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Can anyone show me how I can prove something as simple as $f(z) = z$ is holomorphic but $\bar z$ is not?

Olórin
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1 Answers1

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The former satisfies the Cauchy Riemann equations, the latter does not.

For $f(z)=z$ we have $u=x, v=y$ and easily $u_x=v_y$ and $u_y=-v_x$, but for the $f(z)=\overline{z}$ we have $u=x, v=-y$ so that $u_x\ne v_y$.

Alternatively, you can see it directly

$$\lim_{z\to 0} {\overline{z}-\overline{0}\over z-0}=\lim_{z\to 0} {\overline{z}\over z}=e^{-2i\theta}$$

where $\theta =\arg(z)$, so that the limit does not exist, because we can take $\theta=\theta(t)= t$ for example as a spiral going into the origin, so the value oscillates infinitely. By definition of a derivative, since that limit does not exist, there is no derivative.

Adam Hughes
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  • Is using wirtinger derivative a good way to check in general? – Olórin Feb 08 '15 at 08:52
  • @MathNewb It depends on what machinery you have at your disposal. I included the definition because contradiction a definition is always workable, no matter how much experience you have with the subject. I know many ways to demonstrate $\overline{z}$ is not-holomorphic, and I wanted to make sure I did a simple one because I didn't know how much you knew about the subject. – Adam Hughes Feb 08 '15 at 08:54
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    I know nothing and I want to know more from you but the method you have offered is robust for all cases and that is good enough for me – Olórin Feb 08 '15 at 08:55