There is a theorem that states that if a function $h$ is harmonic on a simply connected domain, there exists a holomorphic function $f$ such that $h = Re f$.
Now, I am having a problem with the statement of this 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...
What we need to prove is not important, I've already done that. The question is about the fact mentioned in the statement: that $h_x - ih_y$ is holomorphic.
The domain is not simply connected. Normally, we need a simply connected domain to prove that $h_x - ih_y$ is holomorphic, because we use the path-independence of an integral of the form $h(z_0) + \int_{z_0}^z (h_x - ih_y)(w)dw$.
Is there another proof that $h_x - ih_y$ is holomorphic in a non-simply connected domain? Or did I misread the statement and this is just an additional condition on our function $h$?
No need for simple connectedness here. The function $$ h'_x - ih'_y $$ (is $C^1$ and) satisfies Cauchy-Riemann's equations.
With $u = h'_x$ and $v = -h'_y$, we have \begin{align} u'_x &= h''_{xx} & v'_y &= -h''_{yy} \\ u'_y &= h''_{xy} & v'_x &= -h''_{yx}. \end{align} Hence $u'_x = v'_y$ (since $h$ is harmonic) and $u'_y = -v'_x$ (since $h$ is $C^2$ and the mixed partials are the same).