If $U$ is simply connected and $u: U\to\Bbb R$ is harmonic then it have a conjugate in $U$

Question

Im not sure if my solution to this exercise is correct:

Let $U\subset \Bbb C$ and $u:U\to\Bbb R$ harmonic. Then $v\in C^1(U,\Bbb R)$ is a conjugate of $u$ if $v_x=-u_y$ and $v_y=u_x$. Show that if $U$ is simply connected then $U$ have a conjugate in $U$. HINT: consider the function $u_x-iu_y$.

I dont understand the hint, and this why I opened this question. What I did: observe that $du=u_x dx+u_y dy$ is closed by definition, then for any fixed $z_0\in U$ there is a path $\Gamma_z$ that join $z_0$ and $z$, because $U$ is simply connected, and we have the identity $$ u(z)=u(z_0)+\int_{\Gamma_z}du=u(z_0)+\int_{\Gamma_z}u_x dx+u_y dy\tag1 $$ that holds because $du$ is exact in $U$. Then the function $$ \tilde u(z):=\int_{\Gamma_z}-u_ydx+u_x dy\tag2 $$ is well defined in $U$, because when $u$ is harmonic then $-u_ydx+u_xdy$ is trivially closed, that is, $-u_{yy}=u_{xx}$. And we have that $d\tilde u=-u_ydx+u_x dy$ by definition, thus $\tilde u_x=-u_y$ and $\tilde u_y=u_x$, as desired.

Questions:

1. The above proof is correct, or there is something inaccurate or not enough justified?

2. Someone knows for what, supposedly, is the hint?

Answer 1

Your proof looks fine.

Now, consider the function $f=u_x-iu_y$. Since $u$ is harmonic, $f$ is holomorphic and therefore analytice. Since $U$ is simply connected,$f$ has a primitive $F$. But then, there's some constant $k$ such that $\operatorname{Re}F=u+k$. Now, you can use $\operatorname{Im}F$ get a conjugate of $u$.