October16 ,2020toOctober21 ,2020 CONSEQUENCESOFTHECAUCHY’SINTEGRALFORMULA ComplexVariables

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MAT334H1-F – LEC0101

Complex Variables

CONSEQUENCES OF THE CAUCHY’S INTEGRAL FORMULA

October 16^th, 2020 to October 21^st, 2020

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 16, 2020 to Oct 21, 2020 1 / 18

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Holomorphic functions are analytic

¹

– 1

Theorem

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic/ℂ-differentiable.

Then𝑓 can belocallyexpressed as a power series in a neighborhood of any point of𝑈. More precisely, if𝐷_𝑟(𝑧₀) ⊂ 𝑈 then

∀𝑧 ∈ 𝐷_𝑟(𝑧₀), 𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

where

𝑎_𝑛= 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤) (𝑤 − 𝑧₀)^𝑛+1d𝑤

and𝛾 ∶ [0, 1] → ℂis defined by𝛾(𝑡) = 𝑧₀+ 𝑟𝑒^{2𝑖𝜋𝑡}and the radius of convergence of this power series is greater than or equal to𝑟.

1I will stop complaining about the termanalytic… Now we know that holomorphic and analytic are synonyms.

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Holomorphic functions are analytic – 2

Proof.By Cauchy’s integral formula 𝑓 (𝑧) = 1

2𝑖𝜋 ∫_𝛾 𝑓 (𝑤) 𝑤 − 𝑧d𝑤

= 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤) 𝑤 − 𝑧₀

1 1 −_𝑤−𝑧^𝑧−𝑧⁰

0

d𝑤

= 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤) 𝑤 − 𝑧₀

+∞

∑𝑛=0( 𝑧 − 𝑧₀ 𝑤 − 𝑧₀)

𝑛

d𝑤 since|𝑧 − 𝑧₀| < |𝑤 − 𝑧₀| = 𝑟

=

+∞

∑𝑛=0( 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤) (𝑤 − 𝑧₀)^𝑛+1d𝑤

)(𝑧 − 𝑧₀)^𝑛

Actually, we should justify the permutation∫ − ∑more carefully. ■

It can be done by noting that for𝜀 > 0there exists𝑁such that if𝑘 ≥ 𝑁then the remainder satisfies

|

+∞

∑𝑛=𝑘( 𝑧 − 𝑧₀ 𝑤 − 𝑧₀)

𝑛

|≤ 𝜀, so that

|𝑓 (𝑧) −

𝑘

∑𝑛=0( 1 2𝑖𝜋 ∫𝛾

𝑓 (𝑤) (𝑤 − 𝑧₀)^𝑛+1d𝑤

)(𝑧 − 𝑧₀)^𝑛

|≤

| 1 2𝑖𝜋 ∫𝛾

𝑓 (𝑤) 𝑤 − 𝑧₀

+∞

∑𝑛=𝑘( 𝑧 − 𝑧₀ 𝑤 − 𝑧₀)

𝑛

d𝑤|≤ 1 2𝜋

|𝑤−𝑧0|=𝑟max |𝑓 |

𝑟 𝜀Length(𝛾) ≤ 𝜀 max

|𝑤−𝑧0|=𝑟|𝑓 |.

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Holomorphic functions are analytic – 3

Corollary

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe a function.

Then𝑓 is holomorphic/analytic/ℂ-differentiable if and only if𝑓 can belocallyexpressed as a power series in a neighborhood of any point of𝑈.

Proof.

⇒: the main theorem from the previous slide.

⇐: if𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛in𝐷_𝑟(𝑧₀)then𝑓 is holomorphic on𝐷_𝑟(𝑧₀)from last week lecture.

■

Corollary

A holomorphic/analytic/ℂ-differentiable function is infinitely many timesℂ-differentiable.

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Holomorphic functions are analytic – 4

The previous results are false forℝ-differentiability.

