10
$\begingroup$

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?

$\endgroup$
  • 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
10
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is a really cool proof, thanks! $\endgroup$ – ZirconCode Jun 26 '17 at 16:58
0
$\begingroup$

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.

$\endgroup$

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.