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Prove limit of complex function 2

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove using limit definition [tex]$\lim_{z \to z_0} (z^2 + c) = z_0^2 +
    c$[/tex].

    2. Relevant equations



    3. The attempt at a solution

    For every [tex]$\varepsilon$[/tex] there should be a [tex]$\delta$[/tex] such that
    [tex]
    \begin{align*}
    \text{if and only if } 0 < |z - z_0| < \delta \text{ then } |(z^2 + c) -
    (z_0^2 + c)| < \varepsilon
    \end{align*}
    [/tex]

    Starting from [tex]$ |(z^2 + c) - (z_0^2 + c)| < \varepsilon$[/tex]
    [tex]
    \begin{align*}
    |(z^2 + c) - (z_0^2 + c)| = |z^2 - z_0^2| = |(z+z_0)(z-z_0)| <
    \varepsilon
    \end{align*}
    [/tex]

    How can I continue from here?
     
  2. jcsd
  3. Jul 15, 2009 #2
    This is the exact same proof as for the real case, I think. The more flexible approach is to first let [tex]\left|z-z_0\right| < 1[/tex] and then apply the triangle inequality (you may need to use one of the variants of the inequality) to get a bound for [tex]\left|z+z_0\right|[/tex] and then choose delta accordingly. If you want to satisfy two inequalities at the same time, delta will be written as the min of two numbers.

    The other way is to apply the triangle inequality directly. Clearly, the [tex]\left|z+z_0\right|[/tex] is the only term that gives us any trouble. Can you rewrite it so that we can use the fact that [tex]\left|z-z_0\right| < \delta[/tex] to our advantage? Hint: you need to add and subtract a term.
     
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