Analytic function in $\mathbb{C}\backslash \{0\}$ that is identically zero 
Suppose $f$ is an analytic function in $\mathbb{C}\backslash \{0\}$ and satisfies the inequality $|f(z)| \leq c|z||Logz|$ ($z \in \mathbb{C}\backslash \{0\}$), for some constant $c >0$. Prove that $f(z)=0$ for every $z \in \mathbb{C}\backslash\{0\}$.

My try:
Note that $f(1)=0$. So its suffice to show that $|f'(z)|=0$. Let $\gamma$ be circular path with center in $z \in \mathbb{C}\backslash \{0\} $ and appropriate radius $r>0$. Using the Cauchy Theorem for derivates we have
\begin{align}
|f'(z)|&=\left| \int_{\gamma}\dfrac{f(\xi)}{(\xi - z)^2}d\xi \right| \\
&\leq \int_{\gamma}\dfrac{|f(\xi)|}{|\xi - z|^2}|d\xi| \\
&\leq c \int_{\gamma}\dfrac{|\xi||Log\xi|}{r^2}|d\xi| 
\end{align}
But now I can't conclude what I want. Is not the correct aproach?
 A: Although your approach seems to fail, you can actually use Cauchy's integral formula for higher orders to make the log radius term insignificant. For instance, use $n=2$ we have
$$|f''(z)|\le C\int_\gamma \frac{r(|\ln r| + |i\theta|)}{r^3}|d\xi|\le 2\pi r \frac{r(|\ln r| + 2\pi)}{r^3}\le C'\frac{\ln r}{r}$$
where $C,C'$ denote appropriate constants. Now let $r\to\infty$.
Caveat: there might be a little problem with the complex $\log$ since it has a "jump" along its "faultline". However, since we only use $\log z$ for our estimating the magnitude of $|f|$, this problem doesn't affect our proof here because one jump contributes at most a $i2\pi=O(1)$ which is thus negligible.
Note: $f$ isn't holomorphic on the closed disk $|z|\le r$, and Cauchy is inapplicable here.
Note 2: per the discussion in the below comments, $0$ is a removable singularity, so the approach still works.
Note 3: my approach actually proves $f$ is either zero or linear, but doesn't exclude the possibility that $f$ is linear (which if we don't specify the choice of the log branch will always persist, see the comment by another user below). However, if you specify that $\log 1=0$, then $f$ must be zero. This isn't implied by my "proof" here, but follows from the user's argument in the comments below (Liouville's theorem and something).
