I have $P(\lambda) = (i\lambda)^m + O(\lambda^{m-1})$ a polynomial in $\lambda$, and $\Gamma$ a contour counterclockwise around the roots of $P$. I need to prove:
$$\frac{1}{2\pi}\int_{\Gamma}\frac{(i\lambda)^k}{P(\lambda)}d\lambda = \delta_{k,m-1} $$
For $k\leq m-1$. Now, I do some thing here that may be wrong... I deform $\Gamma$ to a disc of radius $R$ and make $R$ tend to infinity. For big $R$, $P(\lambda) \approx (i\lambda)^m$ and using Cauchy's differentiation formula I get what I wanted. But I don't know how to formalize this. Could somebody help me?
Thank you very much.
I think I have it... Any comment is welcome.
Deforming the curve is easy, using Cauchy Integral theorem.
Now, I want to prove that $\forall \epsilon \exists R_{\epsilon}$ such that uniformly on $\theta$ and on the coefficients of the polynomial, I have
$$|\frac{P(Re^{i\theta})}{R^me^{im(\frac{\pi}{2}+\theta)}}-1| < \epsilon$$
But I can divide and then I get $|O(R^{-1})|<\epsilon$ which can be done. (Same with the inverse). Thus for $\epsilon_n = \frac{1}{2^n}$ I get $R_n$ and
$$\delta_{m-1,k}(1-\epsilon_n)=\frac{(1 - \epsilon_n)}{2\pi}\int_{D(0,R_n)}\frac{1}{(i\lambda)^{m-k}}d\lambda\leq\frac{1}{2\pi}\int_{\Gamma}\frac{(i\lambda)^{k}}{P(\lambda)}d\lambda$$
$$\frac{1}{2\pi}\int_{\Gamma}\frac{(i\lambda)^{k}}{P(\lambda)}d\lambda \leq \frac{(1 + \epsilon_n)}{2\pi}\int_{D(0,R_n)}\frac{1}{(i\lambda)^{m-k}}d\lambda=\delta_{m-1,k}(1+\epsilon_n)$$