Prove $\int_{0}^{2 \pi} \frac{x \sin (\theta)}{1 - 2 x \cos (\theta) + {x}^{2}} \mathcal{I}\phi(\theta)d \theta = \pi \phi (x)$ for $x \in (- 1 , 1)$. A function $\phi \left(z\right)$ is zero when $z = 0$, and is real when $z$ is real, and is analytic when $| z | \le 1$; if $f \left(x , y\right)$ is the imarginary component of $\phi \left(z\right)$, prove that if $x \in \left(- 1 , 1\right)$,
\begin{align}
 \int_{0}^{2 \pi} \frac{x \sin \left(\theta\right)}{1 - 2 x \cos \left(\theta\right) + {x}^{2}} f \left(\cos \left(\theta\right) , \sin \left(\theta\right)\right) d \theta = \pi \phi \left(x\right) \\
\end{align}
Attempt:
I define $z = {e}^{i \theta}$ and $\mathrm{dz} = \frac{d \theta}{i z}$ s.t.
\begin{align}
 & \int_{0}^{2 \pi} \frac{x \sin \left(\theta\right)}{1 - 2 x \cos \left(\theta\right) + {x}^{2}} f \left(\cos \left(\theta\right) , \sin \left(\theta\right)\right) d \theta & \\
 = & \int_{C} \frac{x \frac{z - {z}^{- 1}}{2 i}}{1 - x \left(z + {z}^{- 1}\right) + {x}^{2}} f \left(z\right) \frac{\mathrm{dz}}{i z} \\
 = & \int_{C} \frac{x \frac{z - {z}^{- 1}}{2 i}}{\left(1 - x z\right) \left(1 - x {z}^{- 1}\right)} f \left(z\right) \frac{\mathrm{dz}}{i z} \\
 = & \int_{C} \frac{x \frac{{z}^{2} - 1}{2 i}}{\left(1 - x z\right) \left(z - x\right)} f \left(z\right) \frac{\mathrm{dz}}{i z} \\
\end{align}
Given that the imaginary component of $\phi \left(z\right) = \frac{\phi \left(z\right) - \overline{\phi \left(z\right)}}{2 i}$ and the complex conjugate is not holomorphic I am not sure how to compute the integral with $f$. I had hoped that I could decompose $f$ into its real and imaginary parts to quote the linearity of the integral,  but I realized that I unfortunately do not know how to evalute the integral with the real component of $\phi \left(z\right)$. I feel that I shoudl be using one of the preconditions on $\phi \left(z\right)$, I was able to prove a restricted version of the problem when $\phi \left(z\right)$ is a linear function.
Anyway, I proceeded by evaluating the integral with $\phi \left(z\right)$ instead of $f \left(z\right)$.
\begin{align}
 & \int_{C} \frac{x \frac{{z}^{2} - 1}{2 i}}{\left(1 - x z\right) \left(z - x\right)} \phi \left(z\right) \frac{\mathrm{dz}}{i z} \\
 = & \frac{- 1}{2} \int_{C} \frac{x \left({z}^{2} - 1\right)}{\left(1 - x z\right) \left(z - x\right) z} \phi \left(z\right) \mathrm{dz} \\
 = & - \pi \left[{\text{Res}}_{z = 0} \frac{x \left({z}^{2} - 1\right)}{\left(1 - x z\right) \left(z - x\right) z} + {\text{Res}}_{z = x} \frac{x \left({z}^{2} - 1\right)}{\left(1 - x z\right) \left(z - x\right) z}\right] \\
 = & - \pi \left[1 - 1\right] = 0 \\
\end{align}
I believe therefore that the integral of the real and imaginary components of $\phi \left(z\right)$ should equal each other up to a possible constant of $i$ when $\phi \left(z\right)$ is analytic on $C$, but I am unsure how to best proceed from $f \left(z\right)$ to $\phi \left(z\right)$.

Note that
$$ f \left(\cos \theta , \sin \theta\right) 
= \Im \phi (e^{i\theta})
= \sum_{k=0}^\infty \frac{\phi^{(k)}(0)}{k!} \sin(k\theta)
$$
Then, integrate
\begin{align}
& \int_{0}^{2 \pi} \frac{x \sin \theta}{1 - 2 x \cos \theta + {x}^{2}} f \left(\cos \theta , \sin \theta\right) d \theta \\
=& \sum_{k=1}^\infty \frac{\phi^{(k)}(0)}{k!} 
\int_{0}^{2 \pi} \frac{x \sin \theta \sin(k\theta)}{1 - 2 x \cos \theta + {x}^{2}} \  d \theta \\
 =& \sum_{k=1}^\infty \frac{\phi^{(k)}(0)}{k!} 
\int_{0}^{2 \pi} \sin(k\theta)\sum_{j=1}^\infty x^j\sin(j\theta)\  d \theta \\
=& \sum_{k=1}^\infty \frac{\phi^{(k)}(0)}{k!}\ \pi x^k=\pi \phi(x)
\end{align}