• Let𝑓 ∶ ℝ → ℝbe defined by𝑓 (𝑥) = {

𝑥²sin(¹𝑥) if𝑥 ≠ 0

0 otherwise

Then𝑓isℝ-differentiable but not𝒞¹.

• Let𝑔 ∶ ℝ → ℝbe defined by𝑔(𝑥) = {

𝑒⁻^𝑥2¹ if𝑥 ≠ 0 0 otherwise Then𝑔isℝ-differentiable, even𝒞^∞, but not analytic at0, i.e. it can’t be expressed as a power series around0:

Indeed∀𝑛 ∈ ℕ_≥0, 𝑔^(𝑛)(0) = 0, so if𝑔were equal to its Taylor series around0then it would be constant equal to0but𝑔is non-zero in any neighborhood of0.

• By a theorem of Borel², given a real sequence(𝑎_𝑛)_𝑛∈ℕ_≥0, there exists a𝒞^∞function defined in a neighborhood of0inℝsuch that∀𝑛 ∈ ℕ_≥0, 𝑓^(𝑛)(0) = 𝑎_𝑛.

Otherwise stated, any real power series is the Taylor expansion of a𝒞^∞function.

If we take𝑎_𝑛= (𝑛!)²then we obtain a power series whose radius of convergence is𝑅 = 0, hence a function with such a Taylor expansion can’t be analytic.

2It generalizes to multivariable functions.

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Continuation of analytic functions – 1

Theorem

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

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

Proof.

𝑧₀

𝑧

Let𝑧 ∈ 𝑈. Since𝑈 is path connected, there exists a𝒞⁰curve𝛾 ∶ [𝑎, 𝑏] → 𝑈from𝛾(𝑎) = 𝑧₀to𝛾(𝑏) = 𝑧.

Since𝑈is open, for every𝑤 ∈ 𝛾([𝑎, 𝑏])there exists𝑟_𝑤> 0such that𝐷_𝑟_𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟₁(𝑤₁), … , 𝐷_𝑟_𝑘(𝑤_𝑘).

By the theorem on Slide 2, if there exists𝑣 ∈ 𝐷_𝑟_𝑖(𝑤_𝑖)such that∀𝑛 ∈ ℕ_≥0, 𝑓^(𝑛)(𝑣) = 0then𝑓 ≡ 0on𝐷_𝑟_𝑖(𝑤_𝑖).

Two consecutive disks intersect (since they cover𝛾), so we conclude using the previous remark disk by disk

from𝑧₀to𝑧. ■

If you attend MAT327, another proof consists in showing that{𝑧 ∈ 𝑈 ∶ ∀𝑛 ∈ ℕ_≥0, 𝑓^(𝑛)(𝑧) = 0}is open, closed, and non-empty, hence equals to𝑈by connectedness.

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Continuation of analytic functions – 2

Corollary

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).

Proof.Then∀𝑛 ∈ ℕ_≥0, (𝑓 − 𝑔)^(𝑛)(𝑧₀) = 0(since𝑓 − 𝑔 ≡ 0on𝐷_𝑟(𝑧₀) ∩ 𝑈).

Hence𝑓 − 𝑔 ≡ 0on𝑈 by the previous theorem. ■

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Continuation of analytic functions – 3

The previous results are false forℝ-differentiability.

Example

Define𝑓 ∶ ℝ → ℝby

𝑓 (𝑥) = {

𝑒⁻^𝑥¹ if𝑥 > 0 0 otherwise then𝑓 isℝ-differentiable (even𝒞^∞). And

• ∀𝑛 ∈ ℕ_≥0, 𝑓^(𝑛)(0) = 0but𝑓 ≢ 0.

• ∀𝑥 ∈ (−∞, 0), 𝑓 (𝑥) = 0but𝑓 ≢ 0.

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Continuation of analytic functions – 4

A common way to construct an analytic function consists in defining it in a ”small” domain and then to extend it to an analytic function with a bigger domain.

By the above result, this analytic continuation, if it exists, is unique.

