$Re(f)=Re(g)$ implies $f(z)=g(z)+ic$ Let $f$ and $g$ be analystic in a region $G$. If $\Re(f)=\Re(g)$ in $G$, then prove
$$f(z)=g(z)+ic \tag{for all $z \in G$}$$
where $c$ is a real constant.
By cauchy riemann we have
$$\int \frac{\partial}{\partial y}Re(g)=-\int\frac{\partial}{\partial x}Re(g)$$
Hence, $Im(f)$ must be some constant?

I would use Cauchy-Riemann to do it like this:
Set
$h(z) = h_R(z) + ih_I(z) = f(z) - g(z); \tag 1$
then 
$\Re(h) = h_R(z) = 0; \tag 2$
now by Cauchy-Riemann
$h_{I, x} = -h_{R,y} = 0 \tag 3$
and
$h_{I, y} = h_{R, x} = 0; \tag 4$
it follows that
$\nabla h_I = 0 \tag 5$
in $G$; thus,
$h_I = c \in \Bbb R, \tag 6$
where $c$ is constant.  So
$f(z) - g(z) = h(z) = h_R(z) + ih_I(z) = 0 + ic = ic, \tag 7$
whence 
$f(z) = g(z) + ic. \tag 8$
