Calculate the integral:
$$I = {1 \over {2\pi i}}\int_r {{{\omega (t) \cdot \overline {\varphi(t)} } \over {t - {z_0}}}} dt$$ where $r$ is a curve on the positive unit circle, ${{z_0}}$ is a point outside the unit circle.
${\omega (z)}$ has a first-order pole ${{z_k}}$ inside the circle and is analytical on the unit cicle.
$\varphi (z)$ is given by the following form: $$\varphi (z){\rm{ = }}{{{a_1}} \over z} + {{{a_2}} \over {{z^2}}} + \cdots + {{{a_n}} \over {{z^n}}},\quad z \ge 1$$ $\varphi (z)$ is analytic outside the unit circle. The complex conjugate of $\varphi (z)$ is as follows: $$\overline {\varphi (z)} {\rm{ = }}{{{{\bar a}_1}} \over {\bar z}} + {{{{\bar a}_2}} \over {{{\bar z}^{\rm{2}}}}} + \cdots + {{{{\bar a}_n}} \over {{{\bar z}^n}}}$$ $\overline {\varphi (z)} $ is an analytic function inside the unit circle. Since ${{z_0}}$ is outside the unit circle and ${z_k}$ is inside the unit circle, according to the residue theorem, we have:
$$\displaylines{ I = {1 \over {2\pi i}}\int_r {{{\omega (z) \cdot \overline {\varphi (z)} } \over {z - {z_0}}}} dz \cr {\rm{ = }}{\mathop{\rm Re}\nolimits} {\rm{s}}({{\omega (z) \cdot \overline {\varphi (z)} } \over {z - {z_0}}},z = {z_k}) \cr = \mathop {\lim }\limits_{z \to {z_k}} ((z - {z_k}) \cdot {{\omega (z) \cdot \overline {\varphi (z)} } \over {z - {z_0}}}) \cr = {{\overline {\varphi ({z_k})} } \over {{z_k} - {z_0}}} \cdot \mathop {\lim }\limits_{z \to {z_k}} ((z - {z_k})\omega (z)) \cr} $$ In the calculation of ${\overline {\varphi ({z})} }$ , I think that it should be computed by substituting $z=z_k$. However, in the paper, it is considered as $$\overline {\varphi ({1 \over {{{\bar z}_k}}})} $$ This has left me feeling confused. I look forward to your response or assistance.
In general $\overline{\varphi}$ is not analytic inside the unit circle. Therefore, the residue theorem is not directly applicable. Instead, you can use that $\overline{z}=1/z$ on $r$, thus you can substitute $\overline{\varphi(z)}$ by $\varphi(1/z)=\overline{a_1}z+\dots+\overline{a_n}z^n=:\psi(z)$ in the integral. The latter function is a polynomial and therefore analytic, hence the residue theorem is applicable. Then, you get $\psi(z_k)$ in your computation which coincides with $\overline{\varphi(1/\overline{z_k})}$ as stated in the paper.