14-Zeroesofanalyticfunctions ComplexVariables

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Complex Variables

14 - Zeroes of analytic functions

Jean-Baptiste Campesato November 4^th, 2020

Reviews from Oct 16 – Zeroes Definition 1.

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic. Let𝑧₀ ∈ 𝑈 be such that𝑓 (𝑧₀) = 0.

We define theorder of vanishing of𝑓 at𝑧₀by𝑚_𝑓(𝑧₀) ≔min{𝑛 ∈ ℕ ∶ 𝑓^(𝑛)(𝑧₀) ≠ 0}. Note that𝑚_𝑓(𝑧₀) > 0since𝑓 (𝑧₀) = 0.

Proposition 2. Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic. Let𝑧₀ ∈ 𝑈 be such that𝑓 (𝑧₀) = 0.

Denote the power series expansion of𝑓 at𝑧₀by𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛. Then𝑚_𝑓(𝑧₀) =min{𝑛 ∈ ℕ ∶ 𝑎𝑛 ≠ 0}.

Proposition 3. Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic. Then𝑧₀is a zero of order𝑛of𝑓 if and only if there exists𝑔 ∶ 𝑈 → ℂholomorphic such that𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)and𝑔(𝑧₀) ≠ 0.

Reviews from Oct 16 – Analytic continuation

Theorem 4. Let𝑈 ⊂ ℂbe adomainand𝑓 ∶ 𝑈 → ℂbe a holomorphic/analytic function.

If there exists𝑧₀∈ 𝑈 such that∀𝑛 ∈ ℕ_≥0, 𝑓^(𝑛)(𝑧₀) = 0then𝑓 ≡ 0on𝑈.

Corollary 5. Let𝑈 ⊂ ℂbe adomainand𝑓 , 𝑔 ∶ 𝑈 → ℂbe holomorphic/analytic functions.

If𝑓 and𝑔coincide in the neighborhood of a point,

i.e. ∃𝑧₀ ∈ 𝑈 , ∃𝑟 > 0, ∀𝑧 ∈ 𝐷_𝑟(𝑧₀) ∩ 𝑈 , 𝑓 (𝑧) = 𝑔(𝑧), then they coincide on𝑈,

i.e.∀𝑧 ∈ 𝑈 , 𝑓 (𝑧) = 𝑔(𝑧).

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Isolated zeroes

It is actually possible to strengthen the previous results.

Theorem 6. Let𝑈 ⊂ ℂbe a domain and𝑓 ∶ 𝑈 → ℂbe a holomorphic/analytic function.

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated^⋆:

if𝑓 (𝑧₀) = 0then there exists𝑟 > 0such that𝐷_𝑟(𝑧₀) ⊂ 𝑈 and∀𝑧 ∈ 𝐷_𝑟(𝑧₀) ⧵ {𝑧₀}, 𝑓 (𝑧) ≠ 0.

Proof. Assume that𝑧₀is a non-isolated zero of𝑓.

We know that𝑓 admits a power series expansion𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛in a neighborhood of𝑧₀.

Assume by contradiction that there exists a smallest𝑛 ∈ ℕ_≥0such that𝑎_𝑛≠ 0then𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where 𝑔is holomorphic and𝑔(𝑧₀) ≠ 0.

For every𝑛 ∈ ℕ_>0, ∃𝑤_𝑛 ∈ (𝐷¹_𝑛(𝑧₀) ∩ 𝑈 ) ⧵ {𝑧⁰}, 𝑓 (𝑤_𝑛) = 0. But then𝑔(𝑤_𝑛) = 0since𝑤_𝑛 ≠ 𝑧₀ Then, since𝑤_𝑛−−−−−→

𝑛→+∞ 𝑧₀, by continuity𝑔(𝑧₀) = lim

𝑛→+∞𝑔(𝑤_𝑛) = 0. Which leads to a contradiction.

