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.


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.

  • $\begingroup$ Thank you very much for your answer. I really want to be able to write proofs that can survive a "nit-picking". So, I must ask more about the nitpicky stuff above. First, I think I understand the part about "you're really using this for a given, fixed $\varepsilon$ to define $\delta$." In other words, you're saying that since $\varepsilon$ is the independent variable in the definition of a limit, then $\delta$ should depend on the choice of $\varepsilon$, and not the other way around. Does that sound right? $\endgroup$ – Mike Bell May 6 '16 at 18:26
  • $\begingroup$ And for the second part, about "the $\delta$ so defined is a new entity, distinct from the $\delta$ that appears above," I don't totally understand this. If (and maybe this is not the right assumption to make) the $\varepsilon$ in my proof is the same as that in the definition of a limit, then will that not require that the $\delta$ that shows up in my proof also be the same $\delta$ that showed up in the definition of the limit? Thank you! $\endgroup$ – Mike Bell May 6 '16 at 18:30
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    $\begingroup$ Hey Mike. Yes, your first part sounds right. Your $\delta$ should be a response to an $\varepsilon$. So, given $\varepsilon$, you can define a $\delta$ such that... whatever. I just wanted to make sure that it was clear that your equation is restricting/defining the $\delta$. $\endgroup$ – Josh Keneda May 7 '16 at 8:30
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    $\begingroup$ The second part isn't really worth worrying about. I was noting that your argument starts with "Assume $|z-z_0|< \delta$," and then you define $\delta$ a few lines later. Either this is a new $\delta$ or it's being defined after already in use. This doesn't actually cause any problems in your argument, but you could consider reorganizing for clarity. You could bring your definition of $\delta$ to the front of your argument, or you could use a different letter when deriving the inequalities. $\endgroup$ – Josh Keneda May 7 '16 at 8:46

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