Homework Help: Limit of a Complex Function

1. Apr 9, 2010

conana

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$\forall\epsilon>0 \exists\delta>0 \forall z (|z-z_0|<\delta\Rightarrow|f(z)-\omega_0|<\epsilon)$$

3. The attempt at a solution

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

Let $$\epsilon>0$$. Then

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

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 $$|x-x_0|<1$$ 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. Apr 9, 2010

rasmhop

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 $\delta < 1$ so if you want say $\delta<\epsilon/2$ (not the right value in this problem), then you just require $\delta < \min(\epsilon/2,1)$ such that $\delta<\epsilon/2$ and $\delta<1$. Of course if you want an explicit value you could just let $\delta=\min(\epsilon/2,1)/2$.

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:
\begin{align*} |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 \end{align*}
If you choose your delta small enough this should work fine.