Is arg(f(z)) continuous when f is analytic? 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!
$\operatorname{Arg}(z)$, 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:


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.
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.
$\operatorname{Arg}(z)$, 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:


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.
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.