Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b, z_0$ denote complex constants. Use the definition of a limit to show that $$\lim_{z \to z_0} (az + b) = az_0 + b.$$

Here is what I have done:

\begin{align*} |az + b - (az_0 + b)| &= |az - az_0 + b - b|\\ &= |a(z - z_0)|\\ &= |a||z - z_0|. \end{align*}

So for a positive number $\epsilon$,

$$|az + b - (az_0 + b)| < \epsilon \text{ whenever } |a||z - z_0| < \epsilon$$

or in other words $|az + b - (az_0 + b)| < \epsilon$ whenever $|z - z_0| < \delta$ where $\delta = \epsilon/|a|$.

Have I proved the statement correctly?

share|cite|improve this question
Looks fine to me! – Martin Wanvik Feb 6 '12 at 18:09
Perfect, excepting one small possible issue you need to avoid in the end... You divide by $|a|$, and what happens if $|a|=0$? – N. S. Feb 6 '12 at 18:13
It looks like a textbook example of a $\delta$-$\epsilon$ proof to me, other than the point raised by N.S. – robjohn Feb 6 '12 at 18:16
So I just make a note of that in my proof?.. where d - e/|a|, |a| != 0 – Jim_CS Feb 6 '12 at 18:25
You can either split the end in two cases: case 1 $|a| \neq 0$, case 2: $|a|=0$, or, pretty standard trick, observe that $|a||z - z_0| < \epsilon$ happens when $|z-z_0| < \frac{\epsilon}{|a|+1}$. – N. S. Feb 6 '12 at 18:40

As mentioned in the comments, your proof works for $a \neq 0$. You can deal with the case where $a = 0$ separately or choose $\epsilon$ in such a way that the proof works for all $a$ (see N.S.'s comment for example). Aside from this your proof is correct.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.