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Consider $u(z)=\ln(|z|^2)=\ln(x^2+y^2)$. I know that $u$ does not have a harmonic conjugate from $\mathbb{C}\setminus\{0\}\to\mathbb{R}$ but playing around with partial derivatives and integrating around the unit circle.

However, I know that a function $u$ has a harmonic conjugate if and only if its conjugate differential $*du$ is exact. This is defined as $*du=-\frac{\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy$.

I calculate this to be $$ *du=\frac{-2y}{x^2+y^2}dx+\frac{2x}{x^2+y^2}dy $$ so I would assume this is not exact. Is there a way to see that easily? Is this how the criterion for existence or nonexistence of a harmonic conjugate is usually applied in terms of the conjugate differential? Thanks.

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up vote 2 down vote accepted

If $*du$ is exact, i.e. if $$\frac{-2y}{x^2+y^2}dx+\frac{2x}{x^2+y^2}dy=df$$ for some function $f$, we would have $$\tag{1}\int_{C}\left(\frac{-2y}{x^2+y^2}dx+\frac{2x}{x^2+y^2}dy\right)=\int_Cdf=0$$ for $C$ being any closed curve. Take $C$ to be the circle centered at the origin with radius $r>0$, i.e. $C$ is given by $(x,y)=(r\cos\theta,r\sin\theta)$, $\theta\in [0,2\pi]$. Then $$\int_{C}\left(\frac{-2y}{x^2+y^2}dx+\frac{2x}{x^2+y^2}dy\right)=\int_0^{2\pi}\left(\frac{-2r\sin\theta}{r^2}d(r\cos\theta)+\frac{2r\cos\theta}{r^2}d(r\sin\theta)\right)$$ $$=\int_0^{2\pi}2(\sin^2\theta+\cos^2\theta)d\theta=4\pi,$$ which contradicts to $(1)$.

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Thanks Paul, this is very clear. –  Dedede May 8 '12 at 18:35
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The components of your $*du$ make up $2\nabla(\arg)$, and it is well known that $\arg$ has no continuous real representative on ${\mathbb C}^*$.

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