That’s a very powerful tool: knowing a function on a ”small” domain determines the function everywhere else³.Holomorphic/analytic functions are very rigid!

⚠The maximal domain may not beℂ: for instance, if we try to extendLog, we won’t be able to do a full turn around the origin since we won’t recover the same values (it increases by2𝑖𝜋).

Example

Let𝑓 ∶ 𝐷₁(0) → ℂbe defined by𝑓 (𝑧) =

+∞

∑𝑛=0

𝑧^𝑛. Then𝑓 coincides with _1−𝑧¹ on𝐷₁(0).

Hence we may extend𝑓 with𝐹 ∶ ℂ ⧵ {1} → ℂdefined by𝐹 (𝑧) = _1−𝑧¹ .

3And we will even weaken the assumptions later this term: it is enough to know the function on a set with a limit point.

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Order of a zero

Definition: order of a zero

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic. Let𝑧₀∈ 𝑈be 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

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic. Let𝑧₀∈ 𝑈be 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}.

Proof.𝑎_𝑛=^𝑓^(𝑛)^(𝑧⁰⁾

𝑛! ■

Proposition

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

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Morera’s theorem

Theorem: Morera’s theorem

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe continuous.

If for every (full) triangle𝑇 lying in𝑈 we have

∫_𝜕𝑇𝑓 = 0then𝑓 is holomorphic/analytic on𝑈.

Proof.Let𝑧₀∈ 𝑈 and𝑟 > 0be such that𝐷_𝑟(𝑧₀) ⊂ 𝑈. We define𝐹 ∶ 𝐷_𝑟(𝑧₀) → ℂby𝐹 (𝑧) =

∫_[𝑧₀_,𝑧]𝑓. Let𝑧, ℎ ∈ ℂbe such that𝑧, 𝑧 + ℎ ∈ 𝐷_𝑟(𝑧₀)then, considering the triangle whose vertices are𝑧₀,𝑧and𝑧 + ℎ, we obtain

∫_[𝑧

0,𝑧]

𝑓 +∫_{[𝑧,𝑧+ℎ]}𝑓 +

∫_{[𝑧+ℎ,𝑧}

0]

𝑓 = 0, i.e. 𝐹 (𝑧 + ℎ) − 𝐹 (𝑧) =

∫_{[𝑧,𝑧+ℎ]}𝑓. Then

|

𝐹 (𝑧 + ℎ) − 𝐹 (𝑧) ℎ − 𝑓 (𝑧)

|=

||

|||

(∫[𝑧,𝑧+ℎ]𝑓 ) − ℎ𝑓(𝑧) ℎ

||

|||

= 1

|ℎ| |∫_{[𝑧,𝑧+ℎ]}(𝑓 (𝑤) − 𝑓 (𝑧)d𝑤

|≤ sup

𝑤∈[𝑧,𝑧+ℎ]

|𝑓 (𝑤) − 𝑓 (𝑧)|

|ℎ| Length([𝑧, 𝑧 + ℎ])

= sup

𝑤∈[𝑧,𝑧+ℎ]

|𝑓 (𝑤) − 𝑓 (𝑧)| −−−→

ℎ→0 0 Hence𝐹 is holomorphic on𝐷_𝑟(𝑧₀)and𝐹^′= 𝑓. Furthermore,𝑓 is holomorphic on𝐷_𝑟(𝑧₀)(and hence at𝑧₀)

as the complex derivative of a holomorphic function. ■

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Theorem: characterizations of holomorphicity/analyticity

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂ. Then the following are equivalent:

1 𝑓 is holomorphic/analytic/ℂ-differentiable, i.e. ∀𝑧₀∈ 𝑈 , lim

ℎ→0

𝑓 (𝑧₀+ ℎ) − 𝑓 (𝑧₀)

ℎ exists.

