The exercise:
Let $h$ be a function harmonic on $\{z\in\mathbb{C}: \rho_1 < |z| < \rho_2\}$. Using the fact that $h_x - ih_y$ is holomorphic, prove that there exist unique constants $(a_n)_{n\in\mathbb{Z}}$ and $b$ with $a_0, b\in\mathbb{R}$, such that \begin{align*} h(z) = Re\left(\sum_{-\infty}^\infty a_n z^n\right) + b\log|z|\qquad (\rho_1 < |z| < \rho_2) \end{align*}
My approach: we know that $g(z) = h_x - ih_y$ is holomorphic in the annulus, but we can't simply integrate it between $z_0$ and $z$ to find $f$ such that $\operatorname{Re} f = h$ because the domain is not simply connected. So, I proceed as follows: decompose $g(z)$ into Laurent series in the annulus, choose any $z_0$ from the annulus and integrate each member of the series: $$\int_{z_0}^z c_n z^n dz.$$ Since for every member except $-1$st, the integral is path-independent and an antiderivative exists, no problem here. As for the member $\frac{c_{-1}}{z}$, the integral from $z_0$ to $z$ depends on the path we choose. Here, I am having a problem with $c_{-1}$. If there was no complex constant, I would just say that the antiderivative is the complex logarithm and its imaginary part $i(\operatorname{Arg} z + 2k\pi)$ changes depending on the path, but the real part $\ln|z|$ is the same. Since we only need the real part, our expression is well-defined and we're done.
But since we have $c_{-1}(\ln|z| + i(\operatorname{Arg} z + 2k\pi)$, the real part would include $2k\pi$ and my argument doesn't work. Is there a reason for $c_{-1}$ to be real? Another point: in my case we extract the $\ln|z|$ and we have no element that becomes $a_0$, so in our case $a_0 = 0$. I guess, I could have $a_0 = \operatorname{Im}(c_{-1})\cdot \operatorname{Arg} z$, but I would have to define the $\operatorname{Log} z$ in an unambiguous manner, and I can't see clearly how...