So, I would like to bound $\int_{C_R} \frac{1}{P(z)} dz$ where $C_R$ is the circle radius R centred at the origin, and $P(z)$ is a polynomial of degree $N=0,1,...,n$ i also want to deduce for what values of N the integral will disappear as $R\rightarrow \infty$
My solution is as follows but i don't know if my argument is rigorous enough.
So in the proof of fundamental theorem of algebra we showed in class that $|\frac{1}{P(z)}|< \frac{2}{|a_n|R^n}$ , parametrised the circle: $\gamma(t)=Re^{it}$ for $t\in [0,2\pi]$, $\gamma'(t)=Rie^{it}$ and let $f(z)=\frac{1}{p(z)}$ then the integral goes as follows:
$$\int_{c_R} f(z)dz < \int_{0}^{2\pi} \frac{2iRe^{it}}{|a_n|R^n}dt = \frac{2i}{|a_n|R^{n-1}} [\frac{e^{it}}{i}]^{2\pi}_{0}$$ which goes to 0 for all values of R.
Which doesn't seem quite right, have I chose the wrong bound? the integral i defined also goes to 0, but I'm thinking for $n=0,1$. Could i say that, on $C_R$, $f(z) < R$ so that would mean q factor of R out and instead of getting $0$ i'd just get the bound $\frac{2i}{|a_n|R^{n-1}}$ which wouldn't go to $0$ for $n=0,1,2$ as $R\rightarrow \infty$
This is a homework question so please don't post a solution, suggestions and identification of errors would be a great help! Thank you!