Let $f(z)$ be a holomorphic function on $\mathbb{C}$. Show that $\overline{f(\bar{z})}$ is holomorphic on $\mathbb{C}$ 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.
 A: $f$ is holomorphic or in other words it's analytic on C. So, pick a $z_0\in \mathbb C$ and write $f(z)=\sum a_n(z-z_0)^n\Rightarrow \overline{f(\bar{z})}=\sum \overline {a_n}(z-\overline {z_0})^n$
A: 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
A: 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.
A: 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.
A: 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.
A: 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  $\bar \Delta \subset \bar\Omega$,
$$\int_{\bar \Delta} \overline{ f (\bar z)} \, dz =0.$$
