$f$ holomorphic on $B(0,R)$ and $f(0)=0$. Can $\Re (f) \le 0$ on $B(0,R)$?

Question

Consider $f$ holomorphic on $B(0,R)$ and $f(0)=0$. Can $\Re (f) \le 0$ for all $z\in B(0,R)$?

Suppose $z=0$ is a zero of $f$ of multiplicity $m\ge 1$: $f(z)=z^mg(z)$, $g$ holomorphic in $B(0,R)$ and $g(0)\ne 0$. I really don't see how to go further with this reasoning. How can I get the real part of $f$ into this proof?

Thanks.

Answer 1

If $f$ is the null function, then you have $\operatorname{Re}\bigl(f(z)\bigr)\leqslant0$ in $B(0,R)$.

Otherwise, that will not occur, by the open mapping theorem. Since $f\bigl(B(0,R)\bigr)$ is an open set to which $0$ belongs, it must contain some open disk $B(0,\varepsilon).$