$\int_{|z|=1} \frac{z^4 + 1}{z^2(\overline{z} - a)(b - \overline{z})} dz$ Let there be numbers ,$$0 < |a| < 1 < |b|$$
calculate the integral :
$$\int_{|z|=1} \frac{z^4 + 1}{z^2(\overline{z} - a)(b - \overline{z})} dz$$
I will provide the solution for this problem and I need some explanation if anyone may help.
On the circle $|z|=1$, we note that $z\bar{z} = 1$, so $\bar{z} = \frac{1}{z}$.
Therefore, we have three points, $0$, $\frac{1}{a}$, and $\frac{1}{b}$. However, since $|a|<1$, we note that $z=\frac{1}{a}$ is not in the given domain,Therefore, we are left with two singular points, $z = \frac{1}{b}$ and $z = 0$.
So we calculate the residue for these points. They did not show the calculation, but they said that $\operatorname{Res}(f,0) = 0$ and $\operatorname{Res}(f,\frac{1}{b}) = \frac{1+b^4}{b^4(b-a)}$.
can someone please explain to me how to culculate such thing with $\bar{z}$ ?
solution:
$\int\limits_{|z|=1} \frac{z^4 +1}{(1-az)(bz-1)} \mathrm{d}z$
$\lim\limits_{z\to 0}\frac{\mathrm{d}}{\mathrm{d}z}(z^2f(z))$ = $\lim\limits_{z\to 0}\frac{\mathrm{d}}{\mathrm{d}z}(z^2\cdot\frac{z^4+1}{(1-az)(bz-1)})$
=$$\lim\limits_{z\to 0}\frac{-4abz^7+5az^6+5bz^6-6z^5+az^2+bz^2-2z}{\left(1-az\right)^2\left(bz-1\right)^2}$$=$0$
$\lim\limits_{z\to \frac{1}{b}}((z-\frac{1}{b})\cdot\frac{z^4+1}{(1-az)(bz-1)})$ =
$\frac{b^4+1}{b^3(b-a)}$

Regarding your calculation: There is no need to compute the residue at $z=0$, and the integral is $2 \pi i$ times the residue of $f$ at $z=1/b$.
For $|z| = 1$ is
$$
\frac{z^4 + 1}{z^2(\overline{z} - a)(b - \overline{z})}
= \frac{z^4 + 1}{( \overline{z}z - az)(bz - \overline{z} z)}
= \frac{z^4 + 1}{(1 - az)(bz - 1)}
$$
and the only singularity of $f(z)=\frac{z^4 + 1}{(1 - az)(bz - 1)}$ in the unit disk is a simple pole at $z=1/b$.
Therefore the integral is equal to
$$
 2 \pi i \operatorname{Res}(f,\frac{1}{b})
= 2 \pi i \lim_{z \to 1/b} (z-\frac 1b)\frac{z^4 + 1}{(1 - az)(bz - 1)} \\
= 2 \pi i \lim_{z \to 1/b} \frac{z^4+1}{b(1-az)} =  2 \pi i \frac{1+b^4}{b^4(b-a)}
$$
