Is the function $f(z)=\frac{(\bar z)^2} z$ analytic at $0$? Is it continuous at $0$ and does it satisfy the Cauchy-Riemann equations? I just want to know: is the function
$$f(z)= \begin{cases}
\frac{(\bar{z})^2}z,&z\ne 0\\
0,&z=0
\end{cases}$$ 
analytic at $0$?, continuous at $0$?, and does it satisfy the Cauchy-Riemann equation at $0$?
I am totally lost how to start. Please help.
 A: The function $f$ is continuous at $0$ because $\vert f(z)\vert=\vert z\vert$ for all $z$ and hence $f(z)\to 0=f(0)$ as $z\to 0$.
It is not differentiable in the complex sense at $0$ because $\frac{f(z)-f(0)}{z-0}=\left(\frac{\bar z}{z}\right)^2$ has no limit as $z\to 0$, $z\neq 0$. (This is equal to $1$ for real $z$ and to $-1$ for $z$ of the fore $\varepsilon\, e^{i\frac\pi4}$, $\varepsilon >0$). 
On the other hand, $f$ does satisfy the Cauchy-Riemann equation at $0$. Indeed, using the $(x,y)$ coordinates we have $f(0,y)=f(iy)=\frac{(-iy)^2}{iy}=-\frac1i y=iy$ for any $y\in\mathbb R\setminus\{ 0\}$, so $\frac{f(0,y)-f(0,0)}{y}=i$ and hence $\frac{\partial f}{\partial y}(0,0)$ exists and is equal to $i$. Likewise, $\frac{\partial f}{\partial x}(0,0)$ exists and is equal to $1$. So we do have $\frac{\partial f}{\partial y}(0,0)=i \, \frac{\partial f}{\partial x}(0,0)$.
A: Just as above in Etienne's post, but here $f$ is not analytic at $z=0$ because it is only differentiable at $z=0$ and nowhere else. So there is no open set centered at $z=0$ where $f$ is differentiable so your function is not analytic at 0. Since the limit quotient above does not exist anywhere else then $f$ is nowhere analytic.
