Doc- Review: Sequence & Series of complex numbers

Review
Sequence & Series of complex numbers
First consider the useful facts

The △ in-equality : If z₁ & z₂ are arbitrary complex no., then 

Proof: 

Taking positive sq. root yields the desired inequality.

Note:

Note equality occur in (i)

Note

Defn A seq. {zn} of complex no. is said to converge to a complex no. z if the seq. {|z_n- z|} of real no. converge to 0

Proof forward Part :
Assume that z_n → z then |zn- z|→ 0  

Converse Part : Assume that Re(z_n) → Re(z)  &  Im(z_n) → Im(z)

consider,

|z_n -  z|= |(x_{n +} iy_n )- (x+ iy)| where, z_n = x_n + iy_n  & z = x + iy

Note: {z_n} can't converge to more than one limit.
If exists is unique.

→ 0 as n→∞ (∵0 ≤|z|≤ 1)

Defⁿ A seq. {zn} of complex no. is called a Cauchy seq. if for each 𝜖 > 0, there exists (N=N(𝜖)) an integer N s.t.

Cauchy Criteria for convergence in complex plane {z_n} converges iff {z_n} is a Cauchy seq.

Forward part: Assume that z_n → z then 

Re(zn) → Re(z)   & Im(z_n) → Im(z)

Proved before
Here {Re(z_n)} & {Im(z_n)} both are Cauchy seq. being convergent seq. of real no.

Converse part: suppose {z_m} is a Cauchy seq. , then using

implies both {Re(z_n) & Im(z_n)} are cauchy seq. of real no. hence both converges,

implies {z_n} converges
Defⁿ The infinite series  of complex nos. is said to be converge if the seq. {S_n},

of paritial sum S is convergent.

from the Cauchy criterion, i.e. "a seq. is conv. iff it is a Cauchy seq", we see that  converges iff {S_n}is a cauchy seq.

For each

from this  it follows that convergence of  (i. e. a neccessary cond for the series  to converge is that z_n →0  as n→ ∞

Remark's in the case of sequence we have, converges to z iff 

converges to Re(z)  converge to Im(z).
A sufficient condⁿ for cgces of

i.e. absolutely convergent series is cgt.

also, as with real series, we say a series  is said to be absolutely convergent if the series  +ve real no. is convergent. Futher using the fact that

Implies that "every absolutely convergent series is convergent." Converse is not true e.g.

 absolutely and hence cgt.

Generalized Cauchy's n^th root test: let   be a series of complex term such that 

(a) if ℓ < 1 then series  cgs abs.
(b) if ℓ >1 then series div.

(c) if  ℓ = 1 the series may or may not conv.

Generalised D' lembert Ratio test:
 then
(a) if L < 1, then the series cgs absolutely
(b) if l > 1, the series div.
(c)if l≤1≤ L, no conclusion.

complex analysis ( Bak &  Newmann)

Topology of the complex plane

1.5 Definition
(i) is called an open disc of radius r centered at z₀ , also called nbhd of z₀.

(ii) circle

= circle with center z₀ , radius r.

(iii) subset is called  If for every , there exists r > 0 . It means that some disc arround z lies entirely in S. for instance, the interior of a circle  the entire complex plane  half plane (Re(z) , Im(z) <0 , Re(z) ) ect. are open sets. An open disc is an open set.
(iv) A set  is s. t. b open iff for each

Note:   is not open

Note:  is open in  but not in  
(v) set S is called closed if its complement


= Coll of points whose neighbourhood have a non empty intersection with both S and  
(ix)  = closure of S

(x) S is bdry iff  for some r > 0.
(xi) S is compact iff S is closed bdd

(xii) set S is said to be disconneced if there exists two disjoint open set s. t.

(xiii) S is s. t. b. disconnected iff S is a union of two non empty disjoit open subsets.
(xiv) S is called connected if it is not disconnected. in other words, S is connected if and only if, each pair of points z₁, z₁ of S can be connected by an arc lying in S

 is calle the line segment with and points z₁ & z₂ and denoted by [z₁,z₂]

∴ if  for each  then the line segment [z₁, z₂] , (where z₁ , z₂ ∈S ) is said to be containe in S.
→ by a polygonal line from z₁ to z_n ... . a finite union of line segments of the form

(z₁ and z_n are said to be polygonally connected)
→ A set S is said to be polygonally connected if any two points of S can be connected by a polygonal line (basically horizonal or verticaly) contained in S.