Where is $f(z) = 2ixy$ holomorphic.
I would work this out using $u(x,y) = 0, v(x,y) = 2xy$
$$u_x = 0 \qquad u_y = 0 \qquad v_x = 2y \qquad v_y = 2x$$
Since $u_x = v_y \Rightarrow x = 0$ and since $u_y = -v_x \Rightarrow y=0$
Thus $f$ is just holomorphic at $0$.
Wikipedia givens an alternative form of Cauchy-Riemann:
First, the Cauchy–Riemann equations may be written in complex form $$i\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$$
If I use this theorem then:
$$i\frac{\partial f}{\partial x} = -2y = 2ix =\frac{\partial f}{\partial y}$$
Then $f$ would be holomorphic on the line $y = -ix$.
Contradiction?
This looks contradictory to me, am I missing something?
Remember that $x$ and $y$ are real numbers here. So there are no solutions to $y=-ix$ other than $x=y=0$. ("The line $y=-ix$" would make sense as a subspace of $\Bbb C^2$, but not in this context.)