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I am attempting to prove that:

$$\lim_{z \rightarrow z_0} (z^2 + c) = z^2_0 + c$$

From the limit definition, we want to show that $0 < |z-z_0| < \delta \implies |(z^2+c)-(z^2_0+c)| < \varepsilon$. We observe that:

\begin{align} |(z^2+c)-(z^2_0+c)| &= |z^2-z^2_0| \\ &= |(z-z_0)(z+z_0)| \\ &= |z-z_0||z+z_0| \\ &< \varepsilon \end{align}

I am unsure as to how to deal with the $|z+z_0|$ term. There are a few thoughts I would like to clarify:

  1. Is it fair to say that if $0 < |z-z_0| < \delta$ holds by premise, then we also have $0 < |z+z_0| < \delta$, since the only thing that changes is the center of the circle on the complex plane? My strategy to finish the proof would then be to write:

$$|z-z_0||z+z_0| < \delta^2$$

Then set $\delta = \sqrt\epsilon$. More generally, does $|z+z_0|$ have any relevance to the inequality $0 < |z-z_0| < \delta$?

  1. If not, then is it legal to fix some radius $R = |z+z_0|$, thus $|z-z_0| R < \varepsilon$ and finishing the proof by setting $\delta = \frac{\varepsilon}{R}$?
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    Let z0 = 1, z = 1.00...01, and you see easily that |z+z0| goes to 2z0, not 0 or delta. – 2017-02-26

2 Answers 2

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You can show by the Triangle inequality that: $$ |z + z_0| = |z - z_0 + 2z_0 | \leq |z - z_0| + 2|z_0| $$ and so $$ |z-z_0||z + z_0| \leq |z - z_0|^2 + 2|z_0||z-z_0| $$ Therefore \begin{align*} |z-z_0| < \delta \implies |z-z_0||z + z_0| &\leq |z - z_0|^2 + 2|z_0||z-z_0| \\ &< \delta^2 + 2|z_0|\delta \\ &< \epsilon \end{align*} where $$ \delta^2 + 2|z_0|\delta - \epsilon= 0 $$ You can prove that for any $\epsilon> 0$, there exists a $\delta > 0$ that solves that equation.

The determinant is: $\Delta = 4|z_0|^2+4\epsilon > 0$, so there is always a solution. Furthermore, the solution $$ \delta = \frac{-2|z_0| + \sqrt{4z_0^2+4\epsilon}}{2} $$ is positive since $4z_0^2+4\epsilon> 4z_0^2$, so $\sqrt{4z_0^2+4\epsilon} > 2|z_0|$

So for any choice of $\epsilon > 0$, we can choose an $\delta > 0$ by the formula above, such that the required inequality holds.

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    How exactly can we justify that $|z-z_0| \leq \varepsilon$? – 2017-02-26
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    My apologies. I switched delta and epsilon by mistake. Corrected my post. – 2017-02-27
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First, $|z-z_0|<\delta$ does not imply that $|z+z_0|<\delta$. To use an example to show this, suppose that $z_0=1$ and $\delta=1$. Then, we see that for $z=3/2$, $|z+z_0|=5/2>\delta$.

We can set, as you suggested, an initial bound. But we set that initial bound for $|z-z_0|$. For example, we require that $|z-z_0|\le 1$. Then, note that

$$|z+z_0|=|z-z_0+2z_0|\le |z-z_0|+2|z_0|<1+2|z_0|$$

Can you finish now?

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    Please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – 2017-04-12