1Holomorphicfunctionsareanalytic Contents 10-ConsequencesofCauchy’sintegralformula ComplexVariables

Complex Variables

10 - Consequences of Cauchy’s integral formula

Jean-Baptiste Campesato

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

Contents

1 Holomorphic functions are analytic 1

2 Continuation of analytic functions 3

3 Order of a zero 4

4 Morera’s theorem 4

5 Characterizations of holomorphicity/analyticity 5

6 Liouville’s theorem 5

7 Analytic logarithm 6

8 Multiplication of complex power series 6

1 Holomorphic functions are analytic

Theorem 1. 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.

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

2𝑖𝜋 ∫_𝛾 𝑓 (𝑤) 𝑤 − 𝑧d𝑤

= 1 2𝑖𝜋 ∫_𝛾

𝑓 (𝑤) 𝑤 − 𝑧₀

1 1 − ^𝑧−𝑧0

𝑤−𝑧₀

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

We need to justify the last equality (i.e. the permutation∫ − ∑).

Let𝜀 > 0. There exists𝑁such that if𝑘 ≥ 𝑁 then 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−𝑧max₀|=𝑟|𝑓 |

𝑟 𝜀Length(𝛾)

≤ 𝜀 max

|𝑤−𝑧₀|=𝑟|𝑓 |

■ Corollary 2. Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe a function.

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

Proof. ⇒: by Theorem1.

⇐: if𝑓 (𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛in𝐷_𝑟(𝑧₀)then𝑓 is holomorphic on𝐷_𝑟(𝑧₀)from last week lecture. ■ Corollary 3. A holomorphic/analytic/ℂ-differentiable function is infinitely many timesℂ-differentiable.

Remark 4. The previous results are false forℝ-differentiability.

• Let𝑓 ∶ ℝ → ℝbe defined by𝑓 (𝑥) = {

𝑥²sin(𝑥¹) if𝑥 ≠ 0

0 otherwise

Then𝑓 isℝ-differentiable but not𝒞¹.

• Let𝑔 ∶ ℝ → ℝbe defined by𝑔(𝑥) = {

𝑒⁻

1

𝑥2 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 Borel^⋆, given a real sequence(𝑎_𝑛)_𝑛∈ℕ

≥0, there exists a𝒞^∞function defined in a neigh- borhood 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.

2 Continuation of analytic functions

Theorem 5. Let𝑈 ⊂ ℂbe adomainand𝑓 ∶ 𝑈 → ℂbe a holomorphic/analytic function.

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

Proof. Let𝑧 ∈ 𝑈. Since𝑈is path connected, there exists a curve𝛾 ∶ [𝑎, 𝑏] → ℂwith𝛾(𝑎) = 𝑧₀and𝛾(𝑏) = 𝑧.

Since𝑈 is 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_𝑟

1(𝑤₁), … , 𝐷_𝑟

𝑘(𝑤_𝑘).

𝑧₀

𝑧

By Theorem1, 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. ■

Remark 6. 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 to𝑈 by connectedness.

Corollary 7. 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. ■

Remark 8. The previous results are false forℝ-differentiability. 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.

⋆It generalizes to multivariable functions.

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 else^⋆.Holomorphic/analytic functions are very rigid!

⚠The maximal domain may not be ℂ: for instance, if we try to extend Log, 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 9. Let𝑓 ∶ 𝐷₁(0) → ℂbe defined by𝑓 (𝑧) =

+∞

∑𝑛=0

𝑧^𝑛. Then𝑓 coincides with ¹

1−𝑧on𝐷₁(0).

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

1−𝑧.

3 Order of a zero

Definition 10. Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

Let𝑧₀∈ 𝑈be such that𝑓 (𝑧₀) = 0. We define theorder of vanishing of𝑓at𝑧₀by𝑚_𝑓(𝑧₀) ≔min{𝑛 ∈ ℕ ∶ 𝑓^(𝑛)(𝑧₀) ≠ 0}. Remark 11. Note that𝑚_𝑓(𝑧₀) > 0since𝑓 (𝑧₀) = 𝑓⁽⁰⁾(𝑧₀) = 0.

Proposition 12. 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}.

Proof. 𝑎_𝑛 = ^{𝑓(𝑛)(𝑧0)}

𝑛! ■

Proposition 13. Let𝑈 ⊂ ℂbe open and𝑓 ∶ 𝑈 → ℂbe holomorphic/analytic.

Then𝑧₀is a zero of order𝑛 ∈ ℕ_>0of𝑓if and only if there exists𝑔 ∶ 𝑈 → ℂholomorphic such that𝑓 (𝑧) = (𝑧−𝑧₀)^𝑛𝑔(𝑧) and𝑔(𝑧₀) ≠ 0.

4 Morera’s theorem

Theorem 14(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) =

∫_[𝑧

0,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.

𝑧₀ 𝑧

𝑧 + ℎ

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

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. ■

5 Characterizations of holomorphicity/analyticity

Theorem 15. 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∫_𝜕𝑇 𝑓 = 0.

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

∫_𝛾𝑓 = 0.

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

Remark 16. When𝑈is simply-connected, we may drop the assumption that the inside of the triangle/curve is included in𝑈in (4) and (5). Furthermore we may also replace ”local primitives/antiderivatives” by ”a primitive/antiderivative” in (6) i.e. there exists𝐹 ∶ 𝑈 → ℂholomorphic s.t.𝐹^′= 𝑓 .

6 Liouville’s theorem

Definition 17. We say that a function𝑓 ∶ ℂ → ℂisentireif it is holomorphic (everywhere) onℂ.

Theorem 18(Liouville’s theorem). A bounded entire function is constant.

Lemma 19(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.

|

𝑓^(𝑛)(𝑧₀) 𝑛! |=

| 1 2𝑖𝜋 ∫_|𝑧−𝑧

0|=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

For the first equality: we know that the𝑛-th coefficient of the power expansion of𝑓 at𝑧₀ is𝑎_𝑛 = 𝑓^(𝑛)(𝑧₀) 𝑛!

and by Theorem1that it is also equal to𝑎_𝑛= 1 2𝑖𝜋 ∫_|𝑧−𝑧

0|=𝑟

𝑓 (𝑤)

(𝑤 − 𝑧₀)^𝑛+1d𝑤. ■

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. ■

Theorem 20(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 = ¹

𝑃 is a entire function and𝑄is bounded since lim

𝑧→+∞𝑄 = 0.

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

7 Analytic logarithm

Theorem 21. 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 exists𝑔 ∶ 𝑈 → ℂ̃ holomorphic/analytic such that𝑔̃^′= ^𝑓′

𝑓 .

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

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

Then, for𝑔 = ̃𝑔 + 𝑤, we have𝑒^𝑔 = 𝑓. ■

Remark 22. Such a function𝑔is not unique! For instance𝑔 + 2𝑖𝜋is another suitable function.

8 Multiplication of complex power series

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

Theorem 23. 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

𝑎_𝑘𝑏_𝑛−𝑘

)(𝑧 − 𝑧₀)^𝑛= (

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛 ) (

+∞

∑𝑛=0

𝑏_𝑛(𝑧 − 𝑧₀)^𝑛 ).

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 series

𝑓 (𝑧) =

+∞

∑𝑛=0

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

𝑎_𝑘𝑏_𝑛−𝑘

■