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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.