Check the continuity and analyticity of complex function $f(z)$ Given function $$\displaystyle f(z)=\left\{\begin{matrix}
\frac{\bar{ z}^2}{z} & \text{if} \;\; z \neq 0\\ 
0 & \text{otherwise}
\end{matrix}\right.$$
I have to  check its continuity and analyticity.
solution i tried- i write the function as $$f(z)=\displaystyle  \frac{r^2e^{-2i\theta}} 
{re^{i\theta}}$$
$$\lim_{r \to 0}f(z)=re^{-3i\theta}\to 0$$
so this is continuous on whole $\mathbb{C}$
For analyticity i solved this as $$\frac{(x-iy)^2}{(x+iy)}=\frac{x^2-y^2-2xiy}{x+iy}$$
after solving this i get $$u=\frac{x^3-xy^2-2xy^2}{x^2+y^2} ,\;\;v=\frac{-2x^2y-x^2y+y^3}{x^2+y^2}$$
now i have to fin the $u_x$ and $v_x$ $u_y$ and $v_y$ ,after doing this it seems very complicated to solve further ,can someone please provide a short method to solve this question,or is there any other method which can make it simple and more understandable.
Please help
Thank you
 A: It is clear that $f$ is continuous for $z \neq 0$ and since $|f(z)| \le |z|$ it is continuous at $z=0$.
Suppose $f$ is analytic, then $g(z) = z^3f(z) = |z|^2 $ would be analytic, but since $g$ is real valued it must be a constant hence a contradiction. Hence $f$ is not analytic.
A: Remember the Wirtinger derivatives $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial \bar z}$. Among their basic properties is that they satisfy the product and quotient rule, further $\dfrac{\partial}{\partial \bar z} \bar z= 1$ and $\dfrac{\partial}{\partial \bar z} z=0$, meaning that when applying $\dfrac{\partial}{\partial \bar z}$ to an algebraic expression in $z$ and $\bar z$, you can treat $z$ as a constant and $\bar z$ as a variable like in old school calculus. Consequently:
$$\dfrac{\partial}{\partial \bar z} \frac{\bar z^2}{z} = 2 \frac{\bar z}{z} \qquad (*)$$
However, another basic use of the Wirtinger derivatives is that a function $f$ is holomorphic in an open set if and only if $\dfrac{\partial}{\partial \bar z} f=0$ in that set (since this condition actually encodes the Cauchy-Riemann equations). So $(*)$ shows that your $f$ is not holomorphic in $\mathbb C \setminus \{0\}$, hence not analytic.
