October23 ,2020andOctober26 ,2020 ISOLATEDSINGULARITIES ComplexVariables

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MAT334H1-F – LEC0101

Complex Variables

ISOLATED SINGULARITIES

October 23^rd, 2020 and October 26^th, 2020

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 23, 2020 and Oct 26, 2020 1 / 10

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Isolated singularities

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

• Either𝑓 is bounded in a neighborhood of𝑧₀,

i.e. ∃𝑀 ∈ ℝ, ∃𝑟 > 0, ∀𝑧 ∈ 𝐷_𝑟(𝑧₀) ∩ 𝑈 , |𝑓 (𝑧)| ≤ 𝑀, then we say that𝑧₀is aremovable singularityof𝑓,

• or lim

𝑧→𝑧₀|𝑓 (𝑧)| = +∞, then we say that𝑧₀is apoleof𝑓,

• otherwise, if none of the above occurs, we say that𝑧₀is anessential singularityof𝑓.

Examples: removable singularities

1 𝑓 (𝑧) = _𝑧^𝑧+𝑖₂₊₁ onℂ ⧵ {0}with𝑧₀= 0.

2 𝑓 (𝑧) = ^sin𝑧_𝑧 onℂ ⧵ {0}with𝑧₀= 0.

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Isolated singularities

• or lim

Examples: poles

1 𝑓 (𝑧) = ¹_𝑧 onℂ ⧵ {0}with𝑧₀= 0.

2 𝑓 (𝑧) = _𝑧₂¹₋₁ on𝐷₁(1)with𝑧₀= 1.

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Isolated singularities

• or lim

Examples: essential singularities

1 𝑓 (𝑧) =cos¹_𝑧 onℂ ⧵ {0}with𝑧₀= 0.

2 𝑓 (𝑧) = 𝑒¹^𝑧 onℂ ⧵ {0}with𝑧₀= 0.

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Isolated singularities

• or lim

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Removable singularities

Theorem: Riemann’s removable singularity theorem

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

1 𝑧₀is a removable singularity of𝑓 (i.e.𝑓 is bounded in a neighborhood of𝑧₀)

2 lim

𝑧→𝑧0

(𝑧 − 𝑧₀)𝑓 (𝑧) = 0

3 𝑓 can be holomorphically/analytically extended on𝑈

(i.e. there exists𝑓 ∶ 𝑈 → ℂ̃ holomorphic/analytic such that𝑓_{|𝑈 ⧵{𝑧}̃ ₀_}= 𝑓)

4 𝑓 can be continuously extended on𝑈

(i.e. there exists𝑓 ∶ 𝑈 → ℂ̃ continuous such that𝑓_{|𝑈 ⧵{𝑧}̃ ₀_}= 𝑓)

Remark

Note that if𝑓admits a continuous extension at𝑧₀then it is holomorphic.

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Removable singularities

Theorem: Riemann’s removable singularity theorem

2 lim

𝑧→𝑧0

(𝑧 − 𝑧₀)𝑓 (𝑧) = 0

(i.e. there exists𝑓 ∶ 𝑈 → ℂ̃ holomorphic/analytic such that𝑓_{|𝑈 ⧵{𝑧}̃

0}= 𝑓)

Proof. 1 ⟹ 2 ⟹ 3 ⟹ 4 ⟹ 1 . The only non-trivial part is 2 ⟹ 3 . Define𝑔 ∶ 𝑈 → ℂby𝑔(𝑧₀) = 0and𝑔(𝑧) = (𝑧 − 𝑧₀)²𝑓 (𝑧)otherwise. Then lim

𝑧→𝑧0

𝑔(𝑧) − 𝑔(𝑧₀) 𝑧 − 𝑧₀ = 0.

