What I’d like to evaluate is the following contour integral:
$\displaystyle \lim_{R \rightarrow \infty} \int_{|z|=R}^{\ } \frac{2018z^{2018} + 2017z^{2017}}{z^{2018} + z^{2017} + 1} dz$,
So what I did is as follows:
Choose $R$ sufficiently large that inside every pole $z_i$ (except $z=\infty$) of $f(z)= \frac{2018z^{2018} + 2017z^{2017}}{z^{2018} + z^{2017} + 1} $ is contained inside the circle $|z|=R$.
And every pole $z_i$ of $f$ is a simple pole, so $Res[f(z)=\frac{g(z)}{h(z)} , z_i] = \frac{g(z_i)}{h’(z_i)}$ provided that $g(z_i) \ne 0$.(clearly which is the case.)
$\displaystyle Res \bigl [ \frac{g(z)}{h(z)} ; z_i \bigr ] = \frac{g(z_i)}{h’(z_i)} = \frac{2018{z_i}^{2018} + 2017{z_i}^{2017}}{2018{z_i}^{2017} + 2017{z_i}^{2016}}=z_i$.
Then the limit value of the above contour integral is
$\displaystyle 2\pi i \sum_{i=1}^{2018} Res \bigl [f(z) ; z_i\bigl ] = 2\pi i \sum_{i=1}^{2018} z_i$.
Considering $h(z)=z^{2018} + z^{2017}+1 = \prod_{i=1}^{2018} (z- z_i)$, $\sum_{i=1}^{2018} z_i = -1$,
Hence $\displaystyle \lim_{R \rightarrow \infty} \int_{|z|=R}^{\ } \frac{2018z^{2018} + 2017z^{2017}}{z^{2018} + z^{2017} + 1} dz = -2\pi i$.
Am I right? I will appreciate if you give me any comment, thank you in advance.