# 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.