source/reference:
https://www.youtube.com/channel/UCaTLkDn9_1Wy5TRYfVULYUw/playlists

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=0𝑧^𝑘, for some 𝑧∈ℂ. having that 𝑆𝑛=1+𝑧+𝑧2+⋯+𝑧𝑛 Finding a closed formula for 𝑆_𝑛 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 𝑆_𝑛=1−𝑧^𝑛+11−𝑧 for 𝑧≠1, and since 𝑧^𝑛+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|^𝑘→0 as 𝑘→∞, thus ∞∑𝑘=0𝑧^𝑘 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=0𝑧^𝑘=11−𝑧 for |𝑧|<1;

For |𝑧|<1 Writing 𝑧=𝑟ℯ^𝑖𝜃, then 𝑧^𝑘=𝑟^𝑘ℯ^𝑖𝑘𝜃=𝑟^𝑘cos(𝑘𝜃)+𝑖𝑟^𝑘sin(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=1𝑖^𝑘𝑘. Does this series converge?

• Note that ∞∑𝑘=1𝑖^𝑘𝑘=∞∑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 𝑖^𝑘=𝑖^2𝑛=(−1)^𝑛 is real. When 𝑘 is odd (i.e. 𝑘 is of the form 𝑘=2𝑛+1), then 𝑖^𝑘=𝑖^2𝑛+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=0𝑧^𝑘 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=0𝑧_𝑘 is a power series with 𝑎_𝑘=1, 𝑧₀=0. It converges for |𝑧|<1.
• ∞∑𝑘=0(−1)^𝑘2^𝑘𝑧^2𝑘=1−𝑧²2+𝑧⁴4−𝑧⁶8+−⋯=∞∑𝑘=0−𝑧²2^𝑘=∞∑𝑘=0𝑤^𝑘, 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

The Radius of Convergence

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=0𝑧^𝑘 has the radius of convergence 1.
• ∞∑𝑘=0(−1)^𝑘2^𝑘𝑧^2𝑘 has radius of convergence 2
• ∞∑𝑘=0𝑘^𝑘𝑧^𝑘 Pick an arbitrary 𝑧∈ℂ\{0}. Observe that |𝑘^𝑘𝑧^𝑘|=(𝑘|𝑧|)^𝑘≥2^𝑘 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=0𝑧^𝑘𝑘^𝑘 Pick an arbitrary 𝑧∈ℂ. Observe that 𝑧^𝑘𝑘^𝑘=|𝑧|𝑘^𝑘≤12^𝑘 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=0𝑧^𝑘 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=0𝑧^𝑘, 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=0𝑧^𝑘=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=0𝑎_𝑘(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=0𝑧^𝑘: Here, 𝑎_𝑘=1, so 𝑎_𝑘𝑎_𝑘+1=1→1 as 𝑘→∞. Thus 𝑅=1.
• ∞∑𝑘=0𝑘𝑧^𝑘: Here, 𝑎_𝑘=𝑘, so 𝑎_𝑘𝑎_𝑘+1→1 as 𝑘→∞. Thus 𝑅=1.
• ∞∑𝑘=0𝑧^𝑘𝑘!: Here, 𝑎_𝑘=1𝑘!, so 𝑎_𝑘𝑎_𝑘+1=(𝑘+1)!𝑘!=𝑘+1→∞ as 𝑘→∞. Thus 𝑅=∞.
• ∞∑𝑘=0𝑧^𝑘𝑘^𝑘: Here, 𝑎_𝑘=1𝑘^𝑘, so 𝑎_𝑘𝑎_𝑘+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|𝑎_𝑘|}=0 then 𝑅=∞.
• If Lim𝑘→∞{^𝑘|𝑎_𝑘|}=∞ then 𝑅=0.

Examples

• ∞∑𝑘=0𝑧^𝑘𝑘^𝑘: Here, 𝑎_𝑘=1𝑘^𝑘, so ^𝑘|𝑎_𝑘|=1𝑘→0 as 𝑘→∞. Thus 𝑅=∞.
• ∞∑𝑘=0𝑘𝑧^𝑘: Here, 𝑎_𝑘=𝑘, so ^𝑘|𝑎_𝑘|=^𝑘𝑘→1 as 𝑘→∞. Thus 𝑅=1.
• ∞∑𝑘=02^𝑘𝑧^𝑘: Here, 𝑎_𝑘=2^𝑘, so ^𝑘|𝑎_𝑘|=2→2 as 𝑘→∞. Thus 𝑅=12.
• ∞∑𝑘=0(−1)^𝑘2^𝑘𝑧^2𝑘: Here, 𝑎_2𝑘=(−1)^𝑘2^𝑘, and 𝑎_2𝑘+1=0, so ^2𝑘|𝑎_2𝑘|=12^1/₂ for 𝑘≥1 and ^2𝑘+1|𝑎_2𝑘+1|=0, and so the sequence ^𝑘|𝑎_𝑘| (𝑘≥1) is 0, 12, 0, 12, 0, 12, 0,⋯ and this sequence does not have a limit. Note: 𝑎_𝑘𝑎_𝑘+1 has no limit either.

The Cauchy Hadamard Criterion

For the series ∞∑𝑘=0(−1)^𝑘2^𝑘𝑧^2𝑘, neither the root test nor the ratio test "work".

Yet, letting 𝑤=𝑧², the series becomes ∞∑𝑘=0(−1)^𝑘2^𝑘𝑤^𝑘, and ^𝑘(−1)^𝑘2^𝑘=12→12 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=0𝑎_𝑘(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)=ℯ^𝑧, then 𝑓^(𝑘)(𝑧)=ℯ^𝑧. Letting 𝑧₀=0, we have 𝑓^(𝑘)(𝑧₀)=ℯ⁰=1 for all 𝑘. Thus 𝑎_𝑘=1𝑘! for all 𝑘, and so ℯ𝑧=∞∑𝑘=0𝑧(𝑘)𝑘! for all 𝑧∈ℂ
• 𝑓(𝑧)=ℯ^𝑧 as above, but now let 𝑧₀=1. Then 𝑓^(𝑘)(𝑧₀)=ℯ¹=ℯ for all 𝑘. Thus 𝑎_𝑘=ℯ𝑘! 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}.