# Thread:

1. ## Continuity of complex function

Note: Sorry if this is the wrong section, the course this is from is 'Complex Analysis', so I assumed it belonged here.

The question
f(z) = im(z)

Where is f continuous?

My attempt
Unless I'm mistaken, we need to show that
$\lim_{z \to z_0} f(z) = f(z_0)$

The limit is $y_0$, and f(z) is also $y_0$. So is the function continuous everywhere, or am I doing this incorrectly?

Thanks.

2. Originally Posted by Glitch
The question
f(z) = im(z) Where is f continuous?
My attempt
Unless I'm mistaken, we need to show that
$\lim_{z \to z_0} f(z) = f(z_0)$
The answer depends level of rigor required.
If you need a $\varepsilon /\delta$ proof here is a suggestion.
If $|z-z_0|<\delta$ then $|y-y_0|\le\sqrt{(x-x_0)^2+(y-y_0)^2}=|z-z_0|$.

3. I don't think we have to use epsilon delta proofs. I'm quite sure my lecturer mentioned that it was non-examinable.

4. An alternative is to use the property

$\displaystyle\lim_{z\to z_0}f(z)=w_0\Leftrightarrow \displaystyle\lim_{z\to z_0}\textrm{Re}(f(z))=\textrm{Re}(w_0)\;\wedge\; \displaystyle\lim_{z\to z_0}\textrm{Im}(f(z))=\textrm{Im}(w_0)$
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