ZEROES OF ANALYTIC FUNCTIONS – 1

MAT334H1-F – LEC0101

Complex Variables

ZEROES OF ANALYTIC FUNCTIONS – 1

Reviews from Oct 16 – Zeroes

Definition: order of a zero

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

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

Reviews from Oct 16 – Analytic continuation

Theorem

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

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

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).

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then 𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 1

It is actually possible to strengthen the previous results.

Theorem

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

Then either𝑓 ≡ 0or the zeroes of𝑓 are isolated¹:

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≠ 0then

𝑓 (𝑧) = (𝑧 − 𝑧₀)^𝑛𝑔(𝑧)where𝑔is 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 , 𝑓^(𝑛)(𝑧 ) = 𝑛!𝑎 = 0and𝑓 ≡ 0on𝑈by the theorem on Slide 3. ■

Isolated zeroes – 2

Corollary

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).

Isolated zeroes – 3

Corollary

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

Let(𝑧_𝑛)_𝑛∈ℕbe a sequence of terms in𝑈which is convergent to𝑧̃in𝑈and such that

∀𝑛 ∈ ℕ, 𝑓 (𝑧_𝑛) = 0.

Then𝑓 ≡ 0on𝑈.

Remark

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).

Isolated zeroes – 4

Homework

Let𝑈 ⊂ ℂbe a domain and𝑓 , 𝑔 ∶ 𝑈 → ℂbe holomorphic/analytic on𝑈.

Prove that if𝑓 𝑔 ≡ 0on𝑈 then either𝑓 ≡ 0or𝑔 ≡ 0.

Homework

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

1 𝑓 (¹𝑛) = 𝑛¹²

2 𝑓 (2𝑛¹) = 𝑓 (2𝑛+1¹ ) = ¹𝑛