Suppose that $h$ is a real-valued harmonic function in some open set. Show that $f=h_y+ih_x$ is analytic.
proof: By assumption, $h$ is harmonic so $h_{xx}+h_{yy}=0$. If $f$ is analytic in the open set, then it must obey the Cauchy-Riemann equations. $$ \frac{\partial s}{\partial x} =\frac{\partial t}{\partial y} ; \ \frac{\partial s}{\partial y}=-\frac{\partial t}{\partial x}$$
So, $f=h_y+ih_x$ can be written as $f=s(x,y)+it(x,y)$ for some s and t. Since $h$ is harmonic on the open set, the mixed partials are equal on the open set, namely, $h_{yx}=h_{xy}$.
Since $h_{xx}+h_{yy}=0$, $f$ is analytic by equality of the mixed partials, i.e. $$ \frac{\partial h_y}{\partial x} =\frac{\partial h_x}{\partial y} ; \ \frac{\partial h_y}{\partial y}=-\frac{\partial h_x}{\partial x} $$
In summary: Since the Cauchy-Riemann equations do not imply analyticity of $f$ (without knowing more about $f$ i.e. continuous partials), we use the fact that $f$ is harmonic on an open set (and therefore has continuous partials in an open set) then unravel the definitions.
Here we are assuming the reader knows a theorem in Ahlfors' Complex analysis:
if $u(x,y)$ and $v(x,y)$ have continuous first-order partials derivatives which satisfy the Cauchy-Riemann differential Equations, then $f(z)=u(z)+iv(z)$ is analytic with continuous derivative $f'(z)$, and conversely.
Are there any suggestions for improvements or critiques? The question feels too easy since we were given that $f$ was harmonic on an open set. Ullrich's book also has this, stated differently, as Proposition 0.1 pg. 5.
Your proof is fine. Maybe it can be streamlined as follows: $h$ is the real part of an analytic function $g=h+iv$. Then, $h_x=v_y$ and $h_y=-v_x$ and $g$ has partial derivatives of all orders. Then, $g_y=f=h_y+iv_y=h_y+ih_x,$ and it satisfies the C-R equations and has partial derivatives of all orders.