Why is a harmonic conjugate unique up to adding a constant?

Question

If $v$ and $v_0$ are harmonic conjugates of $u$, then $u + iv$ and $u + iv_0$ are analytic functions. Then $i(v - v_0)$ is analytic, but how does this imply $v - v_0$ is a constant function?

Answer 1

It does not follow from $v-v_0$ being (real) analytic, if that is what you are asking for. $u+iv$ and $u+iv_0$ are complex analytic, i.e. holomorphic, so they satisfy the Cauchy Riemann differential equations, $u_x = v_y$ an $u_y = -v_x$, the same holds for $v_0$ and $u$. So, consequently, $(v-v_0)_x = 0 = (v-v_0)_y$.

I think you be able to finish the reasoning from here, at least on open connected sets (the statemnent is true for connected sets only).

(Edit: of course, if you assume that $i(v-v_0)$ is a complex analytic function, then obviously the same reasoning applies, so any holomorphic function the real part of which vanishes identically is locally constant)