Hence∀𝑛 ∈ ℕ_≥0, 𝑓^(𝑛)(𝑧₀) = 𝑛!𝑎_𝑛= 0and𝑓 ≡ 0on𝑈 by Theorem4. ■ Corollary 7. Let𝑈 ⊂ ℂbe a domain and𝑓 , 𝑔 ∶ 𝑈 → ℂbe holomorphic/analytic functions.

If𝑓 − 𝑔admits a non-isolated zero

i.e. ∃𝑧₀∈ 𝑈 , ∀𝑟 > 0, ∃𝑧 ∈ (𝑈 ∩ 𝐷𝑟(𝑧₀)) ⧵ {𝑧0}, 𝑓 (𝑧) − 𝑔(𝑧) = 𝑓 (𝑧₀) − 𝑔(𝑧₀) = 0 then𝑓 and𝑔coincide on𝑈,

i.e.∀𝑧 ∈ 𝑈 , 𝑓 (𝑧) = 𝑔(𝑧).

Corollary 8. Let𝑈 ⊂ ℂbe a domain and𝑓 , 𝑔 ∶ 𝑈 → ℂbe holomorphic/analytic functions.

Let(𝑧_𝑛)_𝑛∈ℕbe a sequence of terms in𝑈 which is convergent to𝑧̃in𝑈 and such that∀𝑛 ∈ ℕ, 𝑓 (𝑧_𝑛) = 0.

Then𝑓 ≡ 0on𝑈.

Remark 9. The fact that the limit𝑧 ∈ 𝑈̃ is very important.

Indeed, let𝑓 ∶ ℂ ⧵ {0} → ℂbe defined by𝑓 (𝑧) =sin(^𝜋𝑧). Then𝑓 (¹𝑛) = 0but𝑓 ≢ 0onℂ ⧵ {0}.

Hence, it is possible for the zeroes of𝑓 to accumulate at a point of the boundary of the domain (including

∞, see for instance𝑧_𝑛= 𝜋𝑛for𝑓 =sin).

Homework 10. Let𝑈 ⊂ ℂbe a domain and𝑓 , 𝑔 ∶ 𝑈 → ℂbe holomorphic/analytic on𝑈. Prove that if𝑓 𝑔 ≡ 0on𝑈then either𝑓 ≡ 0or𝑔 ≡ 0.

Homework 11. Let𝑈 = 𝐷₁(0). Find all the holomorphic functions𝑓 ∶ 𝑈 → ℂsatisfying respectively:

1. 𝑓 (¹𝑛) = 𝑛¹²

2. 𝑓 (2𝑛¹) = 𝑓 (2𝑛+1¹ ) = ¹𝑛

⋆Otherwise stated, if you attend MAT327, either𝑓 is constant equal to0or{𝑧 ∈ 𝑈 ∶ 𝑓 (𝑧) = 0}is discrete.

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Reviews from Oct 23 – Poles

Theorem 12. Let𝑈 ⊂ ℂbe open and𝑧₀∈ 𝑈. Assume that𝑓 ∶ 𝑈 ⧵ {𝑧₀} → ℂis holomorphic/analytic. Then TFAE:

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 that lim

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

Definition 13. The integer𝑛 > 0in (2) is uniquely defined and we say that𝑓 admits apole of order𝑛at𝑧₀. Proposition 14. The order of the pole𝑧₀is also:

• The order of vanishing of1/𝑓 at𝑧₀.

• The smallest𝑛such that lim

𝑧→𝑧₀(𝑧 − 𝑧₀)^𝑛+1𝑓 (𝑧) = 0.

The argument principle

Lemma 15(Logarithmic residue).

• If𝑧₀is an isolated zero of𝑓 thenRes(^𝑓

′

𝑓 , 𝑧₀)is the order of𝑧₀.

• If𝑧₀is an isolated pole of𝑓 then−Res(^𝑓

′

𝑓 , 𝑧₀)is the order of𝑧₀. Proof.

• Assume that 𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑚𝑔(𝑧) in a neighborhood of 𝑧₀ where 𝑔 is analytic and 𝑔(𝑧₀) ≠ 0. Then

𝑓^′(𝑧)

𝑓 (𝑧) = 𝑚(𝑧 − 𝑧₀)⁻¹+^𝑔^′^(𝑧)

𝑔(𝑧).

