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Suppose $u$ and $v$ are real valued functions on $\mathbb{C}$. Show that if $v$ is a harmonic conjugate for $u$, then -$u$ is a harmonic conjugate for $v$.

I know I have to use cauchy reumann here. Not sure how to get started. Any hints or help will be greatly appreciate. Thanks in advance.

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    $\begingroup$ What does it mean that one function is a harmonic conjugate of the other? $\endgroup$ – Daniel Fischer Apr 8 '14 at 21:37
  • $\begingroup$ @DanielFischer When there is an analytic function $f$ such that $u$ and $v$ are related by $f=u+iv$, then $u$ and $v$ are harmonic conjugates. $\endgroup$ – User69127 Apr 8 '14 at 22:00
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    $\begingroup$ And $u+iv$ is analytic if and only if $v-iu = (-i)(u+iv)$ is. $\endgroup$ – Daniel Fischer Apr 8 '14 at 22:04
  • $\begingroup$ Sorry for revisited such as old question, but I was just about to post this question until I saw the question had already been posted. I'm wondering, why is $u+iv$ analytic iff $v-iu$ is analytic? $\endgroup$ – user557493 Aug 14 '18 at 10:05
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The easiest way to see this is to realize that if $f(z) = u+iv$ is analytic, then so is $-if(z) = v - iu$.

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  • $\begingroup$ So then $-if(z)=v+(-u)i$ and that's where the $-u$ comes from right? $\endgroup$ – User69127 Apr 8 '14 at 22:02
  • $\begingroup$ Yes, that is where the $-u$ comes from. $\endgroup$ – Stephen Montgomery-Smith Apr 8 '14 at 22:04
  • $\begingroup$ If $f(z)$ is analytic, why is $-if(z)$ also analytic? $\endgroup$ – user557493 Aug 14 '18 at 10:06
  • $\begingroup$ If you multiply an analytic function by a complex number (like $-i$), it remains analytic. Because $d/dz(af(z)) = a f'(z)$. $\endgroup$ – Stephen Montgomery-Smith Aug 14 '18 at 12:32
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Just note that $u+iv$ analytic $\implies$ $(-i)(u+iv)=v-iu$ analytic.

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  • $\begingroup$ All you have to do is to show that if $v$ is the real part of an analytic function, then $-u$ is the imaginary part, right? That's another way of saying $v-iu$ needs to be analytic. You are given that $f= u+iv$ is analytic, and the function $g:z\mapsto -iz$ is analytic, so $v-iu=g\circ f$ is analytic. $\endgroup$ – MPW Apr 8 '14 at 21:49
  • $\begingroup$ Ah i see. So then $-if(z)=v+(-u)i$ and that's where the $-u$ comes from right? $\endgroup$ – User69127 Apr 8 '14 at 22:02
  • $\begingroup$ Yes indeedy!${}{}{}$ $\endgroup$ – MPW Apr 9 '14 at 0:51

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