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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?

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1  
Looks fine to me! – Martin Wanvik Feb 6 '12 at 18:09
2  
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
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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
1  
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.

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