output.to from Sideway
Draft for Information Only

# Content

`Infinite Series of Complex Numbers Infinite Series Example Another Example Absolute Convergence  Example Power Series (Taylor Series) The Radius of Convergence  Examples Analyticity of Power Series  Example Integration of Power Series  Example Ratio Test  Examples The Root Test  Examples The Cauchy Hadamard Criterion Analytic Functions and Power Series  Examples A Corollary`

source/reference:

# Infinite Series of Complex Numbers

## Infinite Series

```DefinitionAn infinite series ∞∑𝑘=0𝑎𝑘=𝑎0+𝑎1+𝑎2+⋯+𝑎𝑛+𝑎𝑛+1 (with 𝑎𝑘∈ℂ) converges to 𝑆 if the sequence of partial sums {𝑆𝑛}, given by 𝑆𝑛=𝑛∑𝑘=0=𝑎𝑘=𝑎0+𝑎1+𝑎2+⋯+𝑎𝑛 converges to 𝑆.```

## Example

Consider 𝑘=0𝑧𝑘, for some 𝑧∈ℂ. having that `𝑆𝑛=1+𝑧+𝑧2+⋯+𝑧𝑛` Finding a closed formula for 𝑆𝑛 in order to find the limit as 𝑛→∞. ```Trick: 𝑆𝑛=1+𝑧+𝑧2+⋯+𝑧𝑛, so 𝑧⋅𝑆𝑛=𝑧+𝑧2+⋯+𝑧𝑛+𝑧𝑛+1, thus 𝑆𝑛−𝑧⋅𝑆𝑛=1−𝑧𝑛+1 ``` Hence 𝑆𝑛=1−𝑧𝑛+11−𝑧 for 𝑧≠1, and since 𝑧𝑛+1→0 as 𝑛→∞ as long as |𝑧|<1. Having that `∞∑𝑘=0𝑧𝑘=11−𝑧 for |𝑧|<1`

For |𝑧|≥1 `TheoremIf a series ∞∑𝑘=0𝑎𝑘 converges then 𝑎𝑘→0 as 𝑘→∞.` If |𝑧|≥1, then |𝑧|𝑘0 as 𝑘→∞, thus 𝑘=0𝑧𝑘 does not converge for |𝑧|≥1. Thus the series diverges for |𝑧|≥1.

Let's now analyze the real and imaginary parts of the equation 𝑘=0𝑧𝑘=11−𝑧 for |𝑧|<1;

For |𝑧|<1 Writing 𝑧=𝑟ℯ𝑖𝜃, then 𝑧𝑘=𝑟𝑘𝑖𝑘𝜃=𝑟𝑘cos(𝑘𝜃)+𝑖𝑟𝑘sin(𝑘𝜃). Thus `∞∑𝑘=0𝑧𝑘=∞∑𝑘=0𝑟𝑘cos(𝑘𝜃)+𝑖∞∑𝑘=0𝑟𝑘sin(𝑘𝜃)` Furthermore, ```11−𝑧=11−𝑟ℯ𝑖𝜃=1−𝑟ℯ−𝑖𝜃(1−𝑟ℯ𝑖𝜃)(1−𝑟ℯ−𝑖𝜃)  =1−𝑟cos 𝜃+𝑖𝑟sin 𝜃1−𝑟ℯ𝑖𝜃−𝑟ℯ−𝑖𝜃+𝑟2=1−𝑟cos 𝜃+𝑖𝑟sin 𝜃1−2𝑟cos 𝜃+𝑟2``` Thus `∞∑𝑘=0𝑟𝑘cos(𝑘𝜃)=1−𝑟cos 𝜃1−2𝑟cos 𝜃+𝑟2 and ∞∑𝑘=0𝑟𝑘sin(𝑘𝜃)=𝑟sin 𝜃1−2𝑟cos 𝜃+𝑟2`

## Another Example

Consider 𝑘=1𝑖𝑘𝑘. Does this series converge?

• Note that 𝑘=1𝑖𝑘𝑘=𝑘=11𝑘 is the harmonic series, which is known to diverge. One way to see this: `∞∑𝑘=11𝑘=1+12+13+14 ≥1/2} +15+16+17+18 ≥1/2} +19+⋯+116 ≥1/2} +⋯`
• But does the series itself (without the absolute values) converge? let's split it up into real and imaginary parts.
• Note: When 𝑘 is even (i.e. 𝑘 is of the form 𝑘=2𝑛), then 𝑖𝑘=𝑖2𝑛=(−1)𝑛 is real. When 𝑘 is odd (i.e. 𝑘 is of the form 𝑘=2𝑛+1), then 𝑖𝑘=𝑖2𝑛+1=𝑖(−1)𝑛 is purely imaginary. Thus ```∞∑𝑘=1𝑖𝑘𝑘=∞∑𝑛=1𝑖2𝑛2𝑛+∞∑𝑛=0𝑖2𝑛+12𝑛+1  =12∞∑𝑛=1(−1)𝑛𝑛+𝑖∞∑𝑛=0(−1)𝑛2𝑛+1 ``` But `∞∑𝑛=1(−1)𝑛𝑛=−1+12−13+14−+⋯` is the alternating harmonic series, which converges.

## Absolute Convergence

`DefinitionA series ∞∑𝑘=0𝑎𝑘 converges absolutely if the series ∞∑𝑘=0|𝑎𝑘| converges.`

Examples

