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Sequences and Limits

Sequences

Consider the following sequences of complex numbers.

1, 1/2, 1/3, 1/4, 1/5, 1/6,…1/n→s
𝑖, 𝑖/2, 𝑖/3, 𝑖/4, 𝑖/5, 𝑖/6,…𝑖/n→s
𝑖, -𝑖/2, 𝑖/3, -𝑖/4, 𝑖/5, -𝑖/6,…𝑖ⁿ/n→s

Unlike sequences of real number, a complex number sequence {sₙ} converges to a limit s if the sequence eventually lies in any (every so small) disk centered at s.

By definition. A sequence {sₙ} of complex numbers converges to s∊ℂ if for every ε>0 there exists an index N≥1 such that |s-s|<ε for all n>N. That is

lim
n→∞sₙ=s

For example,

lim
n→∞1
n=0

lim
n→∞1
np=0 for any 0<p<∞

lim
n→∞c
np=0 for any c∊ℂ, 0<p<∞

lim
n→∞qn=0 for 0<q<1

lim
n→∞zn=0 for |z|<1

lim
n→∞ⁿ√10=1

lim
n→∞ⁿ√n=1

Rules for Limits

• Convergent sequences are bounded
• If {sₙ} converges to s and {tₙ} converges to t, then

  sₙ+tₙ→s+t
  sₙtₙ→st (in particular: asₙ→as for any a∊ℂ)
  sₙ/tₙ→s/t, provided t≠0

For examples

n
n+1=  
1
1+1
n→1 as n→∞

3n²+5
𝑖n²+2𝑖n-1=3+5
n²
𝑖+2𝑖
n-1
n²→3
𝑖=-3𝑖 as n→∞

n²
n+1=  
n
1+1
n→n as n→∞, not bounded

3n+5
𝑖n²+2𝑖n-1=3
n+5
n²
𝑖+2𝑖
n-1
n²→0
𝑖=0 as n→∞

Convergence of Complex Number Sequences

A sequence of complex numbers, {sₙ}, converges to 0 if and only if the sequence {|sₙ|} of absolute values converges to 0.  And a sequence of complex numbers, {sₙ}, with sₙ=xₙ+𝑖yₙ, converges to s=x+𝑖y if and only if xₙ→x and yₙ→y as n→∞.

For example

{𝑖ⁿ
n}=𝑖,-1
2,-𝑖
3,1
4,𝑖
5,-1
6,…→0 as n→∞

Facts about Sequence of Real Numbers

By Squeeze Theorem, suppose that {rₙ}, {sₙ}, and {tₙ} are sequences of real numbers such that rₙ≤sₙ≤tₙ for all n. If both sequences {rₙ} and {tₙ} converge to the same limit, L, then the sequence {sₙ} 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, {𝑖ⁿ
n}

|𝑖ⁿ
n|=|𝑖|ⁿ
n=1
n→0 as n→∞. Thus lim
n→∞𝑖ⁿ
n=0

Let 𝑖ⁿ
n=xₙ+𝑖yₙ,
⇒xₙ={0, n=odd
1/n, n=4k,  
-1/n, n=4k+2, yₙ={0, n=even
1/n, n=4k+1,  
-1/n, n=4k+3
Since -1/n≤xₙ≤1/n, and -1/n≤yₙ≤1/n for all n, the Squeeze theorem implies that
lim
n→∞xₙ=0 and lim
n→∞yₙ=0, hence lim
n→∞𝑖ⁿ
n=0

Limits of Complex Functions

By definition. The complex-valued function f(z) has limit L as z→z₀ if the values of f(z) are near L as z→z. That is

lim
z→z₀f(z)=L if for all ε>0 there exists δ>0 such that |f(z)-L|<ε whenever 0<|z-z₀|<δ.
Where f(z) needs to be defined near z₀ for this definition to make sense, but is not necessary at z₀.

For example,

f(z)=z²-1
z-1,z≠1. Then
lim
z→1f(z)=lim
z→1z²-1
z-1=lim
z→1(z-1)(z+1)
z-1=lim
z→1z+1=2

Let f(z)=Arg z. Then:

lim
z→𝑖Arg z=π
2

lim
z→1Arg z=0

lim
z→-1Arg z=does not exist. since -π<Arg z≤π

Facts about Limits of Complex Functions

• If f has a limit at z₀ then f is bounded near z₀.
• If f(z)→L and g(z)→M as z→z₀ then

  f(z)+g(z)→L+M as z→z₀

  f(z)g(z)→LM as z→z₀

  f(z)/g(z)→L/M as z→z₀ provided that M≠0.

Continuity

By definition. The function f is continuous at z₀, if f(z)→f(z₀) as z→z₀.

f is defined at z₀.

f has a limit as  z→z₀.

The limit equals f(z₀).

Examples:

constant functions

f(z)=z

polynomials

f(z)=|z|

f(z)=P(z)/q(z) wherever q(z)≠0 (p and q are polynomials).