I have a question about a theorem and its proof from the book, Functions of one complex variable(John B. Conway) 3.2 section.
Theorem:
Let $\pmb{G}$ be either the whole plane $\mathbb{C}$ or some open disk. if $ u: \pmb{G} \rightarrow \mathbb{R}$ is a harmonic function then $u$ has a harmonic conjugate.
Proof:
Let $\pmb{G} = \pmb{B}(0; R)$, $0\lt R \leq \infty$ and let $ u: \pmb{G} \rightarrow \mathbb{R}$ be a harmonic function. The proof will be accomplished by finding a harmonic function $v$ such that $u$ and $v$ satisfy the Cauchy-Riemann equation. so define
$\mathit{v}(x,y) = \int^y_0u_x(x,t)dt+\phi(x)$
and determine $\phi$ so that $v_x=-u_x$. differentiating both sides of this equation with respect to $x$ gives
$v_x(x,y)= \int^y_0u_{xx}(x,t)dt+ \phi'(x)$
$=-\int_0^yu_{yy}(x,t)dt+\phi'$
$=-u_y(x,y)+u_y(x,0)+\phi'(x)$
so it must be that $\phi'(x)=-u_y(x,0)$. It is easily checked that $u$ and
$v(x,y)=\int_0^yu_x(x,t)dt - \int_0^xu_y(s,0)ds$
do satisfy the Cauchy-Riemann equations.
So, my question is that where we used an open disk or whole $\mathbb{C}$ to prove this theorem? (according to the statement)
Could someone explain where it went wrong if the region is not simply connected?