Counterexample for a complex analysis proof

Question

I'm having troubles coming up with a counterexample for the following: If $|f(z)|$ is continuous at $z_0$, then the function $f(z)$ is continuous at $z_0$ for complex numbers.

I know I need a $f(z)$ that is discontinuous at that $z_0$, where $|f(z)|$ is not.

Does anyone know of one that would work?

Answer 1

Define $f(z) = z/|z|$ for $z\not= 0$ and $f(0) = 1.$

Answer 2

Let $r(z)$ be a real-valued everywhere discontinuous function, and put $f(z) = e^{ir(z)}$.

Answer 3

Let $f(z) = \begin{cases} 1, & z=z_0 & \\ -1, & \text{otherwise} \end{cases}$. Then $|f(z)| = 1$, is continuous, and $f$ is discontinuous at $z=z_0$.