On continuous extension of harmonic function I am asked to give a specific example of bounded harmonic function defined on $\{|z|<2\}\setminus[-1,1]$ which cannot be continuously extend to $\{|z|<2\}$.
It's well know that if a bounded harmonic function is defined on a punctuated disc with only single point deleted then it can be harmonically extended. I am trying to find some analytic function whose real part is bounded over $\{|z|<2\}\setminus[-1,1]$ while the imaginary part is not, but it does not come to me very quickly. Any comment will be greatly appreciated! Thank you!

Consider $T(z) = \frac{z-1}{z+1}$. $T$ is a Möbius transformation, hence a bijective map of the Riemann sphere. $T$ maps $\widehat{\mathbb{C}}\setminus [-1,\,1]$ onto $\mathbb{C}\setminus(-\infty,\,0]$. $T$ maps the disk $\{\lvert z\rvert < 2\}$ onto the exterior of the disk $\{\lvert w - \frac{5}{3}\rvert \leqslant \frac{4}{3}\}$. Therefore, using the principal branch of the logarithm,
$$f(z) = \log T(z)$$
is a holomorphic function on $\{\lvert z\rvert < 2\}\setminus [-1,\,1]$ whose imaginary part is bounded (by $\pi$) and whose real part is unbounded.
The idea behind it is that when you remove an interval from a domain, you can use a Möbius transformation to map that interval onto the negative real half-axis, and then you have a branch of the logarithm (and roots) defined on the image of the remainder of the domain. The logarithm cannot be continuously extended across any point of the negative real half-axis.
