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!
 A: 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.
A: Hint: Consider the real or imaginary part of $\sqrt{z^2-1}$
This can be well-defined on $\mathbb{C}\setminus[-1,1]$ by computing
$$
\frac12\log(z^2-1)=\frac12\int_{\sqrt2}^z\left(\frac1{w-1}+\frac1{w+1}\right)\,\mathrm{d}w
$$
along any path that doesn't cross $[-1,1]$. The fact that the total residue at $-1$ and $1$ is $1$, means that exponentiating gives a well-defined function. Exponentiating gives $\sqrt{z^2-1}$. Take the real or imaginary part.
A: Consider the holomorphic function
$$f:\mathbb{D}^*\to\mathbb{C}$$
given by
$$f(z)=z+\frac{1}{z}\;.$$
It can be checked that $f(\mathbb{D}^*)=\mathbb{C}\setminus[-2,2]$ and that $f$ is injective (from $\mathbb{D}^*$).
Now, let $g$ be $f^{-1}$. The function $h(z)=g(2z)$ is defined on $\mathbb{C}\setminus[-1,1]$, $h$ is bounded because the image is contained in $\mathbb{D}^*$ and cannot be extended because otherwise it would be constant by Liouville.
