝑖𝜋3+2𝑛𝜋𝑖. Pick 𝑧𝑛=1ln 2+𝑖𝜋3+2𝑛𝜋𝑖 Then 𝑧𝑛→0 as 𝑛→∞. Furthermore: ℯ1𝑧𝑛=ℯln 2+𝑖𝜋3+2𝑛𝜋𝑖  =2ℯ𝑖𝜋3  =212+𝑖32  =1+3𝑖  =𝑤0 for all 𝑛 We thus found 𝑧𝑛 with 𝑧𝑛→0 such that 𝑓(𝑧𝑛)=𝑤0 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 𝑤0∈ℂ with at most one excepton there exists a sequence {𝑧𝑛} with 𝑧𝑛→𝑧0 such that 𝑓(𝑧𝑛)=𝑤0.`

### Example:

𝑓(𝑧)=ℯ1/𝑧 has an essential singularity at 𝑧0=0. Also, 𝑓(𝑧)≠0 for all 𝑧, and so by Picard's theorem, for every 𝑤0≠0 there must exist infnitely many 𝑧𝑛 with 𝑧𝑛→0 such that 𝑓(𝑧𝑛)=𝑤0.

Pick 𝑤0=1 for example. Then 𝑓(𝑧)=𝑤0 if ℯ1/𝑧=1, that is 1𝑧=2𝑛𝜋𝑖 for some 𝑛∈ℤ. Now let 𝑧𝑛=12𝑛𝜋𝑖. Then 𝑧𝑛→0 as 𝑛→∞, and 𝑓(𝑧𝑛)=1 for all 𝑛.

ID: 190500008 Last Updated: 5/8/2019 Revision: 0 Home 5

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