```∞∑𝑘=0𝑧𝑘 converges and converges absolutely for |𝑧|<1. ∞∑𝑘=0𝑖𝑘𝑘 converges, but not absolutely.``` `TheoremIf ∞∑𝑘=0𝑎𝑘 converges absolutely, then it also converges, and ∞∑𝑘=0𝑎𝑘≤∞∑𝑘=0|𝑎𝑘|.`

### Example

If |𝑧|<1, then the series 𝑘=0𝑧𝑘 converge absolutely, so `∞∑𝑘=0𝑧𝑘≤∞∑𝑘=0|𝑧|𝑘` But the left-hand side equals 11−𝑧, and the right-hand side equals 11−|𝑧|, so that `11−𝑧≤11−|𝑧|`

## Power Series (Taylor Series)

```DefinitionA power series (also called Taylor series), centered at 𝑧0∈ℂ, is a series of the form ∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 ```

Example

• 𝑘=0𝑧𝑘 is a power series with 𝑎𝑘=1, 𝑧0=0. It converges for |𝑧|<1.
• 𝑘=0(−1)𝑘2𝑘𝑧2𝑘=1−𝑧22+𝑧44𝑧68+−⋯=𝑘=0−𝑧22𝑘=𝑘=0𝑤𝑘, where 𝑤=−𝑧22. This series converges when |𝑤|<1, and diverges when |𝑤|≥1. Therefore, the original series converges when |𝑧|<2 and diverges when |𝑧|≥2

For what values of 𝑧 does a power series converge?

`TheoremLet ∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 be a power series. Then there exists a number 𝑅, with 0≤𝑅≤∞, such that the series converges absolutely in {|𝑧−𝑧0|<𝑅} and diverges in {|𝑧−𝑧0|>𝑅}. Furthermore, the convergence is uniform in {|𝑧−𝑧0|≤𝑟} for each 𝑟<𝑅`

𝑅 is called the radius of convergence of the power series.

### Examples

• 𝑘=0𝑧𝑘 has the radius of convergence 1.
• 𝑘=0(−1)𝑘2𝑘𝑧2𝑘 has radius of convergence 2
• 𝑘=0𝑘𝑘𝑧𝑘 Pick an arbitrary 𝑧∈ℂ\{0}. Observe that |𝑘𝑘𝑧𝑘|=(𝑘|𝑧|)𝑘≥2𝑘 as soon as 𝑘≥2|𝑧|, thus the series does not converge for any 𝑧≠0. The radius of convergence of this power series is 0!
• 𝑘=0𝑧𝑘𝑘𝑘 Pick an arbitrary 𝑧∈ℂ. Observe that 𝑧𝑘𝑘𝑘=|𝑧|𝑘𝑘12𝑘 as soon as 𝑘≥2|𝑧|. Thus the series converges absolutely for all 𝑧∈ℂ, and so 𝑅=∞!

## Analyticity of Power Series