2 𝑓 ∶ ̃̃ 𝑈 → ℝ²isℝ-differentiable and satisfies the Cauchy–Riemann equations on𝑈.̃

3 𝑓 may be written as a power series𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛on a neighborhood of every𝑧₀∈ 𝑈.

4 𝑓 is continuous and for every (full) triangle𝑇 lying in𝑈we have

∫_𝜕𝑇𝑓 = 0.

5 𝑓 is continuous and for every simple closed curve𝛾on𝑈whose inside is also included in𝑈, we have

∫_𝛾𝑓 = 0.

6 𝑓 admits local primitives/antiderivatives: for every𝑧₀∈ 𝑈 there exists𝐹 ∶ 𝐷_𝑟(𝑧₀) ∩ 𝑈 → ℂ holomorphic for some𝑟 > 0such that𝐹^′= 𝑓 on𝐷_𝑟(𝑧₀) ∩ 𝑈.

When𝑈is simply-connected, we may drop the assumption that the inside of the triangle/curve is included in𝑈in 4 5 furthermore we may also replace ”local primitives/antiderivatives” by ”a primitive/antiderivative” in 6 i.e. there exists 𝐹 ∶ 𝑈 → ℂholomorphic s.t.𝐹^′= 𝑓.

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Liouville’s theorem – 1

Definition

We say that a function𝑓 ∶ ℂ → ℂisentireif it is holomorphic (everywhere) onℂ.

Theorem: Liouville’s theorem

A bounded entire function is constant

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Liouville’s theorem – 2

Lemma: Cauchy’s inequalities

Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic. Let𝑟 > 0. If𝐷_𝑟(𝑧₀) ⊂ 𝑈 then

|𝑓^(𝑛)(𝑧₀)| ≤ 𝑛!

𝑟^𝑛 max

|𝑧−𝑧₀|=𝑟|𝑓 (𝑧)|

Proof.From the theorem on Slide 2

| 𝑓^(𝑛)(𝑧₀)

𝑛! |=

| 1

2𝑖𝜋 ∫_|𝑧−𝑧₀_|=𝑟

𝑓 (𝑤) (𝑤 − 𝑧₀)^𝑛+1d𝑤

|≤

|𝑧−𝑧max₀|=𝑟|𝑓 (𝑧)|

𝑟^𝑛+1

Length(|𝑧 − 𝑧0| = 𝑟)

2𝜋 = max_|𝑧−𝑧

0|=𝑟|𝑓 (𝑧)|

𝑟^𝑛

■ Proof of Liouville’s theorem.

Assume that there exists𝑀 > 0such that∀𝑧 ∈ ℂ, |𝑓 (𝑧)| ≤ 𝑀.

Let𝑧 ∈ ℂ. For𝑟 > 0, by Cauchy’s inequality with𝑛 = 1, we have|𝑓^′(𝑧)| ≤^𝑀_𝑟 −−−−−→

𝑟→+∞ 0.

Hence,∀𝑧 ∈ ℂ, 𝑓^′(𝑧) = 0. Therefore𝑓 is constant onℂ. ■

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Liouville’s theorem – 3

Theorem: d’Alembert–Gauss theorem or the Fundamental Theorem of Algebra

A non-constant polynomial with coeﬀicients inℂadmits a root/zero inℂ.

Proof.Assume that𝑃 is a complex polynomial with no root inℂ.

Then𝑄 = _𝑃¹ is a entire function and𝑄is bounded since lim

𝑧→+∞𝑄 = 0.

Therefore, by Liouville’s theorem,𝑄is constant, and so is𝑃. ■

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Analytic logarithm

Theorem

Let𝑈 ⊂ ℂbe a simply connected domain and𝑓 ∶ 𝑈 → ℂa holomorphic/analytic function which doesn’t vanish, i.e. ∀𝑧 ∈ 𝑈 , 𝑓 (𝑧) ≠ 0.

Then there exists𝑔 ∶ 𝑈 → ℂholomorphic/analytic such that𝑒^𝑔= 𝑓.