We conclude using that ^𝑔_𝑔^′ is holomorphic in a neighborhood of𝑧₀.

• 𝑧₀is a pole of order𝑚of𝑓 if and only if it is a zero of order𝑚of ¹

𝑓. We conclude using that Res(

(1/𝑓 )^′ (1/𝑓 ), 𝑧₀

)= −Res (

𝑓^′ 𝑓 , 𝑧₀

)

■ The previous lemma holds at∞:

Lemma 16.

• If∞is an isolated zero of𝑓 thenRes(^𝑓

′

𝑓 , ∞)is the order of∞.

• If∞is an isolated pole of𝑓 then−Res(^𝑓

′

𝑓 , ∞)is the order of∞.

Proof. ∞is an isolated zero (resp. pole) of order𝑚of𝑓 if and only if0is an isolated zero (resp. pole) of order𝑚of𝑔(𝑧) = 𝑓 (1/𝑧).

Then𝑚 =Res(^𝑔

′

𝑔 , 0) =Res(⁻¹𝑧² 𝑓^′(1/𝑧)

𝑓 (1/𝑧), 0) =Res(^𝑓

′

𝑓 , ∞). ■

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Theorem 17(The argument principle).

Let𝑈 ⊂ ℂbe open. Let𝑆 ⊂ 𝑈be finite. Let𝑓 ∶ 𝑈 ⧵ 𝑆 → ℂbe holomorphic/analytic.

Let𝛾 ∶ [𝑎, 𝑏] → ℂbe piecewise smooth positively oriented simple closed curve on𝑈which doesn’t pass through a zero or a pole of𝑓 and such that its inside is entirely included in𝑈.

Then 1

2𝑖𝜋 ∫_𝛾 𝑓^′(𝑧)

𝑓 (𝑧)d𝑧 = 𝑍_{𝑓 ,𝛾}− 𝑃_{𝑓 ,𝛾} where

• 𝑍_{𝑓 ,𝛾} is the number of zeroes of𝑓 enclosed in𝛾 counted with their multiplicites/orders,

• 𝑃_{𝑓 ,𝛾}is the number of poles of𝑓 enclosed in𝛾 counted with their multiplicites/orders.

Proof. We apply Cauchy’s residue theorem to ^𝑓_𝑓^′ and then we use the above lemma:

1 2𝑖𝜋 ∫_𝛾

𝑓^′(𝑧) 𝑓 (𝑧)d𝑧 =

𝑧∈Inside(𝛾)∑ Res(

𝑓^′ 𝑓 , 𝑧

)=

𝑧zero of∑ 𝑓

Res( 𝑓^′

𝑓 , 𝑧₀ )+

𝑧pole of∑ 𝑓

Res( 𝑓^′

𝑓 , 𝑧₀

)= 𝑍_{𝑓 ,𝛾}− 𝑃_{𝑓 ,𝛾} ■

Remark 18. The value 1 2𝑖𝜋 ∫_𝛾

𝑓^′(𝑧)

𝑓 (𝑧)d𝑧involved in the previous slide is equal to the number of counterclockwise turns made by𝑓 (𝑧)as𝑧goes through𝛾.

Indeed, if we set ̃𝛾(𝑡) = 𝑓 ∘ 𝛾then

∫_𝛾 𝑓^′(𝑧)

𝑓 (𝑧)d𝑧 =

∫_̃𝛾 1 𝑤d𝑤.

Assume for instance that ̃𝛾 ∶ [0, 1] → ℂis defined by ̃𝛾(𝑡) = 𝑧₀+ 𝑟𝑒^{2𝑖𝜋𝑛𝑡}where𝑛 ∈ ℤ.

Then 1 2𝑖𝜋 ∫_̃𝛾

1

𝑤d𝑤 = 𝑛which is the number of counterclockwise turns made by ̃𝛾around𝑧₀. Then the conclusion of the previous statement can be rewritten as

changes of arg(𝑓 (𝑧))as𝑧goes through𝛾

2𝜋 = 𝑍_{𝑓 ,𝛾}− 𝑃_{𝑓 ,𝛾}

That’s why it is calledthe argument principle.

