I am working on an exercise from a graduate first year complex analysis class. The problem states:
let $f: D \to \mathbb{C}$ given by $ f = u + iv$. Assume that $f$ admits an absolutely convergent double power series expansion $$ f(z, \overline{z})= \sum_{n,m = 0}^\infty a_{n \overline{m}} z^n \overline{z}^m $$ and that usual differentiation and integration rules for power series in one variable apply.
i) Under what conditions on the coefficients $a_{n \overline{m}} $ is the function holomorpic in $D$
ii) Under what conditions on the coefficients $a_{n \overline{m}} $ is the function complex harmonic in $D$.
What I have tried so far:
i) f is holomorphic if $\frac{\partial}{\partial \overline{z}} f= 0$, i.e.$$0 = \frac{\partial}{\partial \overline{z}} \sum_{n,m = 0}^\infty a_{n \overline{m}} z^n \overline{z}^m = \sum_{n,m = 0}^\infty \frac{\partial}{\partial \overline{z}} a_{n \overline{m}} z^n \overline{z}^m = \sum_{n,m = 0}^\infty a_{n \overline{m}} z^n (m\overline{z}^{m-1})$$ Where we could pass derivative under the sum since the series is absolutely convergent. Since this has to hold for every $z$ and $\overline{z}$ we must have $a_{n \overline{m}} = 0$ for every $n,m \in \mathbb{N}$.
ii) f is complex harmonic if $Re(f)$ and $Im(f)$ are harmonic. We know that$$ Re(f) = \frac{f+\overline{f}}{2}, \quad Im(f) = \frac{f - \overline{f}}{2i}, \quad \frac{1}{4} \Delta u = \frac{\partial^2}{\partial z \partial \overline{z}} u $$Then\begin{align}0 = \frac{1}{4} \Delta Re(f) & = \frac{\partial^2}{\partial z \partial \overline{z}} \frac{1}{2}\left( \sum_{n,m = 0}^\infty a_{n \overline{m}} z^n \overline{z}^m + \sum_{n,m = 0}^\infty \overline{a_{n \overline{m}}} \overline{z}^n z^m \right) \\\\ &= \frac{1}{2} \sum_{n,m = 0}^\infty a_{n \overline{m}} mn z^{n-1} \overline{z}^{m-1} + \overline{a_{n \overline{m}}} mn \overline{z}^{n-1}z^{m-1}\end{align}So $$a_{n \overline{m}} mn z^{n-1} \overline{z}^{m-1} = - \overline{a_{n \overline{m}}} mn \overline{z}^{n-1}z^{m-1} $$Similarly from the laplacian of the imaginary part we get $$ a_{n \overline{m}} mn z^{n-1} \overline{z}^{m-1} = \overline{a_{n \overline{m}}} mn \overline{z}^{n-1}z^{m-1} $$Then we must have $$ a_{n \overline{m}} = 0$$
I do not think this is correct but have no better ideas. I am looking for feedback and suggestions on how to do this problem.
