Is this function holomorphic in $z = 0$?

Question

Consider the function $$ f(z) = \frac{\mathrm{Im}(z^2)}{|z|^2} \qquad (z \neq 0) $$ with $f(z) = 0$ for $z = 0$. Using $z = x+iy$ we find $$ f(z) = \frac{2xy}{x^2+y^2} \ , $$ so $u = f$ and $v = 0$ for $f = u + iv$. The derivatives of $u$ are: $$ u_x = \frac{2y (y^2 - x^2)}{{(x^2+y^2)}^2} \ , \qquad u_y = \frac{2x (x^2 - y^2)}{{(x^2+y^2)}^2} \ . $$ Then the Cauchy-Riemann equations $u_x = 0$ and $u_y = 0$ imply either $x = y = 0$ or $x = \pm y$, so $f(z)$ is complex differentiable in $z = 0$ and along the lines $x = \pm y$. However, is $f(z)$ holomorphic in $z = 0$? I would say no, because it would have to be complex differentiable in a neighbourhood of $z = 0$, which I don't think it is.

Criticism is welcome, and please point me in the right direction if I'm totally off.

Answer 1

$f$ is not even continuous at $z = 0$, as for $ 0 \neq x \in \mathbb R$, $f(x + ix) = 1$ while $f(0)=0$.

Therefore, $f$ can't be holomorphic at that point.

Answer 2

Along $y=mx$; the limiting value of function is:
$\lim_{x\rightarrow 0} \frac{2xmx}{x^2+(mx)^2}=\frac{2m}{1+m^2}$\which depends on $m$;
so even the value of $\lim_{z\rightarrow 0} f(z)$ fails to exist.