Use $\varepsilon$-$\delta$ Definition of Limits to Show a Statement is True I am trying to show that the following statement is true:
$$\lim_{z\to z_0}\left(z^2 + c\right) = z_0^2+c$$
where $z$, $z_0$, and $c$ are complex. To show this, I am supposed to rely on the following definition of a limit:
$$\lim_{z\to z_0} f(z) = w_0$$
where $w_0$ is also complex, and $f(z)$ is defined in some deleted neighborhood of $z_0$, means that for each positive number, $\varepsilon$, there is a positive number $\delta$, such that
$$\lvert f(z) - w_0 \rvert < \varepsilon \qquad\mathrm{whenever}\qquad 0<\lvert z-z_0 \rvert < \delta.$$
My approach is as follows. Assume $\lvert z-z_0 \rvert < \delta$, and $\varepsilon > 0$.
$$\begin{align} \lvert (z^2+c) - (z_0^2 + c) \rvert &= \lvert z^2-z_0^2\rvert\\&=\lvert (z-z_0)(z+z_0) \rvert\\
&=\lvert z-z_0\rvert \lvert z+z_0 \rvert \\
&= \lvert z-z_0 \rvert \lvert z-z_0 + 2z_0 \rvert \\
& \leq\lvert z-z_0\rvert (\lvert z-z_0 \rvert +\lvert 2z_0 \rvert) \qquad \mathrm{by\,triangle\,inequality}\\
& < \delta^2+2\delta\lvert z_0\rvert
\end{align}$$
Now, if I define $\varepsilon\equiv \delta^2 + 2\delta \lvert z_0 \rvert$, then I can say that 
$$\lvert (z^2+c) - (z_0^2+c)\rvert<\varepsilon \qquad \mathrm{whenever}\qquad \lvert z-z_0\rvert < \delta.
$$
My question is: Did I do anything wrong? This seems much more involved than the previous exercises I've worked through, and so I'm also wondering if there's a better/easier way to solve this problem.
 A: This looks great.  There won't be any significantly better/easier ways to do this problem if you can only work with the basic definition of convergence.  You have a little freedom in deciding which estimates to use in your long string of inequalities, but you'll ultimately have to do something similar in any version of the argument.
If you want me to be nitpicky, I'd say that you should clarify your line "I define $\varepsilon \equiv \delta^2+2\delta|z_0|$," since you're really using this for a given, fixed $\varepsilon$ to define $\delta$.  If you're just starting with proofs, it might also be worth noting that the $\delta$ so defined is a new entity, distinct from the $\delta$ that appears above.  So it might be worth giving this one a different letter unless you're sure that your reader won't mind.  Possible choices of rephrase are: "Choose a $\delta_0>0$ such that $\delta_0^2 + 2\delta_0|z_0| < \varepsilon$" (or equal to $\varepsilon$) or "Making $\delta$ smaller, if necessary, we may suppose that $\delta^2 + 2\delta|z_0|< \epsilon$", if you want to overload the $\delta$ used above.
The nitpicky stuff is pedantic, but it might be worth paying attention to at first.
