$Re(f)=Re(g)$ 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?

Notice that $f-g$ has values in $i\mathbb{R}$ and if $f-g$ is non-constant, then its image must be open in $\mathbb{C}$, which is a contradiction. Whence the result.
