Prove $v$ is harmonic and $\lim_{r \uparrow 1} v(re^{i\theta}) = 0$

Question

Prove that if $v(z) = \mathrm{Im}[(\frac{1+z}{1-z})^2]$, then $v$ is harmonic on the unit disc and $\lim_{r \uparrow 1} v(re^{i\theta}) = 0$ for all $\theta \in [0,2\pi)$. Explain why this does not contradict the maximum principle.

Answer 1

Let $z(t)=\frac{ti}{2+ti}\in D=\{z:|z|<1\}$, where $t>0$, then $$v(z(t))= \mathrm{Im}[(\frac{1+z}{1-z})^2]= \mathrm{Im}[(1+ti)^2]=2t\to \infty \quad (t\to \infty).$$ Therefore $v(z)$ is not continuous and not bounded on the closed disc $\{|z|\le 1\}$. Thus the fact that $ v $ is harmonic on the unit disc and $\lim_{r \uparrow 1} v(re^{i\theta}) = 0$ for all $\theta \in [0,2\pi)$ does not contradict the maximum principle.

Ofcourse if $v(z)$ is harmonic in $D$, continous on the closed disc $\{|z|\le 1\}$ and $\lim_{r \uparrow 1} v(re^{i\theta}) = 0$ for all $\theta \in [0,2\pi)$, then $v(z)=0$ by the maximum principle.