# Link between harmonic and holomorphic functions on a non-simply connected domain.

There is a theorem that states that if a function $h$ is harmonic on a simply connected domain, there exists a holomorphic function $f$ such that $h = Re f$.

Now, I am having a problem with the statement of this exercise:

Let $h$ be a function harmonic on $\{z\in\mathbb{C}: \rho_1 < |z| < \rho_2\}$. Using the fact that $h_x - ih_y$ is holomorphic, prove that...

What we need to prove is not important, I've already done that. The question is about the fact mentioned in the statement: that $h_x - ih_y$ is holomorphic.

The domain is not simply connected. Normally, we need a simply connected domain to prove that $h_x - ih_y$ is holomorphic, because we use the path-independence of an integral of the form $h(z_0) + \int_{z_0}^z (h_x - ih_y)(w)dw$.

Is there another proof that $h_x - ih_y$ is holomorphic in a non-simply connected domain? Or did I misread the statement and this is just an additional condition on our function $h$?

• Remember that Morera's Theorem is local. As long as you have a continuous function, you're set to go. May 4 '15 at 13:26
• That one, I don't understand. If my function $h$ is such that $g(z) = h_x - ih_y$ has a pole in zero, then to solve $f'(z) = g(z)$, I will need to take an integral over some path from $z_0$ to $z$, but it will be path-dependent... May 4 '15 at 13:44
• To show $g$ is holomorphic, you only need a local primitive $f$. You're right that globally $f$ may have periods, but that doesn't stop $f'=g$ from being holomorphic. Try the example of $h(z)=\log |z|$. Then we get, indeed, $g(z)=1/z$. This is holomorphic on the annulus, even though it does not have a global primitive. May 4 '15 at 14:13
• Thanks, Ted! I think I understand why the exercise statement is correct. May 4 '15 at 14:24

No need for simple connectedness here. The function $$h'_x - ih'_y$$ (is $C^1$ and) satisfies Cauchy-Riemann's equations.
With $u = h'_x$ and $v = -h'_y$, we have \begin{align} u'_x &= h''_{xx} & v'_y &= -h''_{yy} \\ u'_y &= h''_{xy} & v'_x &= -h''_{yx}. \end{align} Hence $u'_x = v'_y$ (since $h$ is harmonic) and $u'_y = -v'_x$ (since $h$ is $C^2$ and the mixed partials are the same).
• So, basically, we don't need the simply connectedness for the fact that $g = h_x - ih_y$ is holomorphic, right? And if I want to get $f'(z) = g(z) = h_x(z) - ih_y(z)$, then I need that path integral to define $f(z)$, so I can't use it in my proof, I guess? May 4 '15 at 13:31
• Right, $g=h'_x-ih'_y$ is holomorphic, but doesn't necessarily have an anti-derivative.