$f(z) = z$ for $\leq 1$, and $f(z) = |z|$ for $> 1$. I'm trying to find where $f$ is continuous. I know that $f$ is continuous where $z \leq 1$, since $f$ just maps elements to itself.
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When $|z_0|<1$ or $|z_0|>1$, $f$ is continuous at $z_0$, because the sets $D(0,1)$ and $\{z\in\Bbb C\mid |z|>1\}$ are open sets and the restrictions of $f$ any of these sets is continuous.
When $|z_0|=1$, there are $z$ points arbitrarily close to $z_0$ where $f(z)=z$ and there are $z$ points arbitrarily close to $z_0$ where $f(z)=|z|^2$. So, $f$ is continuous at $z_0$ if and only if $z_0=|z_0|^2$, which happens if and only if $z_0=1$.
José Carlos Santos
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Well, $f(z)=z$ is continuous because, for every $\varepsilon>0$, you can take $\delta=\varepsilon$. And $|z|^2=z\times\overline z$, and therefore it's the product of two continuous functions. – José Carlos Santos Feb 03 '21 at 20:16