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Homework Help: Limit of a Complex Function

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Use epsilon-delta proof to show that [tex]\lim_{z\to z_0}(z^2+c)=z_0^2+c[/tex].

    2. Relevant equations

    [tex]\forall\epsilon>0 \exists\delta>0 \forall z (|z-z_0|<\delta\Rightarrow|f(z)-\omega_0|<\epsilon)[/tex]

    3. The attempt at a solution

    So [tex]f(z)=z^2+c[/tex] and [tex]\omega_0=z_0^2+c[/tex]. In order to write my proof I first need to find delta in terms of epsilon.

    Let [tex]\epsilon>0[/tex]. Then

    [tex]\begin{align*}|f(z)-\omega_0|<\epsilon &\Rightarrow|(z^2+c)-(z_0^2+c)|<\epsilon \\





    And from here I'm kind of stumped. I need to get that z out of the right hand side so that i can choose delta in terms of epsilon only. I know that with functions of a real variable you can restrict the value of [tex]|x-x_0|<1[/tex] or something like that so you can put an upper and lower bound on delta, but I am not sure exactly how to employ this technique since pulling z's out of absolute value sign is a little different that pulling out x's.

    Any help would be greatly appreciated. Thanks in advance.
  2. jcsd
  3. Apr 9, 2010 #2
    This technique still works perfectly fine with complex functions and doesn't depend on pulling out variables of the absolute value sign. You just make sure to choose [itex]\delta < 1[/itex] so if you want say [itex]\delta<\epsilon/2[/itex] (not the right value in this problem), then you just require [itex]\delta < \min(\epsilon/2,1)[/itex] such that [itex]\delta<\epsilon/2[/itex] and [itex]\delta<1[/itex]. Of course if you want an explicit value you could just let [itex]\delta=\min(\epsilon/2,1)/2[/itex].

    Anyway the thing to note is that we would really like an expression in terms of |z-z_0| so we use the triangle inequality:
    |z+z_0||z-z_0| &= |z-z_0 + 2z_0||z-z_0| \\
    &\leq (|z-z_0| + 2|z_0|)|z-z_0| \\
    &< \delta^2 + 2|z_0|\delta
    If you choose your delta small enough this should work fine.
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