October7 ,2020andOctober9 ,2020 POWERSERIES ComplexVariables

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MAT334H1-F – LEC0101

Complex Variables

POWER SERIES

October 7^th, 2020 and October 9^th, 2020

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Oct 7, 2020 and Oct 9, 2020 1 / 20

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Power series: definition

Definition: power series

A(complex) power series centered at𝑧₀∈ ℂis a series of the form

𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

where𝑎_𝑛∈ ℂ.

∑𝑛≥1 𝑛 ,∑

𝑛≥0

(𝑛²+ 1)(𝑧 − 2𝑖)^𝑛,∑

𝑛≥0𝑛!, …

A first natural question is: for which𝑧 ∈ ℂis the series𝑆(𝑧)convergent?

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Power series: definition

Definition: power series

𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

Examples

∑𝑛≥1

(𝑧 − 1)^𝑛 𝑛 ,∑

𝑛≥0

(𝑛²+ 1)(𝑧 − 2𝑖)^𝑛,∑

𝑛≥0

𝑧^𝑛 𝑛!, …

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Definition: power series

𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛

Examples

∑𝑛≥1

(𝑧 − 1)^𝑛 𝑛 ,∑

𝑛≥0

(𝑛²+ 1)(𝑧 − 2𝑖)^𝑛,∑

𝑛≥0

𝑧^𝑛 𝑛!, …

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Radius of convergence – 1

Theorem: radius of convergence

Given a power series𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛there exists a unique𝑅 ∈ [0, +∞]such that

∀𝑧 ∈ ℂ,

⎧⎪

⎪⎨

⎪⎪

⎩

|𝑧 − 𝑧₀| < 𝑅 ⟹ 𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is absolutely convergent

|𝑧 − 𝑧₀| > 𝑅 ⟹ 𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is divergent

We say that𝑅is theradius of convergenceof the power series𝑆(𝑧).

Below is the meaning of𝑅for the extremal values0and+∞:

• 𝑅 = 0means that𝑆(𝑧)is convergent if and only if𝑧 = 𝑧₀.

• 𝑅 = +∞means that𝑆(𝑧)is absolutely convergent for all𝑧 ∈ ℂ.

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Theorem: radius of convergence

Given a power series𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛there exists a unique𝑅 ∈ [0, +∞]such that

∀𝑧 ∈ ℂ,

⎧⎪

⎪⎨

⎪⎪

⎩

|𝑧 − 𝑧₀| < 𝑅 ⟹ 𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is absolutely convergent

|𝑧 − 𝑧₀| > 𝑅 ⟹ 𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is divergent

We say that𝑅is theradius of convergenceof the power series𝑆(𝑧).

Below is the meaning of𝑅for the extremal values0and+∞:

• 𝑅 = 0means that𝑆(𝑧)is convergent if and only if𝑧 = 𝑧₀.

• 𝑅 = +∞means that𝑆(𝑧)is absolutely convergent for all𝑧 ∈ ℂ.

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Radius of convergence – 2

Careful

We can’t conclude anything when|𝑧 − 𝑧₀| = 𝑅.

Indeed, the radius of convergence of the following power series is𝑅 = 1, but:

•

+∞

∑𝑛=0

𝑧^𝑛is divergent for all𝑧 ∈ ℂsuch that|𝑧| = 1.

•

+∞

∑𝑛=0

𝑧^𝑛

𝑛 is divergent for𝑧 = 1but (semi)-convergent for all𝑧 ∈ ℂsuch that|𝑧| = 1and𝑧 ≠ 1.

•

+∞

∑𝑛=0

𝑧^𝑛

𝑛² is absolutely convergent for all𝑧 ∈ ℂsuch that|𝑧| = 1.

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𝑧₀ 𝑅

∑ 𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is ACV

∑ 𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is DV,actually(𝑎_𝑛(𝑧 − 𝑧₀)^𝑛)_𝑛is not even bounded.

Circle of indeterminacy

For𝑧in the circle of indeterminacy|𝑧 − 𝑧₀| = 𝑅, the power series𝑆(𝑧)may be divergent, semi-convergent or absolutely convergent: we can’t conclude...

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Radius of convergence – 4

Proof: existence of 𝑅 for 𝑆(𝑧) = ∑ 𝑎

_𝑛

(𝑧 − 𝑧

₀

)

^𝑛

.

Note that𝐸 = {𝑟 ∈ [0, +∞) ∶ (𝑎𝑛𝑟^𝑛)_𝑛is bounded}is non-empty since0 ∈ 𝐸.

Hence𝑅 =sup𝐸is well-defined (possibly+∞if𝐸is unbounded).

