A contour integral problem from trinity A function $\phi(z)$ is zero when $z=0$, and is real when $x$ is real, and is analytic when $|z| \leq 1$; if $f(x,y)$ is the coefficient of $i$ in $\phi(x+iy)$, prove that if $-1<x<1$,
$\int_{0}^{2 \pi} \frac{x\sin(\theta)}{1-2x\cos(\theta)+x^{2}}f(\cos(\theta),\sin(\theta))d\theta=\pi\phi(x)$.
when repacing $\cos(\theta)$ and $\sin(\theta)$ by $\frac{1}{2}(z+\frac{1}{z})$ and $\frac{1}{2}(z-\frac{1}{z})$,
$\int_{0}^{2 \pi} \frac{x\sin(\theta)}{1-2x\cos(\theta)+x^{2}}f(\cos(\theta),\sin(\theta))d\theta$=$\frac{1}{4i}\oint_{c}\frac{x(z-\frac{1}{z})}{(x-z)(xz-1)}(\phi(z)-\frac{|\phi(z)|^{2}}{\phi(z)})dz$,
$z=e^{i\theta}$, $c$ is the unit circle
then i don't know how to continue to get the final result.

Hints:


*

*Don't bother replacing $f$ with a complicated expression - the rest of the integral is real, so just take an $\mathrm{Im}$ outside and change $f\to\phi$.

*Notice where the poles of the integrand lie (remembering $\phi(0)=0$), in particular whether or not they lie within $C$ (remembering $|x|<1$).

*Use the residue theorem.

