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I'm reading the post Bounded harmonic function is constant .I've some querry related to fiorerb's answer. enter image description here

I'm not getting how to prove,"if $u$ is harmonic then $f$ is analytic.

Related to this problem,i've found a theorem in Brown and Churchill's Complex analysis .Which is like

"A function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $D$ iff $v$ is a harmonic conjugate of $u$"

from this theorem,if $v$ is harmonic conjugate of $u$ only then $f$ is analytic.But, it is nowhere given that $v $ is a harmonic conjugate of $u$.So, how can $f$ be analytic?

Styles
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    For every real-valued harmonic function $u$ defined in the complex plane $\mathbb{C}$ you can construct a so called "harmonic conjugate", a harmonic function $v$ defined in the plane which makes $f = u + iv$ analytic. – Bartosz Malman Oct 21 '17 at 11:43
  • @Malman:got it!!!!!!!!!!!thanks a lot!! – Styles Oct 21 '17 at 12:08
  • I'm curious. C. Falcon's deleted answer is effectively the same thing, only explained in more detail. Yet you rejected his answer, but are satisfied with Malman's? (FYI - when your rep is high enough, you can see deleted posts). – Paul Sinclair Oct 21 '17 at 15:39
  • @PaulSinclair:I did'nt rejected C.Falcon's post.I had some querry related to C.Falcon's post actually,he'd made use of Poincare's lemma,about which i've no knowledge.I was just trying to discuss,but at the very instant he deleted his post. – Styles Oct 21 '17 at 16:19
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    Okay. As I said I was just curious about an apparent disconnect. The extra content of his post, including the reference to Poincare's lemma, was about actually demonstrating that, as Malman put it, "you can construct a so called 'harmonic conjugate'". Otherwise, he was just saying the same thing. – Paul Sinclair Oct 21 '17 at 17:17
  • @PaulSinclair:How can i see deleted posts??? – Styles Oct 21 '17 at 17:27
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    @PKStyles - by earning another 9129 reputation points. – Paul Sinclair Oct 21 '17 at 20:52

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