• Assume that|𝑧 − 𝑧₀| < 𝑅then there exists𝜌such that|𝑧 − 𝑧₀| < 𝜌 < 𝑅and

∃𝑀 > 0, ∀𝑛 ∈ ℕ, |𝑎𝑛𝜌^𝑛| < 𝑀. Then∑

𝑛≥0

|𝑎𝑛(𝑧 − 𝑧₀)^𝑛| = ∑

𝑛≥0

|𝑎𝑛𝜌^𝑛| (|𝑧 − 𝑧₀|

𝜌 )

𝑛

≤ 𝑀∑

𝑛≥0(

|𝑧 − 𝑧₀|

𝜌 )

𝑛

. The last series is convergent (geometric series with ^|𝑧−𝑧_𝜌⁰^|< 1).

Hence𝑆(𝑧) =∑

𝑛≥0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is absolutely convergent.

• Assume that|𝑧 − 𝑧₀| > 𝑅then lim

𝑛→+∞𝑎_𝑛(𝑧 − 𝑧₀)^𝑛≠ 0since(𝑎𝑛(𝑧 − 𝑧₀)^𝑛)_𝑛is not bounded.

Hence𝑆(𝑧) =∑

𝑛≥0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is divergent.

■

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Hence we have that the radius of convergence of𝑆(𝑧) = ∑ 𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is given by 𝑅 =sup{𝑟 ∈ [0, +∞) ∶ (𝑎𝑛𝑟^𝑛)_𝑛is bounded}

=sup

{|𝑧 − 𝑧₀| ∶

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is convergent }

=inf

{|𝑧 − 𝑧₀| ∶

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is divergent }

By a theorem of Hadamard, we also have that𝑅 = 1 lim sup

𝑛→+∞

√|𝑎𝑛 _𝑛|

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Two useful methods to compute the radius of convergence

Theorem: d’Alembert ratio test for power series

Let𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛be a power series such that𝑎_𝑛≠ 0for𝑛big enough.

If the limitℓ ≔ lim

𝑛→+∞| 𝑎_𝑛+1

𝑎_𝑛 |exists or is+∞then the radius of convergence of𝑆(𝑧)is𝑅 = ¹_ℓ.

Theorem: root test for power series

Let𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛be a power series.

If the limitℓ ≔ lim

𝑛→+∞

√|𝑎𝑛 _𝑛|exists or is+∞then the radius of convergence of𝑆(𝑧)is𝑅 = _ℓ¹.

In the two above tests we use the convention ¹

∞ = 0and¹

0= ∞.

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We define𝑆(𝑧) =

+∞

𝑛=𝑛∑₀

(𝑧 − 𝑧₀)^𝑛.

1 What is the radius of convergence𝑅of𝑆(𝑧)?

2 Study𝑆(𝑧)of𝑧such that|𝑧 − 𝑧₀| = 𝑅.

3 Find an expression for𝑆(𝑧)when𝑧 ∈ 𝐷_𝑅(𝑧₀).

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Homework – 1

What are the radii of convergence of the following power series:

1 +∞

∑𝑛=0

𝑎^𝑛𝑧^𝑛(where𝑎 ∈ ℂ)

2 +∞

∑𝑛=0

𝑒^−√𝑛𝑧^𝑛

3 +∞

∑𝑛=0

sin(𝑛)𝑧^𝑛

4 +∞

∑𝑛=0

𝑛²− 𝑛 + 3 2𝑛³+ 𝑛 + 𝜋𝑧^𝑛

5 +∞

∑𝑛=0

ln(𝑛!)𝑧^𝑛

6 +∞

∑𝑛=0(5𝑛 2𝑛)𝑧^𝑛

7 +∞

∑𝑛=0

1

𝑛²+ 1𝑧^2𝑛 (be careful with this one, why?)

8 +∞

∑𝑛=1

1 𝑛3^𝑛𝑧^𝑛

Hint: for the first three, you can simply use that𝑅 =sup{𝑟 ∈ [0, +∞) ∶ (𝑎𝑛𝑟^𝑛)_𝑛is bounded}.

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For question 7 in the previous slide, you can NOT apply the ratio test for power series.

Indeed,

+∞

∑𝑛=0

1

𝑛²+ 1𝑧^2𝑛= 1 + 0𝑧 + 1

2𝑧²+ 0𝑧³+ 1

5𝑧⁴+ 0𝑧⁵+ 1

10𝑧⁶+ 0𝑧⁷+ ⋯.

If we want to rewrite

+∞

∑𝑛=0

1

𝑛²+ 1𝑧^2𝑛=

+∞

∑𝑛=0

𝑎_𝑛𝑧^𝑛then𝑎_𝑛= {

1

(𝑛/2)²+1 if𝑛is even 0 if𝑛is odd . Hence, since𝑎_𝑛= 0for𝑛odd, we can’t compute lim

𝑛→+∞| 𝑎_𝑛+1

𝑎_𝑛 |.

