Given that $f(z)=u(x,y)+iv(x,y)$ and Cauchy-Riemann equations hold for $f$ at a point $z_0=x_0+iy_0$ what can be said about $g(z)=\overline{f(\overline{z})}=u(x,-y)-iv(x,-y)$ at the same point?

I tried to solve that using the chain rule defining $w=-y$ but it leads to a wrong answer.





Since Cauchy-Riemann hold for $f$, $\frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{y}}$ and $\frac{\partial{v}}{\partial{x}}=-\frac{\partial{u}}{\partial{y}}$:



That implies Cauchy Riemann equations don't hold for $g$ at $z_0$ since $w=-y$ which is wrong. I would like to know why the above approach is failing.

I read these related questions but they don't light me up.
let f be a nonconstant analytic function in the domain D, show that the function $g(z)= \overline {f(z)}$ is not analytic in D
Show that $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*:=\{\overline z: z\in D\}$ using Cauchy Riemann equation.
$f(z)$ and $\overline{f(\overline{z})}$ simultaneously holomorphic

  • $\begingroup$ this statement is not correct $$\frac{\partial{g}}{\partial{x}}=\frac{\partial{g}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{g}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=\frac{\partial{u}}{\partial{x}}-i\frac{\partial{v}}{\partial{x}}$$ $\endgroup$ – Guy Fsone Oct 7 '17 at 15:51
  • $\begingroup$ I didn't get it. I thought it was right because: g -> u -> x and g -> v -> x (chain rule "tree" diagram) $\endgroup$ – Zalnd Oct 7 '17 at 16:34
  • $\begingroup$ u is a function not a variable for $g$ $\endgroup$ – Guy Fsone Oct 7 '17 at 16:59
  • $\begingroup$ I think dg/dx = d/dx * ( u(x,-y) - i*v(x,-y) ) also leads to the same equation. $\endgroup$ – Zalnd Oct 7 '17 at 19:49

It seems what you read should light you up some, because you are taking the complex conjugate, which is not holomorphic, in two places. ..

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  • $\begingroup$ If I've understood correctly, I should state that dg/dz = 0. I was looking at the wrong equation. Thanks! $\endgroup$ – Zalnd Oct 9 '17 at 23:06

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