Right, so I'm struggling proving/disproving that for functions $u,v: \mathbb R^2 \to \mathbb R$ if $v$ is a harmonic conjugate of $u$, then $u$ is a harmonic conjugate of $v$ (so the relation is symmetric)...

I've been trying to disprove it so far and I'm not having luck...

Am I correct in saying if a function is holomorphic, then it satifies Cauchy-Riemann equations and that is the best test for holomrophisms?

Also, if a function is indeed holomorphic, is that equivalent to saying it's analytic? Or is there a difference?

The definition of a harmonic function is as follows:

$u(x,y)$ is harmonic if there exists some $v(x,y) : f=u+iv$ is holomorphic, in which case, $v$ is the harmonic conjugate of $u$

If I'm trying to find a counter example to disprove my first statement, do I find 2 functions, $u , v$, substitute them into $f=u+iv$ and show that the Cauchy-Riemann equations don't hold? For some reason, I'm not believing that Cauchy-Riemann equations do test for holomorphism.

If you look at a similar question I posted, some user told me two different functions, and stated that one was holomorphic and one wasn't... but surely the fact that they're independently holomorphic is irrelevant... $f=u+iv$ has to be holomorphic right!?

And finally, if there are two holomorphic functions $u,v$, is it not the case that $f=u+iv$ will also always be holomorphic?

Any help is really really appreciated thanks


1 Answer 1


Your first claim about the harmonic conjugates is correct. In fact, both $u$ and $v$ must satisfy Laplace's equation, which is the PDE $\Delta$$u(x,y)$=$0$. (Here, $\Delta$$\phi(x,y)$ is defined as the trace of the elements in the Hessian matrix, or matrix of second partials. Holomorphic is actually another term synonymous to analytic, yes. However, note that the Cauchy-Riemann equations only give a necessary condition for f(z) to be analytic in some neighborhood around $z_0$, not a sufficient one. For a both sufficient and necessary condition, we need to make sure that the four partial derivatives of the real and imaginary parts of $f= u + iv$ are both continuous and satisfy the Cauchy-Riemann equations themselves. If these sets of conditions are true, along with the Cauchy-Riemann equations, then yes, f is analytic. For a counter-example of this, take $f(z)$ = \begin{cases} \frac{\overline{z}^2}{z} &, z\neq 0 \\ 0 &,z=0 \end{cases} You will see that the CR equations hold, but the complex derivative does not exist at $0$. Hope this helps to point you in the right direction.


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