```TheoremSuppose that ∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 is a power series of radius of convergence 𝑅>0. Then 𝑓(𝑧)=∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 is analytic in {|𝑧−𝑧0|<𝑅} Furthermore, the series can be differentiated term by term, i.e. 𝑓′(𝑧)=∞∑𝑘=1𝑎𝑘⋅𝑘(𝑧−𝑧0)𝑘−1, 𝑓″(𝑧)=∞∑𝑘=2𝑎𝑘⋅𝑘(𝑘−1)(𝑧−𝑧0)𝑘−2, ⋯ In particular, 𝑓(𝑘)(𝑧0)=𝑎𝑘⋅𝑘!, i.e. 𝑎𝑘=𝑓(𝑘)(𝑧0)𝑘! for 𝑘≥0.```

### Example

𝑘=0𝑧𝑘 has the radius of convergence 1, and so by the theorem, `𝑓(𝑧)=∞∑𝑘=0𝑧𝑘 is analytic in {|𝑧|<1}` Taking the derivative and differentiating term by term (as in the theorem), we find `𝑓′(𝑧)=∞∑𝑘=1𝑘𝑧𝑘−1 (=∞∑𝑘=0(𝑘+1)𝑧𝑘)` But also know that 𝑓(𝑧)=11−𝑧, and so 𝑓′(𝑧)=1(1−𝑧)2. Thus `∞∑𝑘=0(𝑘+1)𝑧𝑘=1(1−𝑧)2`

## Integration of Power Series

Note: Power series can similarly be integrated term by term:

```FactIf ∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 has radius of convergence 𝑅, then for any 𝑤 with |𝑤−𝑧0|<𝑅. we have that 𝑤∫𝑧0∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘𝑑𝑧=∞∑𝑘=0𝑎𝑘𝑤∫𝑧0(𝑧−𝑧0)𝑘𝑑𝑧=∞∑𝑘=0𝑎𝑘1𝑘+1(𝑤−𝑧0)𝑘+1 Here, the integral is taken over any curve in the disk {|𝑧−𝑧0|<𝑅} from 𝑧0 to 𝑤.```

### Example

Let's again look at the power series 𝑘=0𝑧𝑘, which has 𝑅=1. Then for any 𝑤 with |𝑤|<1, thus have `𝑤∫𝑧0∞∑𝑘=0𝑧𝑘𝑑𝑧=∞∑𝑘=0𝑤∫𝑧0𝑧𝑘𝑑𝑧=∞∑𝑘=01𝑘+1𝑤𝑘+1=∞∑𝑘=1𝑤𝑘𝑘` Also know that 𝑘=0𝑧𝑘=11−𝑧, hence `𝑤∫𝑧0∞∑𝑘=0𝑧𝑘𝑑𝑧=𝑤∫𝑧011−𝑧𝑑𝑧=−Log(1−𝑧)𝑤｜0=−Log(1−𝑤)` Here, we used that Log z is analytic in ℂ\(−∞,0], hence −Log(1−z) is analytic in ℂ\[1,∞), in particular in {|z|<1}, where it is a primitive of 11−𝑧

We have shown: `𝑤∫𝑧0∞∑𝑘=0𝑧𝑘𝑑𝑧=∞∑𝑘=1𝑤𝑘𝑘 and 𝑤∫𝑧0∞∑𝑘=0𝑧𝑘𝑑𝑧=−Log(1−𝑤)` hence `∞∑𝑘=1𝑤𝑘𝑘=−Log(1−𝑤) for |𝑤|<1` Letting z=1−𝑤 this becomes `Log z=−∞∑𝑘=1(1−z)𝑘𝑘=∞∑𝑘=1(−1)𝑘+1𝑘(z−1)𝑘 for |z−1|<1`

## Ratio Test

Given a power series 𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘, there exists a number 𝑅 with 0≤𝑅≤∞ such that the series converges (absolutely) in {|𝑧−𝑧0|<𝑅} and diverges in {|𝑧−𝑧0|>𝑅}. `Theorem (Ratio Test)If the sequence 𝑎𝑘𝑎𝑘+1 has a limit as 𝑘→∞ then this limit is the radius of convergence, 𝑅, of the poser series ∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘.` Note: "∞" is an allowable limit.

### Examples

• 𝑘=0𝑧𝑘: Here, 𝑎𝑘=1, so 𝑎𝑘𝑎𝑘+1=1→1 as 𝑘→∞. Thus 𝑅=1.
• 𝑘=0𝑘𝑧𝑘: Here, 𝑎𝑘=𝑘, so 𝑎𝑘𝑎𝑘+1→1 as 𝑘→∞. Thus 𝑅=1.
• 𝑘=0𝑧𝑘𝑘!: Here, 𝑎𝑘=1𝑘!, so 𝑎𝑘𝑎𝑘+1=(𝑘+1)!𝑘!=𝑘+1→∞ as 𝑘→∞. Thus 𝑅=∞.
• 𝑘=0𝑧𝑘𝑘𝑘: Here, 𝑎𝑘=1𝑘𝑘, so 𝑎𝑘𝑎𝑘+1=(𝑘+1)𝑘+1(𝑘)𝑘=(𝑘+1)1+1𝑘→∞?? as 𝑘→∞

## The Root Test

