Finding a real-valued harmonic function

Question

Let $h$ and $g$ be two complex-valued functions defined on a domain $\Omega$ such that $\Re h$ and $\Im g$ are harmonic and $\Im h = \Re g$. Show that there exists a real-valued harmonic function of the form $h + \lambda g$ where $\lambda$ is a constant to be determined.

My attempt at the problem is this:

let $h = a + ib$, $g = b + ic$

$a$ and $c$ are harmonic

Then $$h + \lambda g = a + ib + \lambda b + i\lambda c$$ $$ = a + \lambda b + i(b + \lambda c)$$

The laplacian of $h + \lambda g$ should be $0$ so that it is harmonic, but i'm not getting anywhere, i think i'm approaching the problem wrongly.

Any help would be much appreciated

Answer 1

Let $h=u+iv$ and $g=a+ib$, where $u,v,a$ and $b$ are real-valued functions on $\Omega$. Then, by hypothesis, $u$ and $b$ are harmonic and $v=a$.

We have that $h+\lambda g = (u+ia) + \lambda (a+ib)=u+(i+\lambda)a+\lambda i\cdot b$. This should give you an idea of what $\lambda$ is: since we know $u$ and $b$ are harmonic, we should be able to find a $\lambda$ that gets rid of the $a$ term and makes the coefficient on the $b$ term real. It's easy to see that $\lambda =-i$ does the job, and we're left with

$$h-ig=u+b$$

which is the sum of two real-valued harmonic functions and hence real and harmonic.