Suppose that $v$ is a harmonic conjugate of $u$. Show that $-u$ is a harmonic conjugate of $v$.


Using the information, it seems that I can write:
$f(x+iy) := u(x,y) + iv(x,y)$ with $f$ being harmonic
$$\implies \frac{\partial^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2} = 0$$
and $f'(x+iy) = u_x + iv_x$ (since a harmonic function $u$ implies there is a holomorphic function $f$ with real part $u$).

My interpretation of the question is that I want to prove that also:
$f(x+iy) = v(x,y) + i(-u(x,y))$ with
$$\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0.$$
I'm not sure how to do this if so.


We have that $f=u+iv$ is holomorphic and have to show that $g:=v-iu$ is holomorphic.

But this is easy, since $g=-if$.

  • $\begingroup$ So to show $a$ is a harmonic conjugate of $b$, then I need to show there exists a holomorphic function $g$ such that $g(x+iy):= b+ia$? $\endgroup$ – Twenty-six colours Aug 2 '17 at 9:22
  • $\begingroup$ Yes. This is to show. $\endgroup$ – Fred Aug 2 '17 at 9:28
  • $\begingroup$ @Fread's sir, I just read your answer, sir you said, we have to show $g$ is holomorphic, but as you said, $g=-if$ and we know complex linear combination of analytic function are analytic, so $g$ will be holomorphic! (Is am I correct?), Further, can we also show, $u$ is harmonic conjugate of $ -v$? Since,$h= if$ is holomorphic, hence I think $u$ is harmonic conjugate of $-v$ is also true!!(am I correct?) Sir please reply... $\endgroup$ – Akash Patalwanshi Apr 19 '18 at 13:06

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.