The question states
Let $f:\mathbb{C}\to \mathbb{C}$ is given by $\begin{equation}\begin{aligned}f(z)&=\frac{\bar z^2 }{z} for\ z\neq 0\\&=0 for\ z=0\end{aligned}\end{equation}$ then $f$ is
(1) is not continuous at $z=0$
(2)is differentiable but not analytic at $z=0$
(3) is differentiable everywhere
(4) it satisffies CR equations
I tried to check the limit of the function along different path, along $x=0$ the limit was 0 and along $y=0$ also it gave 0, but along $x=y$, I got the limit as a constant term $\frac{-2i}{1+i}$. So since the limit are different along different path it is not continuous at 0. and hence not differentiable.
But how to check for the last option..am I suppose to first convert the equation in $u+iv$ form by substituting $z=x+iy$??This will be too much..
So I check $\frac{df}{d\bar{z}}=0$, then solving it gives me $z=0$ which implies CR equations are satisfied only at z=0. And since the CR equations satisfied only at $z=0$, option 4 also not correct.
Am I correct..
thanks in advance!!