Complex Analysis ( Limits at a point ). We need to prove that $ \lim_{z \to z_{0}}(z^{2}+c)$ = $z_{0}^{2}+c$ , where c is a complex constant , using $\epsilon - \delta$ definition , where $z , z_{0}$ are complex variables.
What I tried : First we say $|(z^{2}+c)-(z_{0}^{2}+c)| < \epsilon $ , and try to find a suitable $\delta$ > 0 for this statement to be true.
So , $|(z^{2}+c)-(z_{0}^{2}+c)| < \epsilon $ , implies ,  $|z^{2}-z_{0}^{2}| < \epsilon $ ,
$=>\ |z+z_{0}||z-z_{0}|< \epsilon $ , 
Now , since $|z+z_{0}| > 0$ ,
$|z-z_{0}|< \frac{\epsilon}{|z+z_{0}|} $ ,
Now we take $\epsilon = \delta*|z+z_{0}|$ ,
Thus , $|(z^{2}+c)-(z_{0}^{2}+c)| < \epsilon $ is true whenever , $0<|z-z_{0}|< \delta $ ,
Is the above proof correct ?
 A: 
So, $|(z^2+c)-(z_0^2+c)| < \epsilon$ implies $|z^2-z_0^2| < \epsilon$.

The statement is correct, but unfortunately it is not helpfut at all. Basically, it says: If the claim is true, then some other statement is true. So what? As long as you haven't shown that the claim is correct, this doesn't help you at all.
If someone asks you to prove that $-1$ equals $1$, you can say

$-1 = 1$ implies $1 = 1$

but this is to no use: We know that $1=1$ is correct, but this does (obviously) not imply that $-1$ equals $1$.
The important thing is that $|(z^2+c)-(z_0^2+c)| < \epsilon$ if and only if $|z^2-z_0^2| < \epsilon$ which is in turn equivalent to
$$|z+z_0| \cdot |z-z_0| < \epsilon. \tag{1}$$

Now we take $\epsilon = \delta \cdot |z+z_0|$.

No, that's not correct. We are given $\epsilon>0$ and we have to find $\delta>0$ such that the inequality $(1)$ is satisfied for all $|z-z_0|<\delta$. If $|z-z_0|<\delta$, then, by the triangle inequality
$$|z-z_0| \cdot |z+z_0| < \delta |z+z_0| \leq \delta ( 2|z_0|+\delta) \leq \delta (2|z_0|+1)$$
for any $\delta \leq 1$. Therefore, $(1)$ is satisfied if
$$\delta (2|z_0|+1) \leq \epsilon.$$
This is equivalent to
$$\delta \leq \frac{\epsilon}{2 |z_0|+1}.$$
Hence, if we choose
$$\delta := \min \left\{1, \frac{\epsilon}{2|z_0|+1} \right\},$$
then $(1)$ is satisfied.
A: The $\delta$ you need to find must be independent of $z$. Note that $$|z+z_0|<|z_0|+|z|<|z_0|+|z_0|+|z-z_0|<2|z_0|+\delta$$ Assume we take $\delta < 1$. Thus we get $$|z+z_0|<2|z_0|+1$$ So $\delta < \frac{\epsilon}{2|z_0|+1}$ is sufficient to prove $|z+z_0||z-z_0|<\epsilon$ In total you can take $\delta = \min\left\{1,\frac{\epsilon}{2|z_0|+1}\right\}$.
