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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?

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    $\begingroup$ Don't forget that $x$ and $y$ are real. $\endgroup$ – Etienne Jan 8 '15 at 23:28
  • $\begingroup$ Also, with the usual definition "holomorphic at $0$" doesn't make sense. The function $f$ is (complex) differentiable at $z=0$, but for $f$ to be holomorphic, it needs to be differentiable on an open set. $\endgroup$ – mrf Jan 9 '15 at 9:09
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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.)

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