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Theorem: convergence of complex sequence

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Oct 11, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    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:
     
  4. Oct 11, 2009 #3
    hi tinytim,
    thanks :smile:,
    ok so here it goes:
    we have 'if and only if' so first im 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 dont know what to do.
     
  5. Oct 11, 2009 #4

    statdad

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    Homework Helper

    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: Oct 11, 2009
  6. Oct 11, 2009 #5
    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]
     
  7. Oct 11, 2009 #6

    statdad

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    Homework Helper

    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.
     
  8. Oct 11, 2009 #7
    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 dont fully understand why these ineqaulities are true?
     
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