Is arg(f(z)) continuous when f is analytic?

Question

My question is same with the title.

When f(z) is analytic(or just continuous), arg(f) is differentiable(or just continuous)?

If not, what restrictions are needed for arg(f) to be continuous?

Thanks!

Answer 1

$\operatorname{Arg}(z)$, if defined in any sensible way on all of $\Bbb C$, isn't continuous, no matter how you slice it.

If $f(z)$ is continuous and never zero, $\operatorname{Arg}(f(z))$ can be continuous. There are two ways to make this work:

1. If there is some continuous, non-self-intersecting curve from the origin to $\infty$ that $f(z)$ never hits, you can use that as the branch cut for $\operatorname{Arg}$, and everything works nicely.

2. Accept the multivalued nature of $\operatorname{Arg}$, and let $\operatorname{Arg}(f(z))$ be multivalued as well. Then it's not strictly speaking a function, but locally it behaves like a function.

In either case, it is continuous. An $\varepsilon$-ball around $\operatorname{Arg}(f(z))$ corresponds to a sector around $f(z)$ in $\Bbb C$. And assuming $f$ is continuous and $f(z)\neq0$, you can always find a $\delta$-ball around $z$ which, when mapped with $f$, lands inside this sector.