Hence𝑔is holomorphic on𝑈. Since𝑔(𝑧₀) = 𝑔^′(𝑧₀) = 0we have𝑔(𝑧) =

+∞

∑𝑛=2

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛= (𝑧 − 𝑧₀)²

+∞

∑𝑛=2

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−2where

𝑧 ∈ 𝐷_𝑟(𝑧₀) ∩ 𝑈for some𝑟 > 0, so that𝑓 (𝑧) =̃

+∞

∑𝑛=2

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−2is a suitable holomorphic extension of𝑓on𝐷_𝑟(𝑧₀) ∩ 𝑈. ■

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Removable singularities

Theorem: Riemann’s removable singularity theorem

2 lim

𝑧→𝑧0

(𝑧 − 𝑧₀)𝑓 (𝑧) = 0

(i.e. there exists𝑓 ∶ 𝑈 → ℂ̃ holomorphic/analytic such that𝑓_{|𝑈 ⧵{𝑧}̃ ₀_}= 𝑓)

Examples

• 𝑓 (𝑧) =^𝑧²⁺¹

𝑧+𝑖 onℂ ⧵ {−𝑖}may be holomorphically extended toℂby𝑓 (𝑧) = 𝑧 − 𝑖.̃

• 𝑓 (𝑧) =^sin^𝑧

𝑧 onℂ ⧵ {0}may be continuous extended at0by setting𝑓 (0) = 1.̃

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Poles

Theorem

1 𝑧₀is a pole of𝑓, i.e. lim

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

2 There exist𝑛 ∈ ℕ_>0and𝑔 ∶ 𝑈 → ℂanalytic such that𝑔(𝑧₀) ≠ 0and𝑓 (𝑧) = 𝑔(𝑧)

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

3 𝑧₀is not a removable singularity of𝑓 and there exists𝑛 ∈ ℕ_>0such thatlim

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

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Poles

Theorem

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Proof: 1 ⟹ ³ Thenlim

𝑧→𝑧0

1

𝑓 (𝑧) = 0and𝑧₀is a removable singularity of1/𝑓, so that we can extend it to a holomorphic functionℎ ∶ 𝑈 → ℂdefined byℎ(𝑧) =

{

1

𝑓 (𝑧) if𝑧 ≠ 𝑧₀ 0 otherwise .

Denote by𝑛 ≔ 𝑚_ℎ(𝑧₀) ∈ ℕ_>0the order of vanishing ofℎat𝑧₀, thenℎ(𝑧) = (𝑧 − 𝑧₀)^𝑛 ̃ℎ(𝑧)where ̃ℎ ∶ 𝑈 → ℂis holomorphic and ̃ℎ(𝑧0) ≠ 0. Thenlim

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) =lim

𝑧→𝑧₀

𝑧 − 𝑧₀

̃ℎ(𝑧) = 0. ■

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Poles

Theorem

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Proof: 3 ⟹ ²

Pick the smallest𝑛such thatlim

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Define𝑔 ∶ 𝑈 ⧵ {𝑧₀} → ℂby𝑔(𝑧) = (𝑧 − 𝑧₀)^𝑛𝑓 (𝑧). Thenlim

𝑧→𝑧0

(𝑧 − 𝑧₀)𝑔(𝑧) = 0and𝑧₀is a removable singularity of 𝑔, so that𝑔may be extended to a holomorphic function𝑔 ∶ 𝑈 → ℂ.

Besides𝑔(𝑧₀) =lim

𝑧→𝑧0

𝑔(𝑧) =lim

𝑧→𝑧0

(𝑧 − 𝑧₀)^𝑛𝑓 (𝑧) ≠ 0by definition of𝑛. ■

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Poles

Theorem

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Proof: 2 ⟹ ¹

Obvious. ■

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Poles

Theorem

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Definition: order of a pole

The integer𝑛 ∈ ℕ_>0in 2 is uniquely defined and we say that𝑓 admits apole of order𝑛at𝑧₀.

We saw in the previous proof that the order of the pole𝑧₀is also:

• The order of vanishing of1/𝑓 at𝑧₀.

