Showing a holomorphic function on $\mathbb C-{0}$ satisfying $|z\frac{du}{dz}|\le c|u(z)| $ is meromorphic I am looking for a proof of the following argument, which comes from a book:

Proposition: Let $u$ be a holomorphic function on $\mathbb C-{0}$, and assuming that there exists $c>0$ such that for all $z\in \mathbb C-{0}$$$|z\frac{du}{dz}|\le c|u(z)| $$ Then $u$ is meromorphic along origin.

The book claims that it is a consequence of Gronwall's inequality:

If $f(t)\ge 0, g(t)\ge 0$ are non-negative functions on $0<t\le t_0$ such that $$f(t)\le a+c\int_t^{t_0}g(s)f(s)ds$$ for some $a\ge0,c\ge 0$, then we have $$f(t)\le a\cdot exp(c\int_t^{t_0}g(s)ds)$$

I tried a little bit but I didn't get anything meaningful with given inequality and the hint. In particular, I don't know what estimate I should get to show that $u$ is meromorphic.
Thanks in advance for any help!  
 A: I'm going to write $f=\mu$, sorry.
Don't know about Gronwall, but it's easy to see from Complex 101 that in fact $f(z)=az^n$ for some integer $n$. Assume $f$ is not identically zero.
Let $$g(z)=\frac{zf'(z)}{f(z)}.$$A priori $g(z)$ is defined only for $z\ne0$ with $f(z)\ne0$. But $g$ is bounded near each zero  of $f$, hence has a removable singularity there. And now by the same argument $g$ has a removable singularity at the origin.
So $g$ is a bounded entire function, hence constant. So $zf'(z)=Cf(z)$. Now if $$f(z)=\sum_{n=-\infty}^\infty a_n z^n$$it follows that $$na_n=Ca_n$$for every $n$; hence $a_n=0$ unless $n=C$. So if $C$ is not an integer then $f=0$,  while if $C=n$ is  an integer then $f(z)=a_nz^n$.
A: Let me point out that the following generalization is actually what I want to prove:

Let $\vec{u}=(u_1,...,u_n)$, with each $u_i$ homolomrphic on $\mathbb C-\{0\}$, satisfying $$|z|\sqrt{\sum_{i=1}^n(\frac{du_i}{dz})^2}\le C\sqrt{\sum_{i=1}^nu_i^2}.$$ then show that $\vec{u}$ is meromorphic along $0$.

To show this, we indeed need the Gronwall's inequality. I learned this proof in Chapter 5 of  Hotta-Takeuchi-Tanisaki's D-modules, Perverse sheaves, and Representation theory.
note that for $z=re^{i\theta}\in \mathbb C^*$, $$|\frac{\partial}{\partial r}u_i(re^{i\theta})|=|\frac{du_i}{dz}(z)|$$
but fixing $r_0>0$, and take any $0<r<r_0$, we have $$\vec{u}(re^{i\theta})-\vec{u}(r_0e^{i\theta})=\int_r^{r_0}\frac{\partial}{\partial s}\vec{u}(se^{i\theta})ds,$$ and hence 
$$ \begin{align} ||\vec{u}(re^{i\theta})|| & \le ||\vec{u}(r_0e^{i\theta})||+\sqrt{\sum_{j=1}^n(\int_r^{r_0}|\frac{\partial}{\partial s}u_j(se^{i\theta})|ds)^2} 
 \\
& \le ||\vec{u}(r_0e^{i\theta})||+\int_r^{r_0}\sqrt{\sum_{j=1}^n|\frac{\partial}{\partial s}u_j(se^{i\theta})|^2}ds \\ &\le ||\vec{u}(r_0e^{i\theta})||+C\int_r^{r_0}\frac{||\vec{u}(se^{i\theta})||}{s}ds\end{align} $$
where $||\cdot||$ is the $l^2$ norm of a vector, and the establishment of the second inequality can be reduced to prove for simple functions, which turns out to be Cauchy-Schwarz inequality.
Now, we can apply apply Gronwall's inequality, which tells us that 
$$\begin{align} ||\vec{u}(re^{i\theta})||&\le ||\vec{u}(r_0e^{i\theta})|| \exp(C\int_r^{r_0}\frac{1}{s}ds) \\ & = ||\vec{u}(r_0e^{i\theta})||\frac{r_0^C}{r^C}\end{align}$$
which implies that $ ||\vec{u}(re^{i\theta})||$ has growth $\frac{1}{r^N}$ for some $N$ large enough, and $r$ small enough. In particular, it shows that the vector $\vec{u}$ is meromorphic along $0$.
