I am doing a course on complex analysis, and a lot of the initial exercises involve showing that a certain function $f:R^{2} \rightarrow R$ is harmonic. The general brute force method of evaluating the Laplacian is extremely tedious at times, for example for the following function given in polar coordinates, I've been searching for a smart way to do the job, in vain: $f(r,\theta)=\frac{1-r^{2}}{1+2rcos\theta+r^{2}}$.

What I tried with this particular problem was trying to find a harmonic conjugate and then try and use the Cauchy Riemann equations to form a holomorphic function, which would then imply (on nice domains) that the real, hence f, and imaginary parts of the function are harmonic. However the integration of the partial derivatives seems to me to be pretty tedious to evaluate. Is there any other "intelligent" way to approach the problem?

P.S.: The Cauchy Riemann equations for a function $f(z)=U(r,\theta)+iV(r,\theta)$ given in polar coordinates are given as: $$\begin{aligned} U_{r}=\frac{1}{r}V_{\theta}\\ U_{\theta}=-rV_{r} \end{aligned}$$


1 Answer 1


$$\begin{split} \frac{1-r^{2}}{1+2r\cos\theta+r^{2}} &=\frac{1-r^2}{1 +2r \cos\theta + r^2\cos^2 \theta + r^2 \sin^2\theta} \\ &= \frac{1-r^2}{(1+r\cos\theta)^2 - (ir\sin\theta)^2}\\ &= \frac{1-z\bar z}{(1+z)(1+\bar z)}\qquad \qquad (z = re^{i\theta})\\ &= \frac 12 \left(\frac{1-z}{1+z} + \frac{1-\bar z}{1+\bar z}\right)\\ &=\text{Re} \left(\frac{1-z}{1+z}\right) \end{split}$$

(Of course I know the answer and make this up; I copied/remembered it since it is a variant of the famous Poisson Kernel)

  • $\begingroup$ Thanks! This is really a nice method for this particular problem. It's really a case by case approach that we need to follow, is it? $\endgroup$
    – Arkya
    Aug 9, 2016 at 16:59
  • $\begingroup$ Yes, @ArkyaChatterjee If the expression is given in polar coordinates, one might try to make up the term $r\cos\theta \pm ir\sin\theta$ since that corresponds to $z, \bar z$. But other than that I do not know of a good general method to deal with a general expression, expect finding the harmonic conjugate by differentiating and then integrating. $\endgroup$
    – user99914
    Aug 9, 2016 at 17:03
  • $\begingroup$ Yeah, I see why that would be the case in general $\endgroup$
    – Arkya
    Aug 9, 2016 at 17:39

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