• The smallest𝑛such thatlim

𝑧→𝑧0

(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

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Poles

Theorem

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Definition: order of a pole

The integer𝑛 ∈ ℕ_>0in 2 is uniquely defined and we say that𝑓 admits apole of order𝑛at𝑧₀. We saw in the previous proof that the order of the pole𝑧₀is also:

• The order of vanishing of1/𝑓 at𝑧₀.

• The smallest𝑛such thatlim

𝑧→𝑧0

(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

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Poles

Theorem

𝑧→𝑧0

|𝑓 (𝑧)| = +∞.

(𝑧 − 𝑧₀)^𝑛 on𝑈 ⧵ {𝑧₀}.

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Examples

• 1is a pole of order1of𝑓 (𝑧) =_𝑧₂¹₋₁=

1 𝑧+1 𝑧−1

• 0is a pole of order3of𝑓 (𝑧) = ¹

𝑧³

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Essential singularities

Not part of MAT334:

Great Picard’s Theorem

Let𝑈 ⊂ ℂbe open,𝑧₀∈ 𝑈and𝑓 ∶ 𝑈 ⧵ {𝑧₀} → ℂbe holomorphic/analytic.

If𝑧₀is an essential singularity of𝑓 then, on any punctured neighborhood of𝑧₀,𝑓 takes all possible complex values, with at most a single exception, infinitely many times.

Example

𝑓 ∶ ℂ ⧵ {0} → ℂdefined by𝑓 (𝑧) = 𝑒¹^𝑧 has an essential singularity at0. It takes the value𝑤 ∈ ℂ ⧵ {0}at𝑧 = Log(𝑤)+2𝑖𝜋𝑛¹ , 𝑛 ∈ ℤ.

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Essential singularities

Not part of MAT334:

Great Picard’s Theorem

Let𝑈 ⊂ ℂbe open,𝑧₀∈ 𝑈and𝑓 ∶ 𝑈 ⧵ {𝑧₀} → ℂbe holomorphic/analytic.

If𝑧₀is an essential singularity of𝑓 then, on any punctured neighborhood of𝑧₀,𝑓 takes all possible complex values, with at most a single exception, infinitely many times.

Example

𝑓 ∶ ℂ ⧵ {0} → ℂdefined by𝑓 (𝑧) = 𝑒¹^𝑧 has an essential singularity at0.

It takes the value𝑤 ∈ ℂ ⧵ {0}at𝑧 = Log(𝑤)+2𝑖𝜋𝑛¹ , 𝑛 ∈ ℤ.

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Spoiler: a first introduction to Laurent series

Assume that𝑧₀is a pole of order𝑛 ∈ ℕ_>0of𝑓.

Then there exists𝑔 ∶ 𝑈 → ℂholomorphic/analytic such that𝑔(𝑧₀) ≠ 0and𝑓 (𝑧) = 𝑔(𝑧) (𝑧 − 𝑧₀)^𝑛 in a neighborhood of𝑧₀.

Since𝑔is analytic at𝑧₀, it may be expressed as a power series in a small neighborhood of𝑧₀: 𝑔(𝑧) =

+∞

∑𝑘=0

𝑎_𝑘(𝑧 − 𝑧₀)^𝑘 and since𝑔(𝑧₀) ≠ 0we know that𝑎₀≠ 0.

Therefore, in a punctured neighborhood of𝑧₀, we may express𝑓 as 𝑓 (𝑧) =

+∞

𝑘≥−𝑛∑

𝑎_𝑘+𝑛(𝑧 − 𝑧₀)^𝑘

= 𝑎₀(𝑧 − 𝑧₀)^−𝑛+ 𝑎₁(𝑧 − 𝑧₀)^−𝑛+1+ ⋯ + 𝑎_𝑛+ 𝑎_𝑛+1(𝑧 − 𝑧₀) + 𝑎_𝑛+2(𝑧 − 𝑧₀)²+ ⋯ Note that the above expression has some negative exponents: it is a first example ofLaurent series, notion that we will study next week.

