Real part of an analytic function I'd like to know how to take the real part (of the power series representation) of an analytic function with the goal of showing the (real part of the analytic) function is real analytic.
I know this may seem trivial but I haven't been able to convince myself that that's the case.
 A: I will show an example with the power series for $e^z,\,z\in\mathbb{C}$. The power series is
$$\sum_{n=0}^{\infty}\frac{z^n}{n!}=\sum_{n=0}^{\infty}\frac{(re^{i\varphi})^n}{n!}.$$
Here $r$ is the magnitude of $z$ and $\varphi$ the phase angle. Note that $\operatorname{Re}(z_1+z_2)=\operatorname{Re}(z_1)+\operatorname{Re}(z_2),$ so
$$\operatorname{Re}\sum_{n=0}^{\infty}\frac{r^ne^{i\varphi n}}{n!}=\sum_{n=0}^{\infty}\operatorname{Re}\frac{r^ne^{i\varphi n}}{n!}.$$
Eulers formula tell us that  $e^{ix}=\cos x+i\sin x,\,x\in\mathbb{R}.$ From now on you should be able to calculate the last steps. Hint: $r^n/n!$ is real so you can first take the real part of $e^{i\varphi n}$ and then multiply it.
A: If anyone is looking for the answer to this question, you need to take into account that if the function is the real part of an analytic function then the function must be infinitely differentiable (as an analytic function is itself infinitely differentiable). Thus, it can be represented as a power series, and thus fulfill the definition of real-analyticity. The one caveat one needs consider, however, is that the power series will be in two dimensions, and the expansion thereof must reflect a two-dimensional Taylor series.
With this in mind, one can show that a harmonic function on an open domain of $\mathbb{R}^2$ is real-analytic (as it can be proven that a harmonic function is the real/imaginary part of an analytic function).