Rouché’s theorem

Theorem 19(Rouché’s theorem – version 1).

Let𝑈 ⊂ ℂbe open,𝑓 , 𝑔 ∶ 𝑈 → ℂbe two holomorphic/analytic functions on𝑈, and𝛾 ∶ [𝑎, 𝑏] → ℂbe a piecewise smooth simple closed curve on𝑈whose inside is also included in𝑈.

Assume that

∀𝑡 ∈ [𝑎, 𝑏], |𝑔(𝛾(𝑡))| < |𝑓 (𝛾(𝑡))|

Then𝑓 and𝑓 + 𝑔have the same number of zeroes inside𝛾, counted with multiplicities.

Proof. For𝑡 ∈ [0, 1], set𝜑_𝑡(𝑧) = 𝑓 (𝑧) + (1 − 𝑡)𝑔(𝑧)andℎ(𝑡) = 1 2𝑖𝜋 ∫_𝛾

𝜑^′_𝑡(𝑧) 𝜑_𝑡(𝑧)d𝑧.

The functionℎis continuous since𝜑_𝑡doesn’t vanish on𝛾, indeed for𝑧 ∈ 𝛾

|𝜑_𝑡(𝑧)| ≥ |𝑓 (𝑧)| + (1 − 𝑡)|𝑔(𝑧)| ≥ |𝑓 (𝑧)| − |𝑔(𝑧)| > 0

Henceℎis a continuous function taking values inℤ(by the principle argument), so it is constant.

Henceℎ(0) = ℎ(1), i.e. 𝑍_{𝑓 +𝑔,𝛾}− 𝑃_{𝑓 +𝑔,𝛾} = 𝑍_{𝑓 ,𝛾}− 𝑃_{𝑓 ,𝛾} by the principle argument.

But these functions have no poles in the inside of𝛾, hence𝑍_{𝑓 +𝑔,𝛾} = 𝑍_{𝑓 ,𝛾}. ■

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Assume that

∀𝑧 ∈ 𝛾, |𝑓 (𝑧) − 𝑔(𝑧)| < |𝑓 (𝑧)|

Then𝑓 and𝑔have the same number of zeroes inside𝛾, counted with multiplicities.

Proof. That’s an immediate consequence of the previous version since𝑧₀ is a zero of order𝑛of 𝑔iff it is a

zero of order𝑛of−𝑔. ■

Assume that

∀𝑧 ∈ 𝛾, |𝑓 (𝑧) + 𝑔(𝑧)| < |𝑓 (𝑧)|

Then𝑓 and𝑔have the same number of zeroes inside𝛾, counted with multiplicities.

Proof. Since𝑧₀is a zero of order𝑛of𝑔iff it is a zero of order𝑛of−𝑔. ■

We already proved the Fundamental Theorem of Algebra (or d’Alembert–Gauss theorem) using Liouville’s theorem (Oct 21): a non-constant complex polynomial admits at least one root.

Here is another proof using Rouché’s theorem.

Theorem 22. A complex polynomial of degree𝑛has exactly𝑛complex roots (counted with multiplicity).

Proof. Assume that𝑃 (𝑧) = 𝑎_𝑛𝑧^𝑛+ 𝑄(𝑧)where𝑄is a polynomial of degree< 𝑛and𝑎_𝑛≠ 0.

If we take𝑅 > 0big enough then|𝑄(𝑧)| < |𝑎_𝑛𝑧^𝑛|on𝛾 ∶ [0, 1] → ℂdefined by𝛾(𝑡) = 𝑅𝑒^{2𝑖𝜋𝑡}.

By Rouché’s theorem,𝑃 (𝑧) = 𝑎_𝑛𝑧^𝑛+𝑄(𝑧)and𝑎_𝑛𝑧^𝑛have the same number of zeroes counted with multiplicity.

■