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


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=𝑛𝑘=0=𝑎𝑘=𝑎0+𝑎1+𝑎2+⋯+𝑎𝑛 converges to 𝑆.


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)+𝑖𝑘=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  =12𝑛=1(−1)𝑛𝑛+𝑖𝑛=0(−1)𝑛2𝑛+1 But 𝑛=1(−1)𝑛𝑛=−1+1213+14−+⋯ is the alternating harmonic series, which converges.

Absolute Convergence

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


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=0𝑎𝑘𝑘=0|𝑎𝑘|.


If |𝑧|<1, then the series 𝑘=0𝑧𝑘 converge absolutely, so 𝑘=0𝑧𝑘𝑘=0|𝑧|𝑘 But the left-hand side equals 11−𝑧, and the right-hand side equals 11−|𝑧|, so that 11−𝑧11−|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


  • 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

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.


  • 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.


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 𝑤𝑧0𝑘=0𝑎𝑘(𝑧−𝑧0)𝑘𝑑𝑧=𝑘=0𝑎𝑘𝑤𝑧0(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.


Let's again look at the power series 𝑘=0𝑧𝑘, which has 𝑅=1. Then for any 𝑤 with |𝑤|<1, thus have 𝑤𝑧0𝑘=0𝑧𝑘𝑑𝑧=𝑘=0𝑤𝑧0𝑧𝑘𝑑𝑧=𝑘=01𝑘+1𝑤𝑘+1=𝑘=1𝑤𝑘𝑘 Also know that 𝑘=0𝑧𝑘=11−𝑧, hence 𝑤𝑧0𝑘=0𝑧𝑘𝑑𝑧=𝑤𝑧011−𝑧𝑑𝑧=−Log(1−𝑧)𝑤0=−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: 𝑤𝑧0𝑘=0𝑧𝑘𝑑𝑧=𝑘=1𝑤𝑘𝑘 and 𝑤𝑧0𝑘=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.


  • 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|}.


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


  • 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.

The Cauchy Hadamard Criterion

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


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 𝑧=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 𝑧=𝑘=0𝑘!(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=𝑑𝑑𝑧𝑘=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


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