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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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    Looks fine to me!2012-02-06
  • 2
    Perfect, excepting one small possible issue you need to avoid in the end... You divide by $|a|$, and what happens if $|a|=0$?2012-02-06
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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.2012-02-06
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    So I just make a note of that in my proof?.. where d - e/|a|, |a| != 02012-02-06
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    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 $|z-z_0| < \frac{\epsilon}{|a|+1}$.2012-02-06
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    Just write "if $a=0$ then there is nothing to prove, so let us assume $a\neq 0$", I think that will be ok with most mathematicians2012-02-06
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    This looks correct.2012-02-06

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