Let $f(z)\ = \frac {z^5} {|z|^4}$ if $z\ \neq 0$ and $f(0) = 0$ Question: Conclude that the partials of u, v exist and that the Cauchy-Riemann equations hold but that $f'(0)$ does not exist. Does this conclusion contradict the Cauchy-Riemann Theorem?
Could some one explain how I can prove this? I was sick and missed an entire week of class, but is this related to the Inverse Function Theorem? 
For the CR equations, the previous question told me to assume that $ u = Re(f)$ and $v = Im(f)$.
 A: $\bullet$ Since you want to check that $f'(z)$ does not exist at $z=0$, seems to me that the problem is to conclude that the Cauchy-Riemann equations hold, but only at $z=0$, as @Dr.MV pointed out in the comments. To see that lets put $f(x+iy)=u(x,y)+iv(x,y)$, hence 
$$
\frac{\partial u}{\partial x}(0,0)= \lim_{x \to 0} \frac{u(x,0)-u(0,0)}{x} = \lim_{x \to 0} \frac{x^5/|x|^4}{x} = 1,
$$
but clearly $\frac{\partial u}{\partial y}(0,0)=0$. Also 
$$
\frac{\partial v}{\partial y}(0,0)= \lim_{y \to 0} \frac{v(0,y)-v(0,0)}{y} = \lim_{y \to 0} \frac{y^5/|y|^4}{y} = 1,
$$
and again it is clear that $\frac{\partial v}{\partial x}(0,0)=0$. Thus indeed C-R equations are satisfied for $z=0$.
$\bullet$ Now, to see that $f'(0)$ does not exist, take $z=te^{i\theta_0}$ with $\theta_0 \in \mathbb{R}$ fixed and $t >0$ such that $t \to 0^+$, in that case 
$$
\lim_{z \to 0} \frac{f(z)-f(0)}{z} = \lim_{t \to 0} \frac{f(te^{i\theta_0})}{te^{i\theta_0}} = \lim_{t \to 0} \frac{(te^{i\theta_0})^5/|te^{i\theta_0}|^4}{te^{i\theta_0}}=\lim_{t \to 0} \frac{t^5e^{i4\theta_0}}{t^5} = e^{i4\theta_0}
$$
However, $\frac{\partial u}{\partial x}(0,0)+i\frac{\partial v}{\partial x}(0,0)=1$,  thus indeed $f'(0)$ can not exist.
$\bullet$ But this conclusion does not contradict the Cauchy-Riemann Theorem! Why? The answer is because the partials $u, v$ are not continuos functions, as Cauchy-Riemann Theorem hypothesis requires.
