Is the function complex differentiable at (0,0)? (in Complex)
$$  g(z) =
\begin{cases}
\frac{z^5}{|z|^4} & \text{if $z \neq 0$} \\
0, & \text{if $z = 0$ }
\end{cases} $$
For the function above, is it differentiable at $z=0$?
I am trying to use following theorem to solve it: 
Let $f(z)=u(x,y)+iv(x,y)$ be defined in some open set G containing the point $z_0$. If the first partial derivates of u and v exist in G, are continuous at $z_0$ and satisfy the C-R equations at $z_0$, then f is differentiable at $z_0$.
I would say, since $g(z)=0$ if $z=0$ that C-R holds at $0$ and that first partial derivates exist in $C$, but I am not sure about the partial derivates being continuous at $0$ (how do I see that)? 
 A: Trying to use Cauchy-Riemann here won't help because the partials are not continuous. In fact this an example of a function which satisfies the Cauchy-Riemann's equations at a point but isn't differentiable there.
To prove that isn't differentiable at the origin, use the definition: $$\lim \limits_{z\to 0}\left(\dfrac{g(z)}z\right)=\lim \limits_{z\to 0}\left(\dfrac{z^4}{|z|^4}\right).$$
Now consider the $\theta$-sublimits $\lim \limits_{\rho \to 0}\left(\dfrac{\left(\rho e^{i\theta}\right) ^4}{\left|\rho e^{i\theta}\right|^4}\right)$, where $\theta\in \mathbb R$. Conclude.
A: Here is one approach:
Note that for $x$ real we have $g(x) = x$, so if $g$ was differentiable, we must have $g'(0) = 1$.
Now check $|{g(z)-g(0) \over z}-1| = | {z^5 \over z|z|^4} -1 | = |  {z^4-|z|^4 \over |z|^4}|$. Can this be made arbitrarily small?
Having done that, this suggests looking at $g(x e^{i {\pi \over 4}})$ (for example) for $x$ real.
A: Consider $g$ as a function from $\mathbb{C}$ to $\mathbb{C}$  where $\mathbb{C} \simeq \mathbb{R}^2$ is a two-dimensional vector space. Let's find the real directional derivatives at $(0,0)$.  For $z\in \mathbb{C}$ and $t >0$ we have 
$$\frac{g(tz) - g(0)}{t}= \frac{(tz)^5}{|t z|^4 t}= \frac{t^5 z^5}{t^5 |z^4|} = (\frac{z}{|z|})^4 \cdot z$$
Therefore the real directional derivative in direction $z$ is $(\frac{z}{|z|})^4 \cdot z$. This function 
$$z \mapsto (\frac{z}{|z|})^4 \cdot z$$
is not an $\mathbb{R}$-linear map. Therefore, $g$ is not real differentiable at $(0,0)$, so neither complex differentiable.
Note that 
$$\frac{\partial g}{\partial x} = 1\\
\frac{\partial g}{\partial y} = i$$
so $\frac{\partial g}{\partial y}= i \cdot \frac{\partial g}{\partial x}$
so the Cauchy-Riemann are equalities hold. The problem is that $g$ is not real differentiable at $(0,0)\ \ \ $  $\tiny{ \text{Cauchy-Riemann:} \frac{\partial g}{\partial (iy)}=  \cdot \frac{\partial g}{\partial x} }$
Obs: Let's check that the map $z \mapsto (\frac{z}{|z|})^4 \cdot z$, $0\mapsto 0$ is not additive. Indeed, $1\mapsto 1$, $i \mapsto i$ but $(1+i) \mapsto -(1+i)$. 
