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How could I use the operator $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ to thow that $f(z)=2ix$ is not an holomorphic function?

I don't know how to proceed. Is anyone could help me at this point?

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A function is holomorphic on an open set $U$ if $\frac{\partial}{\partial \bar z}f(x,y) = 0$ for all $(x,y)\in U$. Noting that $\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$, do you see how to proceed?

The condition $\frac{\partial}{\partial \bar z} f= 0$ is equivalent to the Cauchy-Riemann equations but is conceptually a bit nicer since it immediately shows that $f$ is independent of $\bar z$.

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  • $\begingroup$ Do I have to use $x=\frac{z+\bar{z}}{2}$? $\endgroup$ – user1050421 Mar 6 '16 at 22:09
  • $\begingroup$ yes, equivalently that $z = x+iy$, that's what the exercice supposed $\endgroup$ – reuns Mar 6 '16 at 22:09
  • $\begingroup$ @RobertDavis You can, but you do not have to. That is why I wrote the alternate formulation of $\frac{\partial}{\partial \bar z}$ down. Of course those two relationships are extremely closely related. $\endgroup$ – Cameron Williams Mar 6 '16 at 22:09
  • $\begingroup$ So as it is not holomorphic, it is sufficient to show that $\frac{\partial}{\partial \bar z}f(x,y) \not= 0$, right? $\endgroup$ – user1050421 Mar 6 '16 at 22:10
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    $\begingroup$ My bad. I forgot to put the word "not" in there! $\endgroup$ – Cameron Williams Mar 6 '16 at 22:14

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