Proving the function $g$, defined by an integral, is holomorphic 
Is the function $g$ (defined towards the end of this question) holomorphic?

Let $h:\partial B_1(0)\rightarrow\mathbb{R}$ be continuous. We see that $f:B_1(0)\rightarrow\mathbb{C}$ defined by
$$f(re^{i\varphi}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1+re^{i(\varphi-\theta)}}{1-re^{i(\varphi-\theta)}}h(e^{i\theta})\mathrm{d}\theta$$
is holomorphic, because
\begin{align}
f(re^{i\varphi}) &= \frac{1}{2\pi}\int_{0}^{2\pi}\frac{e^{i\theta}+re^{i\varphi}}{e^{i\theta}-re^{i\varphi}}h(e^{i\theta})\mathrm{d}\theta\\
f(z) &= \frac{1}{2\pi}\int_{0}^{2\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}h(e^{i\theta})\mathrm{d}\theta\\
&= \frac{1}{2\pi}\int_{0}^{2\pi}\frac{e^{i\theta}h(e^{i\theta})}{e^{i\theta}-z}\mathrm{d}\theta + \frac{1}{2\pi}\int_{0}^{2\pi}\frac{zh(e^{i\theta})}{e^{i\theta}-z}\mathrm{d}\theta\\
&= \frac{1}{2\pi i}\int_{\partial B_1(0)}\frac{h(w)}{w-z}\mathrm{d}w + \frac{z}{2\pi i}\int_{\partial B_1(0)}\frac{\frac{h(w)}{w}}{w-z}\mathrm{d}w\\
\end{align}
Thus $f(z) = s(z) + zt(z)$ where $s, t$ are holomorphic, so $f$ is holomorphic.
I have come across a question asking me to prove that $u:B_1(0)\rightarrow \mathbb{R}$ defined by
$$u(re^{i\varphi}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{|e^{i\theta} - re^{i\varphi}|}h(e^{i\theta})\mathrm{d}\theta$$
is harmonic. My approach was to attempt to re-write it as the real part of a holomorphic function, similar to what we have above. However, we see that the denominator of the Poisson Kernel look-alike does not have a "square". If it was squared, the function $u$ would be precisely the real part of the function $f$ I have described above.
Instead, because the denominator is not squared, we have $u = \Re(g)$ where $g$ is defined by
$$g(re^{i\varphi}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1+re^{i(\varphi-\theta)}}{1-re^{i(\varphi-\theta)}}h(e^{i\theta})|e^{i\theta}-re^{i\varphi}|\mathrm{d}\theta$$
I've played around with it, but I can't find a way of showing that $g$ is holomorphic. (I don't even know if it is, because anything involving complex norms often isn't holomorphic.) Is $g$ holomorphic? Any help will be appreciated!
 A: tl;dr the question had a typo.

The function $u:B_1(0)\rightarrow \mathbb{R}$ defined by
$$u(re^{i\varphi}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{|e^{i\theta} - re^{i\varphi}|}h(e^{i\theta})\mathrm{d}\theta$$
Is not generally harmonic, and so the corresponding function $g:B_1(0)\rightarrow \mathbb{C}$ defined by
$$g(re^{i\varphi}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1+re^{i(\varphi-\theta)}}{1-re^{i(\varphi-\theta)}}h(e^{i\theta})|e^{i\theta}-re^{i\varphi}|\mathrm{d}\theta$$
is not generally holomorphic.
Proof
Let $h:\partial B_1(0)\rightarrow\mathbb{R}$ be continuous. We know that the Dirichlet problem on the unit ball has a unique solution, given by
$$\hat u(re^{i\varphi}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{|e^{i\theta} - re^{i\varphi}|^2}h(e^{i\theta})\mathrm{d}\theta$$
However, using the fact that $\lim_{r\nearrow 1}\frac{1-r^2}{|e^{i\theta} - re^{i\varphi}|} = 0$, we can use elementary inequalities to show that $\overline u(x):= \cases{u(x),\ x\in B_1(0)\\ h(x),\ x\in \partial B_1(0)}$ is continuous on the closed ball, and it clearly agrees with $h$ on the ball. Hence if $u$ is harmonic on the ball, it must be equal to $\hat u$ using uniqueness of the solution to the dirichlet problem.
Now by simply setting $h$ to be a constant function, we see that the two integrals do not agree, so $u$ is not harmonic. It follows that $g$ is not holomorphic (although the harmonicity of $u$ was the property of interest).
