Sufficient for analyticity

Question

Let $f$ be complex valued function such that $f(z)=U(x,y)+iV(x,y)$, Here $U(x,y)=x^2-y^2$ is given, Now here it is easy to find out $V(x,y)=2xy + C$, $C$ is any arbitrary complex constant, Now my doubt is that is there unique $V(x,y)$ for making $f$ analytic or is there infinite number of $V(x,y)$ (for the arbitrary complex constant) for making $f$ analytic? I am here little bit confused. Please what is right! Thank you.

Answer 1

Since the functions $U(x,y)=x^2-y^2$ and $V(x,y)=2xy+C$ satisfy the Cauchy Riemann equations, thus $f=U+iV$ is analytic for all constant $C$.

Note that $$f=x^2-y^2+i(2xy+C)=(x+iy)^2+iC$$ i.e. $f=z^2+C'$ which is analytic, of course.