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The problem at hand is this:

letting $u, v : \mathbb{R}^2 \rightarrow \mathbb{R}$ be two functions, assume $v$ is the harmonic conjugate of $u$ on $\mathbb{R}^2$. Show that $uv$ is the imaginary part of an entire function and conclude that $uv$ is harmonic.

This is my take on the question, that i'm not very sure of:

$u$ and $v$ are harmonic conjugates, therefore a function $g = u + iv$ is holomorphic and we have the Cauchy Reimann equations (1):

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

let $uv$ be a the imaginary part of a function $f$, and $a$ be its real part. So we have $f = a + iuv$ which should be holomorphic. Therefore, it should satisfy the Cauchy Reimann equations (2):

$$\frac{\partial a}{\partial x} = \frac{\partial (uv)}{\partial y} = v\frac{\partial u}{\partial y} + u \frac{\partial v}{\partial y}=v\frac{\partial u}{\partial y} + u\frac{\partial u}{\partial x}$$ $$\frac{\partial a}{\partial y} = -\frac{\partial(uv)}{\partial x} = - v\frac{\partial u}{\partial x} -u\frac{\partial v}{\partial x} = -v\frac{\partial u}{\partial x} + u\frac{\partial u}{\partial y}$$

Where I have replaced equations (1) in (2) to get all the derivatives is terms of u.

To see if this is satisfied I computed the laplacian of $a$, and making all the developments I got $\frac{\partial^2 a}{\partial x^2} + \frac{\partial^2 a}{\partial y^2} = 0$.

This is the part I'm not sure of, I concluded that $a$ is therefore harmonic and that the Cauchy reimann equations are indeed satisfied, therefore $f$ is indeed holomorphic over the whole plane (entire), thus $uv$ is harmonic.

Is this work remotely okay? Any help or comment is much appreciated.

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    $\begingroup$ Can you express the real and imaginary parts of $g(z)^2$? $\endgroup$ – Hagen von Eitzen Feb 12 '17 at 12:46
  • $\begingroup$ I see..just needed a bit of intuition. Thank you $\endgroup$ – user368063 Feb 12 '17 at 12:56
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I think you tried the hard way. I am not sure either about your procedure. I think it's ok, but I try another way.

You are dealing with the composition of holomorphic functions $f(g(x+iy))$. In your case $g(x+iy)=u(x,y)+iv(x,y)$, holomorphic by hypothesis, and $f(x+iy)=a(x,y)+ixy$ with $a$ making $f$ holomorphic. First of all, the composition of holomorfic functions is holomorfic (here). Next try to impose the Cauchy-Riemann equations to $f$ and, if succeed, you have two holomorphic functions:

$$\frac{\partial a}{\partial x} = \frac{\partial (xy)}{\partial y} =x $$

$$a=\frac{x^2}{2}+f(y)$$

$$\frac{\partial a}{\partial y} = -\frac{\partial(xy)}{\partial x} =y$$

$$a=-\frac{y^2}{2}+g(x);\;g(x)=\frac{x^2}{2}$$

Succeed! so,

$$a(x,y)=(1/2)(x^2-y^2);\;f(x+iy)=(1/2)(x^2-y^2)+ixy$$

And

$$f(u+iv)=(1/2)(u^2-v^2)+iuv;\; f(u+iv)=(1/2)(u+iv)^2$$

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