I have to compute this integral and I don't have any idea how to get further on:
$$\frac{1}{2 \pi i} \int_{\mid z \mid = 1} \frac{6z^{98}}{23z^{99}-2z^{81}+z^4-7}dz$$
I tried Rouché to maybe exclude some singularities from $B(0,1)=\{ z \in \mathbb{C} \mid \, |z| <1\}$ and then apply the residue theorem with no success. Thanks for help !
My Rouché try:
$f(z) := 23z^{99}-2z^{81}+z^4-7 ~~~~~~g(z) := 23 z^{99}$
$ \Rightarrow \forall \,z$ with $ |z| = 1 :~~|f(z)-g(z)|=|-2z^{81}+z^4-7| \leq 2+1+7 = 10 <23=|g(z)| \leq |g(z)|+|f(z)| $
$ \Rightarrow$ All zeroes of $f$ lie in $B(0,1)$
The denominator has no zeros in $\{z:\lvert z\rvert \geqslant 1\}$, so we can use Cauchy's integral theorem to shift the contour to a large circle,
$$\int_{\lvert z\rvert = 1} \frac{6z^{98}}{23z^{99} -2z^{81} + z^4 -7}\,dz = \int_{\lvert z\rvert = R} \frac{6z^{98}}{23z^{99} -2z^{81} + z^4 -7}\,dz$$
for any $R \geqslant 1$. We can easily compute the limit
$$\lim_{R\to\infty} \frac{1}{2\pi i}\int_{\lvert z\rvert = R} \frac{6z^{98}}{23z^{99} -2z^{81} + z^4 -7}\,dz$$
with the parametrisation $z = R e^{i\varphi}$.