Suppose that $f$ is an analytic function in $\mathbb{C}-\{0\}$ and that $|f(z)| \le c|z||\log(z)|$, for some constant $c>0$. Prove that $f(z)=0$ for all $z$. ($\log(z)$ is the principal logarithm of $z$.)
My attempt:
I know that $f$ has a removable singularity at the origin, since $$\lim_{z\rightarrow 0}f(z)=0$$
Therefore $$f(z)=a_1z+a_2z^2+...$$
and $$\frac{f(z)}{z}=a_1+a_2z+...$$
But $|\log(z)|$ grows at a slower rate than any polynomial, so $a_n=0$, $\forall n\ge2$, and $f(z)=a_1z$.
After this i got stuck. Any hints?