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Isolated singularities – 1

In the above proofs, we used in an essential manner that the function𝑓 was holomorphic in a punctured neighborhood of𝑧₀, i.e. that there exists𝑟 > 0such that𝑓 is holomorphic on

𝐷_𝑟(𝑧₀) ⧵ {𝑧₀} = {𝑧 ∈ ℂ ∶ 0 < |𝑧 − 𝑧0| < 𝑟}

Therefore, when we will work with functions having several singularities, we will need to assume that they areisolated.

Formally, let𝑈 ⊂ ℂbe open,𝑆 ⊂ 𝑈 be thesingular locusand𝑓 ∶ 𝑈 ⧵ 𝑆 → ℂbe holomorphic.

We need that if𝑧₀∈ 𝑆 then𝑓 is holomorphic on𝐷_𝑟(𝑧₀) ⧵ {𝑧₀}for a small𝑟 > 0.

Otherwise stated, that there exists a small disk centered at𝑧₀which doesn’t contain another singular point.

To summarize,𝑆needs to satisfy∀𝑧₀∈ 𝑆, ∃𝑟 > 0, 𝐷_𝑟(𝑧₀) ∩ 𝑆 = {𝑧₀}.

If you take MAT327, it simply means that𝑆is discrete in𝑈.

We will only handle isolated singularities, we won’t study wilder singular loci.

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Isolated singularities – 2

Example

The function𝑓 ∶ ℂ ⧵ {±1} → ℂdefined by𝑓 (𝑧) = _𝑧₂¹₋₁ is holomorphic and has 2 isolated singularities at−1and+1.

Non-Example

Let𝑓 ∶ ℂ ⧵ ({𝜋𝑛¹ ∶ 𝑛 ∈ ℤ} ∪ {0}) → ℂbe defined by𝑓 (𝑧) =cot¹_𝑧. Then0is not an isolated singularity of𝑓:

ℜ ℑ

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Singularity at ∞ – 1

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

Then∞is an isolated singularity of𝑓 (i.e.𝑓 is defined in a neighborhood of∞but not at∞).

1 Either𝑓 is bounded in a neighborhood of∞,

i.e.∃𝑀 ∈ ℝ, ∃𝑟 > 0, ∀𝑧 ∈ 𝑈 , |𝑧| > 𝑟 ⟹ |𝑓 (𝑧)| ≤ 𝑀, then we say that∞is aremovable singularityof𝑓,

2 or lim

𝑧→∞|𝑓 (𝑧)| = +∞, then we say that∞is apoleof𝑓,

3 otherwise, if none of the above occurs, we say that∞is anessential singularityof𝑓.

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

Hence, if we set𝑔(𝑧) = 𝑓 (¹𝑧)then the type of singularity of𝑓 at∞coincides with the type of singularity of𝑔at0.

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Singularity at ∞ – 1

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

Then∞is an isolated singularity of𝑓 (i.e.𝑓 is defined in a neighborhood of∞but not at∞).

1 Either𝑓 is bounded in a neighborhood of∞,

i.e.∃𝑀 ∈ ℝ, ∃𝑟 > 0, ∀𝑧 ∈ 𝑈 , |𝑧| > 𝑟 ⟹ |𝑓 (𝑧)| ≤ 𝑀, then we say that∞is aremovable singularityof𝑓,

2 or lim

𝑧→∞|𝑓 (𝑧)| = +∞, then we say that∞is apoleof𝑓,

3 otherwise, if none of the above occurs, we say that∞is anessential singularityof𝑓. Recall that the inversionℂ → ̂̂ ℂ,𝑧 ↦ ¹_𝑧, swaps0and∞.

Hence, if we set𝑔(𝑧) = 𝑓 (¹𝑧)then the type of singularity of𝑓 at∞coincides with the type of singularity of𝑔at0.

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Singularity at ∞ – 2

Non-Example

Let𝑓 ∶ ℂ ⧵ {𝜋𝑛 ∶ 𝑛 ∈ ℤ} → ℂbe defined by𝑓 (𝑧) =cot𝑧.

Then∞is not an isolated singularity of𝑓:

ℜ ℑ