Prove $\int_{|z|=1}\frac{P_n(z)}{z^{n+1}(z-a)}dz=0$ I've been solving problems from my Complex Analysis course, and I want to make sure if what I think what I think may be the path to solution is correct. The problem says:

Prove that $$\int_{|z|=r}\frac{P_n(z)}{z^{n+1}(z-a)}dz=0$$ if $P_n$ is a
polynomial of degree less or equal to $n$, and $a\in D(0,r)$.

The notation $D(0,r)$ refers to the open disk with center $0$ and radius $r$. My attemp:
Given the fact that $P_n$ is a polynomial of degree at much $n$, I've considered using Cauchy's Formula for derivatives, since this implies $(P_n(z))^{(n)}=n!$ (and it seemed to work well with Cauchy's integral formula, but now I doubt it's usefulness),  but the problem I have is that I'm unsure if this case verifies the hypothesis of the Formula. First (and blindly assuming I am able to use this formula here) I tried to use as my $f(z)$ in Cauchy's Formula $P_n(z)/(z-a)$, doing this
$$\int_{|z|=r}\frac{P_n(z)}{z^{n+1}(z-a)}dz=\int_{|z|=r}\frac{P_n(z)/(z-a)}{z^{n+1}}dz=\frac{2\pi i}{n!}\text{Ind}_{|z|=r}(0)\left(\frac{P_n}{z-a}\right)^{(n)}(0),$$
but clearly it did not work since the result is not $0$ ($\text{Ind}(0)=1$) and I believe this formula can't be used here because $P_n(z)/(z-a)$ is not holomorphic in $z=a$.
Trying the other way, to be said, using as $f(z) $ the function $P_n(z)/z^{n+1}$ does not seem useful either, and the same problem appears since it's not holomorphic in $z=0$.
Could you give me any hint on how to reorganize my expression in order to correctly use Cauchy? Any help or hint will be appreciated, thanks in advance.
 A: Denote $I(r) = \int_{|z|=r}\frac{P_n(z)}{z^{n+1}(z-a)}dz$ for $r > a$. The crucial fact is that the integral does not depend on $r$, so that
$$
 I(r) =  \lim_{\rho \to \infty} I(\rho) = 0
$$
because of the limit on the degree on $P$.
The independence on $r$ follows from the residue theorem
$$ I(r) = 2 \pi i \bigl(\operatorname{Res}(f, 0) + \operatorname{Res}(f, a)\bigr)
$$
with $f(z) = \frac{P_n(z)}{z^{n+1}(z-a)}$, but also from a generalized version of Cauchy's integral theorem: If $\gamma_r$ denotes the circle with radius $r$ (in mathematically positive orientation) then
$$
  \int_{\gamma_{r_1}} f(z) \, dz =  \int_{\gamma_{r_2}} f(z) \, dz 
$$
for $a < r_1 < r_2$ because $f$ is holomorphic in a neighbourhood of the annulus $r_1 \le |z| \le r_2$, and
$$
 \operatorname{Ind}(\gamma_{r_1} - \gamma_{r_2}, z) = 0
$$
for all $z$ with $r_1 < |z| < r_2$.
Yet another option is to calculate the integral using the residue at infinity:
$$ 
I(r) = -2\pi i \operatorname{Res}(f, \infty) = 2\pi i \operatorname{Res}\left( \frac 1z f(\frac 1z)\right)
= 2\pi i \operatorname{Res}\left( \frac{z^{n+1}P(1/z)}{1-az^2}\right) = 0
$$
because $\frac{z^{n+1}P(1/z)}{1-az^2}$ has a removable singularity at $z=0$. Of course this is nothing but the result of substituting $z=1/w$ in the original integral.