```Theorem (Root Test)If the sequence {𝑘|𝑎𝑘|} has a limit as 𝑘→∞ then 𝑅=1 Lim𝑘→∞{𝑘|𝑎𝑘|}.```

Note:

• If Lim𝑘→∞{𝑘|𝑎𝑘|}=0 then 𝑅=∞.
• If Lim𝑘→∞{𝑘|𝑎𝑘|}=∞ then 𝑅=0.

### Examples

• 𝑘=0𝑧𝑘𝑘𝑘: Here, 𝑎𝑘=1𝑘𝑘, so 𝑘|𝑎𝑘|=1𝑘→0 as 𝑘→∞. Thus 𝑅=∞.
• 𝑘=0𝑘𝑧𝑘: Here, 𝑎𝑘=𝑘, so 𝑘|𝑎𝑘|=𝑘𝑘→1 as 𝑘→∞. Thus 𝑅=1.
• 𝑘=02𝑘𝑧𝑘: Here, 𝑎𝑘=2𝑘, so 𝑘|𝑎𝑘|=2→2 as 𝑘→∞. Thus 𝑅=12.
• 𝑘=0(−1)𝑘2𝑘𝑧2𝑘: Here, 𝑎2𝑘=(−1)𝑘2𝑘, and 𝑎2𝑘+1=0, so 2𝑘|𝑎2𝑘|=121/2 for 𝑘≥1 and 2𝑘+1|𝑎2𝑘+1|=0, and so the sequence 𝑘|𝑎𝑘| (𝑘≥1) is `0, 12, 0, 12, 0, 12, 0,⋯` and this sequence does not have a limit. Note: 𝑎𝑘𝑎𝑘+1 has no limit either.

For the series 𝑘=0(−1)𝑘2𝑘𝑧2𝑘, neither the root test nor the ratio test "work".

