Is the function $f(z) = z$ continuous on $|z| = 1$?

Question

Is the function $f(z) = z$ continuous on $|z| = 1$? Here $z$ is a complex number.

I'm scratching my head here as $z_{0}$ is a modulus.

How do I calculate this?

Answer 1

Since $f(z)=z$ is continuous everywhere, it is continuous on the circle $|z|=1.$

Answer 2

In the basic topological definition, a function $f : X \to Y$ between two topological spaces $X$ and $Y$ is continuous if the preimage of any open set $U \subset Y$ is an open set $\textrm{preim}_f(U) \subset X$. Since $f$ is the identity map on $\mathbb{C}$, it is continuous on all of $\mathbb{C}$. This is easy to show since the preimage of any subset $U \in \mathbb{C}$ under $f$ is the set itself, $\textrm{preim}_f(U) = U$, so if $U$ was open, so is $\textrm{preim}_f(U) = U$. You can essentially prove that the identity is always continuous between two identical topological sets. But then this argument also holds for the circle $|z|=1$, since $f$ is also the identity on the circle.

If you're not familiar with topology, forget this answer and parametrize the circle as $z = e^{i\phi}$.