If $u,v$ are harmonic and satisfy C-R on a set with a limit point, must $u +iv$ be analytic? Let $\Omega \subseteq \mathbb{C}$ be a domain, and suppose that $u,v$ are real-valued harmonic functions defined on $\Omega$. Furthermore, let $W \subseteq \Omega$ be the set on which $u,v$ satisfy the Cauchy-Riemann equations. 
If $W$ has a limit point in $\Omega$, does that mean that $u + iv$ is an analytic function on $\Omega$?
This problem appears on an old complex analysis qualifying exam. My hunch is that the answer is no. The counterexample I'm trying to construct is as follows: we take $\Omega =$ some annulus about the origin, and then look at $ u(x,y) = \ln|z| = \ln(x^2 + y^2)$ and $ v(x,y) = \operatorname{Arg} z$ as harmonic functions on this annulus. The catch would be that $\ln|z|$ and $\operatorname{Arg} z$ satisfy the C-R equations everywhere on this annulus, but $\ln|z| + i \operatorname{Arg} z$ cannot be analytic on the full annulus.
The only trouble I'm having is that I cannot see how to define $\operatorname{Arg} z$ as an harmonic function on the full annulus $\Omega$. I have tried messing around with the formula $\operatorname{Arg} z = \arctan (\frac{x}{y}),$ but of course this formula runs into trouble where $y = 0$. So perhaps the statement is true and $u + iv$ must be analytic?
Hints or solutions are greatly appreciated
 A: It seems that $u + iv$ must be analytic. The harmonicity of $u$ and $v$ ensures that $f= u_x - iu_y$,$ g= v_y + iv_x$ are analytic on $\Omega$ (one can check that these two functions satisfy the C-R equations). Since $f$ and $g$ agree on a set with a limit point, they must agree on all of $\Omega$, hence $u$ and $v$ satisfy the C-R equations on all of $\Omega$. So, indeed, $u + iv$ is analytic on all of $\Omega$.
A: I general I think you need the continuity of the partial derivatives $f_x$ and $f_y$ and take care of the notion of analyticity.
One can prove that $f_x$ and $f_y$ exist in a neighborhood of a point $z$ in the domain of $f$ and the partial derivatives are continuous at $z$ with $f_y=if_x$, then $f$ is differentiable at $z$.
A pathological example is given by $f(z)=|z|^2=x^2+y^2$, which has continuous partial derivatives for all $z$, but is differentiable iff $if_y=f_x$, i.e. only at $z=0$ (i.e. $f$ is not analytic at $0$ because there exists no neighborhood $U$ of $0$ s.t. $f$ is diff. on $U$)
Even more "locally", a function can satisfy the C.R. equations at a given point but not differentiability at that point.
