How find this limits $\varlimsup_{a\to 0,a\in C}\left|\frac{|f(z+a)|-|f(z)|}{|a|}\right|$ Question:

let $f:C\to C$ is Holomorphic functions,and ,find this value
  $$\varlimsup_{a\to 0,a\in C}\left|\dfrac{|f(z+a)|-|f(z)|}{|a|}\right|$$

My idea:
I guess,this limit is $|f'(z)|$? 
becasue if this is real function,we know this is $$|f'(x)|=\lim_{a\to 0}\left|\dfrac{f(x+a)-f(x)}{a}\right|$$
My guess is true?if wrong,then How find this limits,..
 A: Yes, the $\varlimsup$ is $\lvert f'(z)\rvert$. We can write
$$f(z+a) = f(z) + a\cdot f'(z) + a^2 \cdot g(a)$$
where $g$ is continuous at $0$ (actually it's holomorphic, but we only need the boundedness near $0$; well, even less).
That yields the inequalities
$$\lvert f(z)\rvert - \lvert af'(z)\rvert - \lvert a^2g(a)\rvert \leqslant \lvert f(z+a)\rvert \leqslant \lvert f(z)\rvert + \lvert af'(z)\rvert + \lvert a^2g(a)\rvert,$$
and hence
$$\bigl\lvert \lvert f(z+a)\rvert - \lvert f(z)\rvert\bigr\rvert \leqslant \lvert a\rvert\cdot \lvert f'(z)\rvert + \lvert a\rvert^2 \lvert g(a)\rvert,$$
from which division by $\lvert a\rvert$ gets us
$$\left\lvert \frac{\lvert f(z+a)\rvert-\lvert f(z)\rvert}{\lvert a\rvert}\right\rvert \leqslant \lvert f'(z)\rvert + \underbrace{\lvert a\rvert\cdot\lvert g(a)\rvert}_{\xrightarrow{a\to 0} 0}$$
and therefore
$$\varlimsup_{a\to 0} \left\lvert \frac{\lvert f(z+a)\rvert-\lvert f(z)\rvert}{\lvert a\rvert}\right\rvert \leqslant \lvert f'(z)\rvert.$$
If $f'(z) = 0$, we're done, since the $\varlimsup$ is certainly non-negative. If $f'(z) \neq 0$, choose a $\varphi \in \mathbb{R}$ such that $e^{i\varphi} f'(z)\overline{f(z)}$ is a non-negative real number. Then consider $a$ of the form $a = re^{i\varphi}$ with $r > 0$. For such $a$, we have
$$\lvert f(z)+ af'(z)\rvert = \lvert f(z)\rvert + \lvert af'(z)\rvert,$$
and hence
$$\lvert f(z) + af'(z)\rvert - \lvert a^2 g(a)\rvert \leqslant \lvert f(z+a)\rvert \leqslant \lvert f(z)+ af'(z)\rvert + \lvert a^2 g(a)\rvert$$
yields
$$\lvert f'(z)\rvert - \lvert ag(a)\rvert \leqslant \frac{\lvert f(z+a)\rvert - \lvert f(z)\rvert}{\lvert a\rvert} \leqslant \lvert f'(z)\rvert + \lvert a g(a)\rvert$$
for these $a$, which shows
$$\varlimsup_{a\to 0} \left\lvert \frac{\lvert f(z+a)\rvert-\lvert f(z)\rvert}{\lvert a\rvert}\right\rvert \geqslant \lvert f'(z)\rvert.$$
