October16 ,2020toOctober21 ,2020 CONSEQUENCESOFTHECAUCHY’SINTEGRALFORMULA ComplexVariables

MAT334H1-F – LEC0101

Complex Variables

CONSEQUENCES OF THE CAUCHY’S INTEGRAL FORMULA

Holomorphic functions are analytic

– 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) = 𝑧 + 𝑟𝑒^{2𝑖𝜋𝑡}and the radius of convergence of this power

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 |.

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. ■

+∞ 𝑧 − 𝑧 ^𝑛

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.

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.

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

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.

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.

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.

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.

2

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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.

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_𝑟𝑤(𝑤) ⊂ 𝑈. Since𝛾([𝑎, 𝑏])is compact we may assume that it is covered by finitely many of these disks𝐷_𝑟1(𝑤₁), … , 𝐷_𝑟𝑘(𝑤_𝑘).

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

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

from𝑧 to𝑧. ■

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. ■

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. ■

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.

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).

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.

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.

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.

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

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) =

∫_[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.

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) =

∫_[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.

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.

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.

𝑧₀

𝑧 𝑧 + ℎ

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 |𝑓 (𝑤) − 𝑓 (𝑧)| −−−→

ℎ→0 0

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.

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

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. ■

First equality: we know that the𝑛-th coefficient of the power expansion of𝑓at𝑧₀is𝑎_𝑛= 𝑓^(𝑛)(𝑧₀) 𝑛! and by the theorem on Slide 2 that it is also equal to𝑎_𝑛= 1

2𝑖𝜋 ∫

𝑓 (𝑤) (𝑤 − 𝑧)^𝑛+1d𝑤.

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.

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. ■

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. ■

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

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. ■

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

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. ■

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

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. ■

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

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),

+∞ 𝑛

𝑛 +∞

𝑛 +∞

𝑛

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

𝑎_𝑘𝑏_𝑛−𝑘

■

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

𝑎_𝑘𝑏_𝑛−𝑘

■

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

𝑎_𝑘𝑏_𝑛−𝑘

■