October26 ,2020andOctober28 ,2020 LAURENTSERIESANDRESIDUES ComplexVariables

MAT334H1-F – LEC0101

Complex Variables

LAURENT SERIES AND RESIDUES

October 26^th, 2020 and October 28^th, 2020

Informal presentation

Recall that a function𝑓 ∶ 𝑈 → ℂ(𝑈 ⊂ ℂopen) is holomorphic/analytic if and only if𝑓 can be described as a power series

𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛 in a neighborhood of every point𝑧₀∈ 𝑈.

We are going to generalize this property in order to study𝑓 in the neighborhood of an isolated singularity𝑧₀(i.e.𝑓 is not defined at𝑧₀but is holomorphic in a punctured neighborhood of𝑧₀), for that purpose we are going to work with Laurent series allowing negative exponents:

𝑓 (𝑧) =

+∞

𝑛=−∞∑

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

Intuitively, the more we need negative exponents in the above expression, the wilder is the singularity.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 2 / 14

Laurent’s theorem

Theorem: Laurent’s theorem

Let𝑧₀∈ ℂand0 ≤ 𝑟 < 𝑅 ≤ +∞. Set𝑈 = {𝑧 ∈ ℂ ∶ 𝑟 < |𝑧 − 𝑧0| < 𝑅}and let𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic then

∀𝑧 ∈ 𝑈 , 𝑓 (𝑧) =

∞ 𝑛=−∞∑

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

where𝑎_𝑛= 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤)

(𝑤 − 𝑧₀)^𝑛+1d𝑤and𝛾 ∶ [0, 1] → ℂis defined by𝛾(𝑡) = 𝑧₀+ 𝜌𝑒^{2𝑖𝜋𝑡}with𝜌 ∈ (𝑟, 𝑅).

We call such a series (a”power series”with exponents inℤ) aLaurent series.

• If𝑅 = +∞then𝑈is the complement of the closed disk centered at𝑧₀and of radius𝑟, hence it is a neighborhood of∞.

• If𝑟 = 0then𝑈is a punctured open disk centered at𝑧₀and of radius𝑅.

• If𝑟 = 0and𝑅 = +∞then𝑈 = ℂ ⧵ {𝑧₀}.

Example:

The function𝑓 ∶ ℂ ⧵ {0, 1} → ℂdefined by𝑓 (𝑧) = _{𝑧(1−𝑧)}¹ is holomorphic onℂ ⧵ {0, 1}and has two isolated singularities, namely0and1.

• For0 < |𝑧| < 1,𝑓 (𝑧) = 1 𝑧+ 1

1 − 𝑧= 𝑧⁻¹+

+∞

∑𝑛=0

𝑧^𝑛=

+∞

𝑛=−1∑ 𝑧^𝑛.

• For0 < |𝑧 − 1| < 1,𝑓 (𝑧) = − 1 𝑧 − 1+1

𝑧 = −(𝑧 − 1)⁻¹+

+∞

∑𝑛=0

(−1)^𝑛(𝑧 − 1)^𝑛=

+∞

𝑛=−1∑

(−1)^𝑛(𝑧 − 1)^𝑛.

• For1 < |𝑧|,𝑓 (𝑧) = −1 𝑧²

𝑧

𝑧 − 1 = −1 𝑧²

1

1 −¹_𝑧 = −1 𝑧²

+∞

∑𝑛=0

𝑧^−𝑛=

−2 𝑛=−∞∑

(−1)𝑧^𝑛

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 4 / 14

Proof of Laurent’s theorem – 1

Lemma

Let𝑧₀∈ ℂand0 ≤ 𝑟 < 𝑅 ≤ +∞. Set𝑈 = {𝑧 ∈ ℂ ∶ 𝑟 < |𝑧 − 𝑧0| < 𝑅}and let𝑔 ∶ 𝑈 → ℂbe holomorphic/analytic.

Let𝜌₁, 𝜌₂∈ ℝbe s.t.𝑟 < 𝜌₁< 𝜌₂< 𝑅. Let𝛾₁, 𝛾₂∶ [0, 1] → ℂbe defined by𝛾_𝑘(𝑡) = 𝑧₀+ 𝜌_𝑘𝑒^{2𝑖𝜋𝑡}. Then∫_𝛾1𝑔(𝑧)d𝑧 =

∫_𝛾2𝑔(𝑧)d𝑧.

Proof.

𝑧₀𝛾₁ 𝛾₂

We consider two simple closed curves as in the drawing (in red and blue), then by Cauchy’s integral theorem:

0 = 0 + 0 =

∫_𝑟𝑒𝑑𝑔(𝑧)d𝑧 +

∫_{𝑏𝑙𝑢𝑒}𝑔(𝑧)d𝑧

=∫_−𝛾

1

𝑔(𝑧)d𝑧 +

∫_𝛾

2

𝑔(𝑧)d𝑧

= −∫_𝛾

1

𝑔(𝑧)d𝑧 +

∫_𝛾

2

𝑔(𝑧)d𝑧 Hence

∫_𝛾

1

𝑔(𝑧)d𝑧 =

∫_𝛾

2

𝑔(𝑧)d𝑧.

