Is the function $V (z)=\int_{ \mathbb {T} }\frac {e^{it}+z}{e^{it}-z}d\mu (e^{it})$ analytic on the unit disc $|z|<1$? Let $\mathbb {T}$ be the unit circle in the complex plane and let $\mathbb{D}$ denote the open unit disc.
Let $\mu$ be a complex Borel measure on  $\mathbb {T}$ .
Is the function $V$ defined as $$V (z)=\int_{ \mathbb {T} }\frac {e^{it}+z}{e^{it}-z}d\mu (e^{it})\;\;  \;\;z\in \mathbb{D}  $$ analytic on $\mathbb{D}$? If yes, then How?
 A: $\bullet\;$ First show, for each fixed $t$, the function
$$
\frac {e^{it}+z}{e^{it}-z}
$$
is analytic on the open unit disk.
$\bullet\;$ Then find what is needed to verify this calculation: if $\gamma$ is a closed contour contained in the open unit disk, then
$$
\int_\gamma V (z)\;dz=
\int_{ \mathbb {T} }\int_\gamma \frac {e^{it}+z}{e^{it}-z}\;dz\;d\mu (e^{it}) = \int_{\mathbb T} 0\;d\mu(e^{it}) = 0.
$$
$\bullet\;$ Conclude that $V(z)$ is analytic.
A: Let 
$$F(z,t)= \sum_{n=-\infty}^{\infty} z^{|n|}e^{-i|n|t} =\sum_{n=0}^{\infty} z^{n}e^{-int}+\sum_{n=1}^{\infty} z^{n}e^{-int}  \\= \frac{1}{1-ze^{-it}}+ \frac{ze^{-it}}{1-ze^{-it}} =\frac {e^{it}+z}{e^{it}-z}$$
$z\mapsto F(z,t)$ is analytic on $\mathbb D$. Since $\mathbb T $ is compact, as  $\mu$ is a Borel measure  and hence finite on very compact sets in particular $\mu(\Bbb T)<\infty$. For fixed $z\in \Bbb D$ the map $t\mapsto F(z,t)$ is bounded on $\Bbb{T}$ and thus $V (z)\in L^1(\Bbb T)$
$$V (z)=\int_{ \mathbb {T} }\frac {e^{it}+z}{e^{it}-z}d\mu (e^{it}) = \sum_{n=-\infty}^{\infty} z^{|n|}\int_{ \mathbb {T} }e^{-i|n|t}d\mu (e^{it}) \\= \mu(\mathbb T) + \sum_{n=1}^{\infty} \left(2\int_{ \mathbb {T} }e^{-int}d\mu (e^{it})\right)z^{n}  \;\;  \;\;z\in \mathbb{D}$$
that is 
$$V (z)=\mu(\mathbb T) + \sum_{n=1}^{\infty}  a_nz^{n}  \;\;  \;\;z\in \mathbb{D}$$
with $$a_n  =\left(2\int_{ \mathbb {T} }e^{-int}d\mu (e^{it})\right) \qquad\text{and}\qquad \mu(\mathbb T) =\int_{ \mathbb {T} }d\mu (e^{it}) $$