One way is to directly work with the usual ratio test (for series, not power series): if𝑧 ≠ 0then

𝑛→+∞lim

||

1

(𝑛+1)²+1𝑧^2(𝑛+1)

1 𝑛²+1𝑧^2𝑛

||

= lim

𝑛→+∞|

𝑛²+ 1 (𝑛 + 1)²+ 1𝑧²

|= |𝑧|²

Hence, by the usual ratio test, the series is absolutely convergent when|𝑧| < 1and divergent when|𝑧| > 1. Therefore the radius of convergence is1.

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Sum of two power series – 1

Theorem

Let𝑆_𝐴=

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛and𝑆_𝐵=

+∞

∑𝑛=0

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

We define𝑆_𝐴+𝐵(𝑧) ≔

+∞

∑𝑛=0

(𝑎_𝑛+ 𝑏_𝑛)(𝑧 − 𝑧₀)^𝑛and denote its radius of convergence by𝑅_𝐴+𝐵. Then

1 𝑅_𝐴+𝐵≥min(𝑅_𝐴, 𝑅_𝐵).

2 If𝑅_𝐴≠ 𝑅_𝑏then𝑅_𝐴+𝐵=min(𝑅_𝐴, 𝑅_𝐵).

3 If|𝑧 − 𝑧₀| <min(𝑅_𝐴, 𝑅_𝐵)then𝑆_𝐴+𝐵(𝑧) = 𝑆_𝐴(𝑧) + 𝑆_𝐵(𝑧), i.e.

+∞

∑𝑛=0

(𝑎_𝑛+ 𝑏_𝑛)(𝑧 − 𝑧₀)^𝑛=

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛+

+∞

∑𝑛=0

𝑏_𝑛(𝑧 − 𝑧₀)^𝑛.

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Careful

When𝑅_𝐴= 𝑅_𝐵we may have𝑅_𝐴+𝐵= 𝑅_𝐴= 𝑅_𝐵or𝑅_𝐴+𝐵> 𝑅_𝐴= 𝑅_𝐵.

1 𝑆_𝐴(𝑧) =

+∞

∑𝑛=0

𝑧^𝑛and𝑆_𝐵(𝑧) =

+∞

∑𝑛=0

𝑛𝑧^𝑛are of radii1, and𝑆_𝐴+𝐵(𝑧) =

+∞

∑𝑛=0

(1 + 𝑛)𝑧^𝑛is of radius 1.

2 𝑆_𝐴(𝑧) =

+∞

∑𝑛=0

+∞

∑𝑛=0

(2^−𝑛− 1)𝑧^𝑛are of radii1, and𝑆_𝐴+𝐵(𝑧) =

+∞

∑𝑛=0

2^−𝑛𝑧^𝑛is of radius 2.

3 𝑆_𝐴(𝑧) =

+∞

∑𝑛=0

+∞

∑𝑛=0

−𝑧^𝑛are of radii1, and𝑆_𝐴+𝐵(𝑧) =

+∞

∑𝑛=0

0𝑧^𝑛is of radius+∞.

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Product of two power series – 1

Theorem

Let𝑆_𝐴=

+∞

∑𝑛=0

+∞

∑𝑛=0

We define𝑆_𝐴𝐵(𝑧) ≔

+∞

∑𝑛=0(

𝑛

∑𝑘=0

𝑎_𝑘𝑏_𝑛−𝑘

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

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 ).

Careful

𝑆_𝐴𝐵(𝑧)≠

+∞

∑𝑛=0

(𝑎_𝑛𝑏_𝑛)(𝑧 − 𝑧₀)^𝑛: we need to use theCauchy productas defined above!

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Theorem

Let𝑆_𝐴=

+∞

∑𝑛=0

+∞

∑𝑛=0

We define𝑆_𝐴𝐵(𝑧) ≔

+∞

∑𝑛=0(

𝑛

∑𝑘=0

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

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 ).

Careful

𝑆_𝐴𝐵(𝑧)≠

+∞

∑𝑛=0

(𝑎_𝑛𝑏_𝑛)(𝑧 − 𝑧₀)^𝑛: we need to use theCauchy productas defined above!

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Product of two power series – 2

Careful

We may have𝑅_𝐴𝐵 =min(𝑅_𝐴, 𝑅_𝐵)or𝑅_𝐴𝐵>min(𝑅_𝐴, 𝑅_𝐵), even when𝑅_𝐴≠ 𝑅_𝐵.

𝑆_𝐴(𝑧) =

+∞

∑𝑛=0

𝑧^𝑛is of radius1and𝑆_𝐵(𝑧) = 1 − 𝑧is of radius+∞but𝑆_𝐴𝐵(𝑧) = 1is of radius+∞.