■

Proof of Laurent’s theorem – 2

Proof of Laurent’s theorem.

𝑧₀𝛾₁ 𝛾₂ 𝑧

For𝑤 ∈ 𝛾₁,

| 𝑤 − 𝑧₀

𝑧 − 𝑧₀|< 1.

For𝑤 ∈ 𝛾₂,

| 𝑧 − 𝑧₀ 𝑤 − 𝑧₀|< 1.

Let𝑟 < 𝜌₁< |𝑧| < 𝜌₂< 𝑅and𝛾₁, 𝛾₂∶ [0, 1] → ℂdefined by 𝛾_𝑘(𝑡) = 𝑧₀+ 𝜌_𝑘𝑒^{2𝑖𝜋𝑡}.

We consider two simple closed curves as in the drawing (in red and blue).

Then by Cauchy’s integral formula and theorem

0 + 2𝑖𝜋𝑓 (𝑧) =

∫_𝑟𝑒𝑑 𝑓 (𝑤) 𝑤 − 𝑧d𝑤 +

∫_{𝑏𝑙𝑢𝑒} 𝑓 (𝑤) 𝑤 − 𝑧d𝑤

= −∫_𝛾1 𝑓 (𝑤) 𝑤 − 𝑧d𝑤 +

∫_𝛾2 𝑓 (𝑤) 𝑤 − 𝑧d𝑤

= −∫_𝛾1 𝑓 (𝑤) 𝑧₀− 𝑧

1 1 −^𝑤−𝑧0

𝑧−𝑧0

d𝑤 +∫_𝛾2 𝑓 (𝑤) 𝑤 − 𝑧₀

1 1 −^𝑧−𝑧0

𝑤−𝑧0

d𝑤

=∫

𝛾1

𝑓 (𝑤) 𝑧₀− 𝑧 ∑_𝑛≥0(

𝑤 − 𝑧₀ 𝑧₀− 𝑧 )

𝑛

d𝑤 +∫

𝛾2

𝑓 (𝑤) 𝑤 − 𝑧₀∑

𝑛≥0( 𝑧 − 𝑧₀ 𝑤 − 𝑧₀)

𝑛

d𝑤

=∑

𝑛<0(∫_𝛾1 𝑓 (𝑤) (𝑤 − 𝑧0)^𝑛+1d𝑤

)(𝑧 − 𝑧₀)^𝑛+∑

𝑛≥0(∫_𝛾2 𝑓 (𝑤) (𝑤 − 𝑧0)^𝑛+1d𝑤

)(𝑧 − 𝑧₀)^𝑛

Then we can replace𝛾₁and𝛾₂by𝛾thanks to the previous lemma.

■

We need to justify the∑ − ∫permutation of the last equality: that’s exactly the same proof as for the power expression of a holomorphic function (see lecture from October 16).

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 6 / 14

Laurent series and isolated singularities – 1

Proposition

Let𝑈 ⊂ ℂbe open and𝑧₀∈ 𝑈. Assume that𝑓 ∶ 𝑈 ⧵ {𝑧₀} → ℂis holomorphic/analytic.

Let𝑅 > 0be such that𝐷_𝑅(𝑧₀) ⊂ 𝑈. Then𝑓 admits a Laurent series expansion centered at𝑧₀on 𝐷_𝑅(𝑧₀) ⧵ {𝑧₀}(apply Laurent’s theorem with𝑟 = 0)

𝑓 (𝑧) =

∞ 𝑛=−∞∑

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛 Then

1 𝑧₀is a removable singularity iff∀𝑛 < 0, 𝑎_𝑛= 0

2 𝑧₀is a pole iff∃𝑚 ≥ 1, 𝑎_−𝑚≠ 0and∀𝑛 < −𝑚, 𝑎_𝑛= 0 (𝑚is the order of the pole𝑧₀)

3 𝑧₀is an essential singularity if and only if for infinitely many𝑛 ∈ ℕ,𝑎_−𝑛≠ 0.

Proof.

1 𝑓 coincides on𝐷_𝑅(𝑧₀) ⧵ {𝑧₀}with∑^∞_𝑛=0𝑎_𝑛(𝑧 − 𝑧₀)^𝑛holomorphic at𝑧₀.

2 𝑓 coincides on𝐷_𝑅(𝑧₀) ⧵ {𝑧₀}with^{∑∞𝑛=0𝑎𝑛−𝑚(𝑧−𝑧0)𝑛}

(𝑧−𝑧₀)^𝑚 .