Proof. Since𝑓 doesn’t vanish, ^𝑓^′

𝑓 is holomorphic on𝑈.

Then, since𝑈is simply connected, ^𝑓_𝑓^′ admits a complex primitive/antiderivative, i.e. there exists

̃

𝑔 ∶ 𝑈 → ℂholomorphic/analytic such that𝑔^′̃ = ^𝑓_𝑓^′.

Then(𝑓 𝑒^{− ̃}^𝑔)^′= (𝑓^′− ̃𝑔^′𝑓 )𝑒^{− ̃}^𝑔= 0and therefore𝑓 𝑒^{− ̃}^𝑔= 𝐾is constant since𝑈is connected.

Since𝐾 ≠ 0, there exists𝑤 ∈ ℂsuch that𝑒^𝑤= 𝐾.

Then, for𝑔 = ̃𝑔 + 𝑤, we have𝑒^𝑔= 𝑓. ■

Careful

Such a function𝑔 is not unique! For instance𝑔 + 2𝑖𝜋is another suitable function.

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Multiplication of complex power series – 1

We are going to prove the following theorem, but using results related to analytic functions.

Theorem

Let𝑆_𝐴(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛and𝑆_𝐵(𝑧) =

+∞

∑𝑛=0

𝑏_𝑛(𝑧 − 𝑧₀)^𝑛be two power series of radii𝑅_𝐴and𝑅_𝐵.

We define𝑆_𝐴𝐵(𝑧) ≔

+∞

∑𝑛=0(

𝑛

∑𝑘=0

𝑎_𝑘𝑏_𝑛−𝑘

)(𝑧 − 𝑧₀)^𝑛and denote its radius of convergence by𝑅_𝐴𝐵. Then

1 𝑅_𝐴𝐵≥min(𝑅_𝐴, 𝑅_𝐵).

2 If|𝑧 − 𝑧₀| <min(𝑅_𝐴, 𝑅_𝐵)then𝑆_𝐴𝐵(𝑧) = 𝑆_𝐴(𝑧)𝑆_𝐵(𝑧), i.e.

+∞

∑𝑛=0(

𝑛

∑𝑘=0

)(𝑧 − 𝑧₀)^𝑛= (

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛 ) (

+∞

∑𝑛=0

𝑏_𝑛(𝑧 − 𝑧₀)^𝑛 ).

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Multiplication of complex power series – 1

Proof.We know that𝑆_𝐴and𝑆_𝐵are holomorphic on𝑧 ∈ ℂsuch that|𝑧 − 𝑧₀| <min(𝑅_𝐴, 𝑅_𝐵).

Then𝑓 (𝑧) = 𝑆_𝐴(𝑧)𝑆_𝐵(𝑧)is holomorphic on|𝑧 − 𝑧₀| <min(𝑅_𝐴, 𝑅_𝐵), so it can be written as a power series𝑓 (𝑧) =

+∞

∑𝑛=0

𝑐_𝑛(𝑧 − 𝑧₀)^𝑛on|𝑧 − 𝑧₀| <min(𝑅_𝐴, 𝑅_𝐵)(particularly its radius of convergence is at leastmin(𝑅_𝐴, 𝑅_𝐵)).

Then

𝑐_𝑛= 𝑓^(𝑛)(𝑧₀) 𝑛! = 1

𝑛!

𝑛

∑𝑘=0

𝑛!

(𝑛 − 𝑘)!𝑘!𝑆_𝐴^{(𝑛−𝑘)}(𝑧₀)𝑆_𝐵^(𝑘)(𝑧₀)by Leibniz rule

=

𝑛

∑𝑘=0

𝑆_𝐴^{(𝑛−𝑘)}(𝑧₀) (𝑛 − 𝑘)!

𝑆_𝐵^(𝑘)(𝑧₀) 𝑘!

=

𝑛

∑𝑘=0

𝑎_𝑛−𝑘𝑏_𝑘=

𝑛

∑𝑘=0

■