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?


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 $\bar z$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ My bad. I forgot to put the word "not" in there! $\endgroup$ – Cameron Williams Mar 6 '16 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.