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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 𝑎_𝑘

Note: The coefficients 𝑎_𝑘 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)  =−12
11−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 𝑎_𝑘 Continued

Recall: For a Taylor series, 𝑓(𝑧)=∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘, |𝑧−𝑧0|<𝑅, the 𝑎_𝑘 can be calculated via 𝑎_𝑘=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 𝑎_𝑘, but it is useful to estimate 𝑎_𝑘. 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=1𝑧²−𝑧²2!+𝑧⁴4!−+⋯=−12!+𝑧²4!−+⋯ No negative powers of 𝑧!
• 𝑓(𝑧)=cos 𝑧𝑧⁴=1𝑧⁴1−𝑧²2!+𝑧⁴4!−+⋯=1𝑧⁴−12!1𝑧²+14!−𝑧²6!+−⋯ Finitely many negative powers of 𝑧!
• 𝑓(𝑧)=cos1𝑧=1−12!1𝑧²+14!1𝑧⁴−16!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:

Removable Singularities

Recall: 𝑧₀ is a removable singularity of 𝑓 if its Laurent series, centered at 𝑧₀ satisfies that 𝑎_𝑘=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 𝑎_𝑘=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 𝑎_𝑘≠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)=ℯ^1/𝑥→∞ as 𝑥→0 from the right and 𝑓(𝑧)=ℯ^1/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 {𝑧_𝑛} with 𝑧_𝑛→𝑧₀ such that 𝑓(𝑧_𝑛)→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)=ℯ^1/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 𝑧_𝑛 with 𝑧_𝑛→0 such that 𝑓(𝑧_𝑛)=𝑤₀.

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