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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?

  • 0
    This looks correct.2012-02-06

1 Answers 1

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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.