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Since $f(z)$ is holomorphic, I used Cauchy-Riemann equations and got $u_x = v_y ,\ u_y = -v_x$

Then I wanted to check if Cauchy-Riemann equations are satisfied for $\overline{f(\bar{z})}$

It does.

But then I realised that Cauchy-Riemann equations say that if differentiable at $z$, then satisfies CREs at $z$. And not the other way around.

So does that mean basically my proof is wrong? How would I prove this then?

Thanks.

Satvik Saha
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user120227
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  • Unless I'm grossly mistaken, differentiable at $z$ and the CR condition at $z$ enjoy an 'if and only if' relationship. – The Count Jul 26 '19 at 01:46

6 Answers6

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$f$ is holomorphic or in other words it's analytic on C. So, pick a $z_0\in \mathbb C$ and write $f(z)=\sum a_n(z-z_0)^n\Rightarrow \overline{f(\bar{z})}=\sum \overline {a_n}(z-\overline {z_0})^n$

Haha
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A function that is real differentiable (or $C^1$) and satisfies Cauchy-Riemann's equations is holomorphic. Since your $f$ is holomorphic, it follows that its real and imaginary parts are smooth and there is no problem using Cauchy-Riemann as you describe.

An alternative solution is to compute the derivative directly. Let $g(z) = \overline{f(\bar z)}$. Then \begin{align} \lim_{h\to 0} \frac{g(z+h)-g(z)}{h} &= \lim_{h \to 0} \frac{\overline{f(\bar z + \bar h)}-\overline{f(\bar z)}}{h} \\ &= \lim_{h \to 0} \overline{\,\frac{{f(\bar z + \bar h)}-{f(\bar z)}}{\bar h}\,} \\ &= \overline{\,\lim_{s \to 0} \frac{{f(\bar z + s)}-{f(\bar z)}}{s}\,} = \overline{f'(\bar z)} \end{align} (with $s=\bar h$). Hence $g$ is $\mathbb{C}$-differentiable and thus holomorphic

mrf
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  • Thanks for your reply. What do you mean when you say a function is real differentiable? Does that mean: if f = u+iv then f is differentiable with respect to u ? – user120227 Nov 03 '14 at 22:13
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    No, I mean that $u$ and $v$ are differentiable as maps $\mathbb{R}^2 \to \mathbb{R}^2$. (See http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions) – mrf Nov 03 '14 at 22:16
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If $g(z)=\overline{f(\overline{z})}$, then we have that $$ \frac{g(z+h)-g(z)}{h}=\frac{\overline{f(\overline{z+h})}-\overline{f(\overline{z})}}{h} =\frac{1}{h}\Big(\overline{f(\overline{z+h})-f(\overline{z})}\Big)=\overline{\frac{1}{\overline{h}}\Big(f(\overline{z+h})-f(\overline{z})}\Big)\to \overline{f'(\overline{z})}. $$ Indeed, $g$ is holomorphic.

Yiorgos S. Smyrlis
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If $f$ is differentiable as a function on $\mathbb{R}^2$, we do have equivalence between complex-differentiability and Cauchy-Riemann equations holding. See link.

ir7
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Since $f(z)$ is holomorphic on $\Omega$, it is analytic in $\Omega$: $\forall a \in \Omega, \forall r > 0$ such that $\overline{D}(a, r) \subset \Omega, \exists (a_n)_{n \geq 0} \subset \mathbb{C}$ such that $$f(z) = \sum_{i=0}^\infty a_n(z-a)^n\, \forall z \in D(a, r).$$ Therefore $$\begin{align*} \overline{f(\overline{z})} &= \overline{\sum_{i=0}^\infty a_n(\overline{z}-a)^n}\\ &=\sum_{i=0}^\infty \overline{a_n(\overline{z}-a)^n}\\ &=\sum_{i=0}^\infty \overline{a_n}\overline{(\overline{z}-a)^n}\\ &=\sum_{i=0}^\infty \overline{a_n}\overline{(\overline{z}-a)}^n\\ &=\sum_{i=0}^\infty \overline{a_n}(z-\overline{a})^n. \end{align*}$$ Hence $\overline{f(\overline{z})}$ is analytic, which means it is holomorphic.

Laurent Hayez
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An old question, but for completeness sake... A different, and often very convenient, method from those in other answers here, is to use a converse of Cauchy's theorem, Morera's theorem (theorem 10.17 of Rudin's Real and Complex Analysis): a continuous complex function $f$ is holomorphic on an open set $\Omega$ if (and only if) $$ \int_\Delta f \, dz =0$$ for every closed triangle $\Delta \subset \Omega$. In the case here, taking the complex conjugate of the integral, and making the 'change of variables' $z\to \bar z$, one gets, for every $\bar \Delta \subset \bar\Omega$, $$\int_{\bar \Delta} \overline{ f (\bar z)} \, dz =0.$$

peter a g
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