Prove that the functions $f(z)$ and $\overline{f(\overline{z})}$ are simultaneously holomorphic.

I take this to mean that $f(z)$ is holomorphic if and only if $\overline{f(\overline{z})}$ is holomorphic.

Let $g(z)=\overline{f(\overline{z})}$. Note that $\overline{g(\overline{z})}=f(z)$. So it suffices to prove that if $f(z)$ is holomorphic, then $g(z)$ is holomorphic.

Write $f(z)=u(z)+iv(z)$. Since $f(z)$ is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations:

$$\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.$$

We have $g(z) = u(\overline{z})+i(-v(\overline{z}))$. To prove that $g(z)$ is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations:

$$\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.$$

How can we obtain this from the above relations?

  • 2
    $\begingroup$ Use the chain rule. $\endgroup$ – Andrés E. Caicedo Aug 24 '13 at 2:46
  • $\begingroup$ See my response at math.stackexchange.com/questions/470597/…. $\endgroup$ – Ted Shifrin Aug 24 '13 at 3:04
  • $\begingroup$ @AndresCaicedo Usually the chain rule in one variable is $\dfrac{df(g(x))}{dx} = f'(g(x))\cdot g'(x)$. What would be the form here? I'm thinking of $\dfrac{\partial v(x,-y)}{\partial y} = (\dfrac{\partial v(x,-y)}{\partial (-y)})(\dfrac{d(-y)}{dy}) = -\dfrac{\partial v(x,-y)}{\partial (-y)}$. Is that written correctly? $\endgroup$ – PJ Miller Aug 24 '13 at 3:09

I know that I am not answering your final question, but anyway... if $g(z)=\overline{f(\overline z)}$, then

$$\begin{align*} g^\prime(a)=&\,\lim_{z\to a}\frac{g(z)-g(a)}{z-a}\\[2mm] =&\,\lim_{z\to a}\frac{\overline{f(\overline z)}-\overline{f(\overline a)}}{z-a}\\[2mm] =&\,\lim_{z\to a}\frac{\overline{f(\overline z)-f(\overline a)}}{\overline{\ \overline{z-a}\ }}\\[2mm] =&\,\lim_{z\to a}\overline{\,\Biggl[\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}\Biggr]}\\[2mm] =&\,\overline{\lim_{z\to a}\,\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}}\\[2mm] =&\,\overline{\lim_{w\to\overline a}\,\frac{f(w)-f(\overline a)}{w-\overline a}}\\[2mm] =&\overline{f^\prime(\overline a)}\,. \end{align*}$$

Thus, $g$ is holomorphic. The converse is proved similarly (or you can use the fact that the transformation $f\mapsto g$ is idempotent, that is, you return to your original function $f$ when applied twice).

  • $\begingroup$ I think this is a better approach than using the Cauchy-Riemann equations. It is better to just stay with the complex arguments when you can, instead of going to x + iy. The reason is that the x + iy computations tend to be more complicated, so it is easier to make arithmetic mistakes. Of course, sometimes this is the only way to do the problem; but it's always worth looking for a simpler alternative. $\endgroup$ – Betty Mock Aug 24 '13 at 4:17
  • $\begingroup$ @matematicos, you have used $z\rightarrow a$ then $w=\overline z\rightarrow a$ won't it converge to $\overline a$ $\endgroup$ – User Mar 9 '16 at 9:53
  • $\begingroup$ @Human You are right. Correcting now. $\endgroup$ – Matemáticos Chibchas Mar 10 '16 at 18:47

Proof based in holomorphic $\implies$ locally power series: $$f(z) = \sum_{n=0}^\infty a_n (z-c)^n,$$ $$ \tilde f(z) = \overline{f(\overline{z})} = \overline{\sum_{n=0}^\infty a_n (\overline{z}-c)^n} = \sum_{n=0}^\infty\overline{a_n (\overline{z}-c)^n} = \sum_{n=0}^\infty\overline{a}_n(z-\overline{c})^n, $$ where the third equality is true because the continuity of conjugation.

  • $\begingroup$ This is a really cool proof, thanks! $\endgroup$ – ZirconCode Jun 26 '17 at 16:58

Morera's theorem is an extremely useful tool when one wants to prove a function is holomophic but doesn't want to look at derivatives explicitly. In this case, let $\gamma$ be a simple closed curve.

$\int _\gamma \overline{f(\overline{z})} dz = \overline{\int _\gamma f(\overline{z}) dz} = \int _\gamma f(\overline{z}) d\overline{z} = -\int_{\overline{\gamma}} f(u) du$ where $\overline{\gamma}$ is the curve $\gamma$, reflected with respect to the x-axis and the - comes from the corresponding change of the orientation from counterclockwise to clockwise.

Now, by the above inequality and Morera's theorem: $f(z)$ is holomorphic iff its integral over an arbitrary simple closed curve is $0$ iff $\overline{f(\overline{z})}$ is holomorphic, which we wanted to prove.


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