Yet, letting 𝑤=𝑧2, the series becomes 𝑘=0(−1)𝑘2𝑘𝑤𝑘, and 𝑘(−1)𝑘2𝑘=1212 as 𝑘→∞, so the new series converges for |𝑤|<2. Thus the original series converges for |𝑧|<2. Is there a formula that finds this?

```Fact (Cauchy-Hadamard)The radius of convergence of the poser series ∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 equals 𝑅=1 Lim sup𝑘→∞ 𝑘|𝑎𝑘|```

## Analytic Functions and Power Series

Recall: 𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘 is analytic in {|𝑧−𝑧0|<𝑅}, where 𝑅 is the radius of convergence of the poser series, Now: ```TheoremLet 𝑓:𝑈→ℂ be analytic and let {|𝑧−𝑧0|<𝑟}⊂𝑈. Then in this disk, 𝑓 has a power series representation 𝑓(𝑧)=∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘, |𝑧−𝑧0|<𝑟, where 𝑎𝑘=𝑓(𝑘)(𝑧0)𝑘!, 𝑘≥0. The radius of convergence of this power series is 𝑅≥𝑟```

### Examples

`𝑓(𝑧)=∞∑𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘, |𝑧−𝑧0|<𝑟, where 𝑎𝑘=𝑓(𝑘)(𝑧0)𝑘!, 𝑘≥0.`
• 𝑓(𝑧)=ℯ𝑧, then 𝑓(𝑘)(𝑧)=ℯ𝑧. Letting 𝑧0=0, we have 𝑓(𝑘)(𝑧0)=ℯ0=1 for all 𝑘. Thus 𝑎𝑘=1𝑘! for all 𝑘, and so `ℯ𝑧=∞∑𝑘=0𝑧(𝑘)𝑘! for all 𝑧∈ℂ`
• 𝑓(𝑧)=ℯ𝑧 as above, but now let 𝑧0=1. Then 𝑓(𝑘)(𝑧0)=ℯ1=ℯ for all 𝑘. Thus 𝑎𝑘=𝑘! for all 𝑘, and so `ℯ𝑧=∞∑𝑘=0ℯ𝑘!(𝑧−1)𝑘 for all 𝑧∈ℂ`
• 𝑓(𝑧)=sin 𝑧 is analytic in ℂ. Let 𝑧0=0. Then ```𝑓(𝑧)=sin 𝑧, 𝑓(0)=0 𝑓′ (𝑧)=cos 𝑧, 𝑓′(0)=1 𝑓″(𝑧)=−sin 𝑧, 𝑓″(0)=0 𝑓(3)(𝑧)=−cos 𝑧, 𝑓(3)(0)=−1 𝑓(4)(𝑧)=sin 𝑧, 𝑓(4)(0)=0 ⋯ ``` Thus ```sin 𝑧=0+11!𝑧+02!𝑧2+−13!𝑧3+04!𝑧4+15!𝑧5+⋯  =𝑧−𝑧33!+𝑧55!−𝑧77!+𝑧99!−+⋯  =∞∑𝑘=0(−1)𝑘(2𝑘+1)!𝑧2𝑘+1```
• 𝑓(𝑧)=cos 𝑧 is analytic in ℂ. Let 𝑧0=0. Then ```cos 𝑧=𝑑𝑑𝑧sin 𝑧=𝑑𝑑𝑧∞∑𝑘=0(−1)𝑘(2𝑘+1)!𝑧2𝑘+1  =∞∑𝑘=0(−1)𝑘(2𝑘+1)!(2𝑘+1)𝑧2𝑘  =∞∑𝑘=0(−1)𝑘(2𝑘)!𝑧2𝑘  =1−𝑧22!+𝑧44!−𝑧66!+𝑧88!−+⋯ ```

## A Corollary

Note: The theorem implies that an analytic function is entirely determined in a disk by all of its derivatives 𝑓(𝑘)(𝑧0) at the center 𝑧0 of the disk.

`CorollaryIf 𝑓 and 𝑔 are analytic in {|𝑧−𝑧0|<𝑟} and if 𝑓(𝑘)(𝑧0)=𝑔(𝑘)(𝑧0) for all 𝑘, then 𝑓(𝑧)=𝑔(𝑧) for all 𝑧 in {|𝑧−𝑧0|<𝑟}.`

ID: 190500003 Last Updated: 3/5/2019 Revision: 0

Home 5

Management

HBR 3

Information

Recreation

Culture

Chinese 1097

English 337

Computer

Hardware 149

Software

Application 198

Manim 113

Numeric 19

Programming

Web 283

Unicode 494

HTML 65

CSS 58

ASP.NET 145

OS 389

Python 19

Knowledge

Mathematics

Algebra 25

Geometry 21

Calculus 67

Complex Analysis 21

Engineering

Mechanical

Rigid Bodies

Statics 92

Dynamics 37

Control

Physics

Electric 27