Question about limit of a harmonic function

complex-analysisharmonic functionslimitsmaximum-principle

I am trying assignments of complex analysis and need help in this particular question.

Let $u(z)=\Im\frac{(1+z^2)^2}{(1-z^2)^2}$.

  1. Show that $u$ is harmonic in $U$, the unit disk centered at the origin.
  2. Show that $\lim\limits_{r\to 1} u(r e^{i\theta}) = 0$ for all $\theta$. Why doesn't this contradict the maximum modulus principle for harmonic functions?

For 1. I proved that it is harmonic by the definition that sum of partial derivatives wrt both $x$ and $y$ is $0$. But is there any other way to prove that it is harmonic as using the definition involves a lot of calculations?

For 2. I tried writing $z=r \sin\theta +i \cos\theta$ and putting $r = 1$ but I don't get zero; instead I get $\frac{i\sin\theta}{ 1+\cos\theta}$.

If I use the maximum modulus principle, I get LHS $=0$ and RHS $=\int_{0}^{2\pi} f(r e^{i\theta}$). I don't understand what contradiction one should expect if limit given above tends to $0$ and why there must be no contradiction?

I request you to kindly shed some light on this.

Best Answer

For $a)$; the real and the imaginary part of an holomorphic function are harmonic. Thus, it is enough to prove that $\frac{(1+z^2)^2}{(1-z^2)^2}$ on $U$.

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