Link:http://output.to/sideway/default.asp?qno=190300018 Complex Analysis Complex Number Sequences and Limits source/reference: Sequences and LimitsSequencesConsider the following sequences of complex numbers. 1, 1/2, 1/3, 1/4, 1/5, 1/6,β¦1/nβs π, π/2, π/3, π/4, π/5, π/6,β¦π/nβs π, -π/2, π/3, -π/4, π/5, -π/6,β¦πβΏ/nβs Unlike sequences of real number, a complex number sequence {sβ} converges to a limit s if the sequence eventually lies in any (every so small) disk centered at s. By definition. A sequence {sβ} of complex numbers converges to sββ if for every Ξ΅>0 there exists an index Nβ₯1 such that |s-s|<Ξ΅ for all n>N. That is lim
nββsβ=s
For example, lim nββ1 n=0 lim nββ1 np=0 for any 0<p<β lim nββc np=0 for any cββ, 0<p<β lim nββqn=0 for 0<q<1 lim nββzn=0 for |z|<1 lim
nβββΏβ10=1
lim
nβββΏβn=1
Rules for Limits
For examples n n+1= 1 1+1 nβ1 as nββ 3nΒ²+5 πnΒ²+2πn-1=3+5 nΒ² π+2π n-1 nΒ²β3 π=-3π as nββ nΒ² n+1= n 1+1 nβn as nββ, not bounded 3n+5 πnΒ²+2πn-1=3 n+5 nΒ² π+2π n-1 nΒ²β0 π=0 as nββ Convergence of Complex Number SequencesA sequence of complex numbers, {sβ}, converges to 0 if and only if the sequence {|sβ|} of absolute values converges to 0. And a sequence of complex numbers, {sβ}, with sβ=xβ+πyβ, converges to s=x+πy if and only if xββx and yββy as nββ. For example {πβΏ n}=π,-1 2,-π 3,1 4,π 5,-1 6,β¦β0 as nββ Facts about Sequence of Real NumbersBy Squeeze Theorem, suppose that {rβ}, {sβ}, and {tβ} are sequences of real numbers such that rββ€sββ€tβ for all n. If both sequences {rβ} and {tβ} converge to the same limit, L, then the sequence {sβ} has no choice but to converge to the limit L as well. By theorem. A bounded, monotone sequence of real numbers converges. For example, Complex Number Sequences, {πβΏ |πβΏ n|=|π|βΏ n=1 nβ0 as nββ. Thus lim nββπβΏ n=0 Let πβΏ n=xβ+πyβ, βxβ={0, n=odd 1/n, n=4k, -1/n, n=4k+2, yβ={0, n=even 1/n, n=4k+1, -1/n, n=4k+3 Since -1/nβ€xββ€1/n, and -1/nβ€yββ€1/n for all n, the Squeeze theorem implies that lim nββxβ=0 and lim nββyβ=0, hence lim nββπβΏ n=0 Limits of Complex FunctionsBy definition. The complex-valued function f(z) has limit L as zβzβ if the values of f(z) are near L as zβz. That is lim
zβzβf(z)=L if for all Ξ΅>0 there exists Ξ΄>0 such that |f(z)-L|<Ξ΅ whenever 0<|z-zβ|<Ξ΄.
Where f(z) needs to be defined near zβ for this definition to make sense, but is not necessary at zβ.
For example, f(z)=zΒ²-1 z-1,zβ 1. Then lim zβ1f(z)=lim zβ1zΒ²-1 z-1=lim zβ1(z-1)(z+1) z-1=lim zβ1z+1=2 Let f(z)=Arg z. Then: lim zβπArg z=Ο 2 lim
zβ1Arg z=0
lim
zβ-1Arg z=does not exist. since -Ο<Arg zβ€Ο
Facts about Limits of Complex Functions
ContinuityBy definition. The function f is continuous at zβ, if f(z)βf(zβ) as zβzβ. f is defined at zβ. f has a limit as zβzβ. The limit equals f(zβ). Examples: constant functions f(z)=z polynomials f(z)=|z| f(z)=P(z)/q(z) wherever q(z)β 0 (p and q are polynomials). |
Sideway BICK Blog 18/03 |