# Harmonic conjugates and holomorphic functions

I've been researching complex analysis recently and come across a couple of questions that I'm a little confused about. This is one of them.

Let $u:U\to \mathbb R$ be a harmonic function on an open subset of $\mathbb C$. Show $v$ is a harmonic conjugate of u if and only if $u$ is a harmonic conjugate of $-v$.

I know I need to apply the Cauchy-Riemann equations but I'm a bit stuck on the logic of the problem, since if a function satisfies Cauchy-Riemann, it is not necessarily holomorphic is it?

Thanks

• In fact, if a continuous complex function satisfies C-R in an open domain, then it is holomorphic there. – Jonathan Y. Nov 4 '13 at 12:42
• how can I apply this to harmonic functions – Tom Nov 4 '13 at 12:47
• By noting that $u+iv$ is holomorphic in a domian iff $v-iu = -i(u+iv)$ is. – Jonathan Y. Nov 4 '13 at 12:49
• @JonathanY. Is that all? You don't have to assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable? – bof Nov 4 '13 at 12:56
• It depends on your starting point. As the question is currently phrased, both $u,v$ are a priori taken to be harmonic (hence $C^\infty$). But one could simply consider continuous $u,v:\Omega\to\mathbb{R}^2$ which have partials in $\Omega$ and satisfy C-R, and the rest follows (in fact, one could even slightly relax the requirement that $\Omega$ is open, as described in the link above). – Jonathan Y. Nov 4 '13 at 13:40

## 1 Answer

You don't have to resort to the Cauchy-Riemann equations if you know a couple of facts.

1. $v(x,y)$ is a harmonic conjugate of $u(x,y)$ in $U$ if and only if the function $f(z)=u+iv$ is analytic in $U$.

2. A constant multiple of a differentiable function is differentiable. Therefore, if $f(z)$ is analytic in $U$, so is any constant multiple of $f(z)$.

So, if $v$ is a harmonic conjugate of $u$, take the analytic function $f(z)=u+iv$ and multiply it by the constant $i$, and get the analytic function $if(z)=-v+iu$ showing that $u$ is a harmonic conjugate of $-v$.