Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $(1)$.

share|cite|improve this answer
Thanks Paul, this is very clear. – Dedede May 8 '12 at 18:35

The components of your $*du$ make up $2\nabla(\arg)$, and it is well known that $\arg$ has no continuous real representative on ${\mathbb C}^*$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.