- Prove that the function u defined by
\[ u (x, y) = \operatorname{Re} \left(\frac {i + z}{i - z}\right) \quad \text { and } \quad u (0, 1) = 0 \]
is harmonic in the unit disc and vanishes on its boundary. Note that u is not bounded in D.
Solution
Let \[ F(z)=\frac{i+z}{i-z}. \] Its only pole is at \(z=i\), which lies on \(\partial\mathbb D\), so \(F\) is holomorphic in \(\mathbb D\). Hence \[ u(z)=\Re F(z) \] is harmonic in \(\mathbb D\).
If \(|z|=1\) and \(z\ne i\), then \(\bar z=1/z\). A direct computation gives \[ \overline{F(z)} =\frac{-i+\bar z}{-i-\bar z} =\frac{-iz+1}{-iz-1} =-\frac{i+z}{i-z} =-F(z). \] Thus \(F(z)\) is purely imaginary and \[ u(z)=\Re F(z)=0 \] on \(\partial\mathbb D\setminus\{i\}\). With the stipulated boundary value \(u(i)=u(0,1)=0\), the boundary values vanish everywhere.
The function is not bounded in the disc. Along the radius \(z=ri\), \(0<r<1\), \[ u(ri)=\Re\frac{i+ri}{i-ri}=\frac{1+r}{1-r}\longrightarrow+\infty \qquad(r\uparrow1). \] So \(u\) is harmonic in \(\mathbb D\), has zero boundary values, but is unbounded.