In this example, intuitively: 1

1 − 𝑧(1 − 𝑧) = 1

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Theorem

Let𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛be a power series of radius of convergence𝑅.

Then𝑆is holomorphic/analytic¹/ℂ-differentiable on𝐷_𝑅(𝑧₀)and moreover

∀𝑧 ∈ 𝐷_𝑅(𝑧₀), 𝑆^′(𝑧) =

+∞

∑𝑛=1

𝑛𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−1=

+∞

∑𝑛=0

(𝑛 + 1)𝑎_𝑛+1(𝑧 − 𝑧₀)^𝑛

whose radius of convergence is also𝑅.

1I really don’t like the choice in the textbook…

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Differentiability of power series – 2

The previous result contains several things:

1 The power series

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛and

+∞

∑𝑛=1

𝑛𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−1have same radius of convergence,

2 The function𝑆 ∶ 𝐷_𝑅(𝑧₀) → ℂdefined by𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is holomorphic on𝐷_𝑅(𝑧₀),

3 For all𝑧 ∈ 𝐷_𝑅(𝑧₀), 𝑆^′(𝑧) =

+∞

∑𝑛=1

𝑛𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−1.

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Differentiability of power series – 3

Corollary

Let𝑆(𝑧) =

+∞

∑𝑛=0

Then𝑆is infinitely many timesℂ-differentiable on𝐷_𝑅(𝑧₀)and moreover

∀𝑧 ∈ 𝐷_𝑅(𝑧₀), 𝑆^(𝑘)(𝑧) =

+∞

∑𝑛=𝑘

𝑛!

(𝑛 − 𝑘)!𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−𝑘 whose radius of convergence is also𝑅.

If𝑆(𝑧) =∑

𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is a power series with radius of convergence𝑅 > 0, then𝑎_𝑛= 𝑆 (𝑧₀) 𝑛! .

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Differentiability of power series – 3

Corollary

Let𝑆(𝑧) =

+∞

∑𝑛=0

Then𝑆is infinitely many timesℂ-differentiable on𝐷_𝑅(𝑧₀)and moreover

∀𝑧 ∈ 𝐷_𝑅(𝑧₀), 𝑆^(𝑘)(𝑧) =

+∞

∑𝑛=𝑘

𝑛!

(𝑛 − 𝑘)!𝑎_𝑛(𝑧 − 𝑧₀)^𝑛−𝑘 whose radius of convergence is also𝑅.

(Important) Corollary

If𝑆(𝑧) =

+∞

∑𝑛=0

𝑎_𝑛(𝑧 − 𝑧₀)^𝑛is a power series with radius of convergence𝑅 > 0, then𝑎_𝑛= 𝑆^(𝑛)(𝑧₀) 𝑛! .

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Example: power series expansion of the complex exponential

1 The radius of convergence of𝑆(𝑧) =

+∞

∑𝑛=0

𝑧^𝑛 𝑛! is+∞.

Indeed, we may conclude with the ratio test since lim

𝑛→+∞

||

1 (𝑛+1)!

1 𝑛!

||

= lim

𝑛→+∞

1 𝑛 + 1 = 0.

2 For all𝑧 ∈ ℂ,𝑆^′(𝑧) = 𝑆(𝑧).

Indeed,𝑆^′(𝑧) =

+∞

∑𝑛=1

𝑛𝑧^𝑛−1 𝑛! =

+∞

∑𝑛=1

𝑧^𝑛−1 (𝑛 − 1)!= 𝑆(𝑧).

3 For all𝑧 ∈ ℂ,𝑒^𝑧=

+∞

∑𝑛=0

𝑧^𝑛 𝑛!. Indeed, ^𝜕

𝜕𝑧(𝑒^−𝑧𝑆(𝑧)) = −𝑒^−𝑧𝑆(𝑧) + 𝑒^−𝑧𝑆^′(𝑧) = −𝑒^−𝑧𝑆(𝑧) + 𝑒^−𝑧𝑆(𝑧) = 0.

Thence𝑒^−𝑧𝑆(𝑧) = 𝜆is constant sinceℂis connected.

We evaluate at𝑧 = 0to obtain that𝜆 = 1, so that𝑆(𝑧) = 𝑒^𝑧.

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Differentiability of power series – 5

Example: power series expansion of the complex cosine

For𝑧 ∈ ℂ, we have

cos(𝑧) = 𝑒^𝑖𝑧+ 𝑒^−𝑖𝑧 2

= 1 2 (

+∞

∑𝑛=0

(𝑖𝑧)^𝑛 𝑛! +

+∞

∑𝑛=0

(𝑖𝑧)^𝑛 𝑛! )

=

+∞

∑𝑛=0

(−1)^𝑛 𝑧^2𝑛 (2𝑛)!