3 if𝑧₀is not a removable singularity nor a pole then it is an essential singularity. ■

Laurent series and isolated singularities – 2

Example

For𝑧 ∈ ℂ ⧵ {0},𝑒^1𝑧 =

+∞

∑𝑛=0

(𝑧⁻¹)^𝑛 𝑛! =

0 𝑛=−∞∑

𝑧^𝑛 (−𝑛)!. Hence𝑒^1𝑧 has an essential singularity at0.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 8 / 14

Singularity at ∞

Theorem

Let𝑈 ⊂ ℂbe an open neighborhood of infinity, i.e. there exists𝑟 > 0such that {𝑧 ∈ ℂ ∶ |𝑧| > 𝑟} ⊂ 𝑈. Let𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

Then𝑓 admits a Laurent series expansion centered at0on{𝑧 ∈ ℂ ∶ |𝑧| > 𝑟}(apply Laurent’s theorem with𝑅 = +∞)

𝑓 (𝑧) =

∞ 𝑛=−∞∑

𝑎_𝑛𝑧^𝑛 Then

1 ∞is a removable singularity iff∀𝑛 > 0, 𝑎_𝑛= 0

2 ∞is a pole iff∃𝑚 ≥ 1, 𝑎_𝑚≠ 0and∀𝑛 > 𝑚, 𝑎_𝑛= 0 (𝑚is the order of the pole at∞)

3 ∞is an essential singularity if and only if for infinitely many𝑛 ∈ ℕ,𝑎_𝑛≠ 0.

Proof.Recall that the inversionℂ → ̂̂ ℂ,𝑧 ↦ ¹_𝑧, swaps0and∞. ■

Principal part

Definition

Theprincipal partof a Laurent series 𝑓 (𝑧) =

+∞

𝑛=−∞∑

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

is the part consisting only of the negative exponents

−1 𝑛=−∞∑

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 10 / 14

Residues – 1

Definition: residue

Let𝑧₀be an isolated singularity of𝑓. Denote the Laurent expansion of𝑓 around𝑧₀by 𝑓 (𝑧) =

+∞

𝑛=−∞∑

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

Then theresidue of𝑓 at𝑧₀is the coefficient of(𝑧 − 𝑧₀)⁻¹, i.e. Res(𝑓 , 𝑧₀) ≔ 𝑎₋₁.

Proposition

Assume that𝑓 is holomorphic on𝐷_𝑟(𝑧₀) ⧵ {𝑧₀}for some𝑟 > 0then Res(𝑓 , 𝑧₀) = 1

2𝑖𝜋 ∫_𝛾𝑓 (𝑤)d𝑤 where𝛾 ∶ [0, 1] → ℂis defined by𝛾(𝑡) = 𝑧₀+ 𝜌𝑒^{2𝑖𝜋𝑡}with𝜌 ∈ (0, 𝑟).

Note that the above integral doesn’t depend on the choice of𝜌by the above lemma.

Residues – 2

Example

In a punctured neighborhood of0, we have 𝑒^𝑧− 1

𝑧⁴ = 1 𝑧³ + 1

2𝑧² + 1 6𝑧+ 1

24+ 𝑧 120+ ⋯ HenceRes(^𝑒

𝑧−1

𝑧⁴ , 0) = ¹6.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 12 / 14

Residues – 3

Proposition: how to compute residues

• If𝑧₀is a pole of order1of𝑓 thenRes(𝑓 , 𝑧₀) = lim

𝑧→𝑧₀(𝑧 − 𝑧₀)𝑓 (𝑧).

• If𝑧₀is a pole of order𝑘of𝑓 thenRes(𝑓 , 𝑧₀) = ℎ^(𝑘−1)(𝑧₀)

(𝑘 − 1)! whereℎ(𝑧) = (𝑧 − 𝑧₀)^𝑘𝑓 (𝑧).

• If𝑓 is holomorphic at𝑧₀and𝑔has a zero of order1at𝑧₀thenRes (

𝑓 𝑔, 𝑧₀

)= 𝑓 (𝑧₀) 𝑔^′(𝑧₀).

• Assume that𝑧₀is an isolated zero of𝑓 then the order of vanishing of𝑓 at𝑧₀isRes (

𝑓^′ 𝑓 , 𝑧₀).

Homework

• Prove the above identities!

• Compute some residues (i.e. practice exercises from the textbook!).

Residue at ∞

Definition: residue at ∞

Assume that𝑓 is holomorphic/analytic in a neighborhood of infinity, i.e. on{𝑧 ∈ ℂ ∶ |𝑧| > 𝑟}for some𝑟. We define theresidue of𝑓 at∞by

Res(𝑓 , ∞) ≔Res (

−1 𝑧²𝑓 (1

𝑧 ), 0)

Proposition

Assume that𝑓 is holomorphic on{𝑧 ∈ ℂ ∶ |𝑧| > 𝑟}for some𝑟 > 0then Res(𝑓 , ∞) = − 1

2𝑖𝜋 ∫_𝛾𝑓 (𝑤)d𝑤 where𝛾 ∶ [0, 1] → ℂis defined by𝛾(𝑡) = 𝑧₀+ 𝜌𝑒^{2𝑖𝜋𝑡}with𝜌 > 𝑟.

The sign is due to the fact that𝛾is not positively oriented: it is considered as the boundary of the complement of the disk since that’s where is located∞.

Homework:prove the proposition (you will see where does−_𝑧¹₂ come from).

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 26, 2020 and Oct 28, 2020 14 / 14