Is it analytic ? $f(z)=z^2\bar{z}$ Please find the question and its solution.



My answer gone wrong dont know why, I think the answer marked in the question is wrong.
My try:
I solved for CR equations and replaced x=y=0 and found CR equations satisfied. But not sure why in the solution it says CR equations doesn't satisfy.
Please let me know what should be the correct answer ?
 A: You are right: since $u_x(0,0)=u_y(0,0)=v_x(0,0)=v_y(0,0)=0$, $(0,0)$ is a solution of the Cauchy-Riemann equations (and, since the partial derivatives are continuous at $(0,0)$,  $f$ differentiable at $0$).
A: The problem is poorly phrased. It awkwardly mixes real and complex differentiability as well as pointwise and local notions. Let me try to clear things up.
As a map $\mathbb R^2\to\mathbb R^2$ the map $f(z)=z^2\bar z$ has polynomials as component functions. In particular, it is real differentiable everywhere and even real analytic (meaning that all components have convergent real power series expansions near all points in the domain).
However, the Cauchy-Riemann equations are only satisfied at $z=0$. So as a map $\mathbb C\to\mathbb C$,  $f$ is complex differentiable at $z=0$ but not at any $z\ne0$. This means that $f$ is not complex analytic at zero. Otherwise it would have a convergent complex power series expansion in a neighborhood of zero, and would thus be complex differentiable in that neighborhood.
A: Apart from the continuity of all partial derivatives $u_x, v_x, u_y, v_y$ along with $u$ and $v$ everywhere, note that $u_x=v_y \forall x,y$
But $u_y=-v_x$ holds for $xy=0\implies x=0$ or $y=0$ i.e. on coordinate axes. So

*

*CR-equations hold at $z=0$ and the test says that $f(z)$ is differentiable at $z=0$ (not analytic though).


I would say that the question is poorly formatted as all the options stand incorrect.

