Showing that the real and imaginary part of a holomorphic function are $\mathrm{C}^{\infty}$ This is a fairly easy question. Let me first describe the setting:
Let $f$ be defined on a domain $\Omega \subseteq \mathbb{C}$. Define the two functions
\begin{equation}
\widetilde{u}(x,y) = \Re(f(x+iy))\\
\widetilde{v}(x,y) = \Im(f(x+iy))
\end{equation}
so that of course
\begin{equation}
f(x+iy) = \widetilde{u}(x,y)+i\widetilde{v}(x,y).
\end{equation}
Let
\begin{equation}
\widetilde{\Omega} = \{ (x,y) \in \mathbb{R}^{2}: x+iy \in \Omega \}.
\end{equation}
Two facts are in order:


*

*$f \in \mathrm{Hol}(\Omega)$ if and only if $\widetilde{u}$ and $\widetilde{v}$ are differentiable and satisfy the Cauchy-Riemann equations in $\widetilde{\Omega}$;

*if $f \in \mathrm{Hol}(\Omega)$, then $f$ is analytic in $\Omega$. In particular, $f \in \mathrm{C}^{\infty}(\Omega)$.
From these facts I want to prove that if $f \in \mathrm{Hol}(\Omega)$, then $\widetilde{u},\widetilde{v} \in \mathrm{C}^{\infty}(\widetilde{\Omega})$.
The way I see it:
Using the C-R equations, we can write
\begin{equation}
f’(x+iy) = \frac{\partial \widetilde{u}}{\partial x}(x,y) -i \frac{\partial \widetilde{u}}{\partial y}(x,y) = \frac{\partial \widetilde{v}}{\partial y}(x,y) +i \frac{\partial \widetilde{v}}{\partial x}(x,y).
\end{equation}
Using (1), knowing by (2) that $f’$ is holomorphic, we see that $\widetilde{u},\widetilde{v}$ have differentiable (hence continuous) first partial derivatives, therefore they’re $\mathrm{C}^{1}(\widetilde{\Omega})$.
Iterating using the formula above, we see that $f^{(n)}$ can be written in $2^{n}$ different ways in which all the $n$-th partial derivatives of $\widetilde{u},\widetilde{v}$ figure, the mixed derivatives figuring with all their possible permutations. To clarify what I mean let me write down the case $n=2$, using a slimmer notation:
\begin{equation}
f’’ = \widetilde{u}_{xx} -i \widetilde{u}_{xy} = - \widetilde{u}_{yy} -i \widetilde{u}_{yx} = \widetilde{v}_{yx} -i \widetilde{v}_{yy} = \widetilde{v}_{xy} +i \widetilde{v}_{xx}.
\end{equation}
So, again by (1) and (2), at the $n$-th step of iteration we can conclude that $\widetilde{u},\widetilde{v} \in \mathrm{C}^{n-1}(\widetilde{\Omega})$. Therefore, $\widetilde{u},\widetilde{v} \in \mathrm{C}^{\infty}(\widetilde{\Omega})$.
Now, my question: I’m just looking for alternative/cleaner ways to get this result. Thanks to all of you who will take their time to answer.
 A: Suppose $f = u + \sqrt{-1}v$ is holomorphic. Then the CR equations can be used to show that $u$ and $v$ are harmonic, i.e., $u_{xx} + u_{yy} =0$ and $v_{xx} + v_{yy}=0$ (this is a good undergraduate exercise). Then use elliptic regularity theory to assert that harmonic functions are smooth. 
Does this help?
A: Suppose that $f = u + \sqrt{-1} v$ for some $C^1$-functions $u,v : \tilde{\Omega} \to \mathbb{R}$. Writing $f$ as a sum of convergent power series, we have $$f(z) = \sum_{k=0}^{\infty} c_k (z-w)^k,$$ where $c_k = \frac{1}{k!}\frac{\partial^k f}{\partial z^k}$ and $w$ is some fixed point in $\tilde{\Omega}$. Note that we are appealing to the Cauchy integral formula here in order to write $f$ as the above series. (You should read the proof of this thoroughly - I would suggest Alexander Isaev's book Twenty-One Lectures on Complex Analysis.) 
If you believe the above series representation (with the coefficients given as above) then we see that $\frac{\partial^k f}{\partial z^k}$ is defined for all $k \in \mathbb{N}_0$. Do you then believe that the partial derivatives of $u$ and $v$ are so defined for all orders? (If not: Write $\frac{\partial f}{\partial z}$ in terms of its real and imaginary parts). 
Warning: I rushed this.
