Theorem: convergence of complex sequence

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Homework Help Overview

The discussion revolves around the convergence of a sequence of complex numbers, specifically examining the conditions under which a sequence converges based on the convergence of its real and imaginary parts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the proof of convergence using the properties of real and imaginary parts, with some attempting to apply the triangle inequality and δ-ε definitions. Questions arise regarding the implications of inequalities and the structure of the proof.

Discussion Status

There is an active exploration of the proof structure, with participants offering insights into the relationships between the convergence of complex sequences and their real components. Some guidance has been provided regarding the use of inequalities, but there remains uncertainty about the application of these concepts.

Contextual Notes

Participants are working under the assumption that they need to demonstrate the equivalence of convergence for complex sequences and their respective real sequences, which involves specific definitions and properties of limits.

sara_87
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Homework Statement



show that a sequence zn (of complex numbers) converges if and only if it's real and imaginary parts converge.

Homework Equations





The Attempt at a Solution



if the real parts converge and the imaginary parts converge then using linearity, we have:
lim(real + imaginary)= lim(real) + lim(imaginary)

I think this is a wrong proof (and surely incomplete). does anyone have any ideas?
 
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Hi sara_87! :smile:

Just use an ordinary δ,ε proof, with δ = δ1 + iδ2 and ε = ε1 + iε2, and remember |a + ib| = √(a2 + b2) :wink:
 
hi tinytim,
thanks :smile:,
ok so here it goes:
we have 'if and only if' so first I am going to 'try' and prove it in the first direction:
let zn be a convergent sequence (and converges to zc), then:
for all [tex]\epsilon>0[/tex] there exists [tex]\delta[/tex] such that [tex]\left|z_n-z_c\right|<\epsilon[/tex] for all n>[tex]\delta[/tex].
let zc=a+ib:
[tex]\left|z_n-(a+ib)\right|=\left|z_n+(-a-ib)\right|\leq\left|z_n\right|+\left|(-a-ib)\right|=sqrt(a^2+b^2)[/tex]
(i used the triangle inequality)
but now i don't know what to do.
 
Think about this:

If [tex]z_1 =e + f i[/tex] and [tex]z_2 = g + h i[/tex] are complex numbers, then

[tex] |e - g | \le |z_1 - z_2|[/tex]

and

[tex] |f - h| \le |z_1 - z_2|[/tex]
 
Last edited:
i agree, but why would this help here, do i need to use this to put
[tex]\left|z_n\right|+\sqrt{a^2+b^2}[/tex] in terms of something that could be my [tex]\epsilon[/tex]
 
Suppose you want to show that

[tex] z_n = x_n + y_ni \to z = x+yi[/tex]
if, and only if

[tex] x_n \to x \text{ and } y_n \to y[/tex]

If you know the complex sequence converges then the inequalities will show the two real sequences converge. If the real sequences converge, you can show the complex one does as well.
 
oh right i see,
so first, to prove in the first direction:
let zn be a convergent sequence, then
lim(zn)=lim(xn)+i*lim(yn)

zn-zc=(xn-x)+ i (yn-y)
where zc=(x,y) is the limit of the sequence zn.
so:
[tex]\left|x_n-x\right|\leq\left|z_n-z_c\right|[/tex] and [tex]\left|y_n-y\right|\leq\left|z_n-z_c\right|[/tex]
is this right? buti don't fully understand why these ineqaulities are true?
 

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