Integral of Reciprocal Of Polynomial Always $0$? 
Prove that $$\int_{|z|=r}\dfrac{1}{P(z)}=0$$ where $P(z)$ is a polynomial with degree $2$ or higher, and all the zeros of $P(z)$ are contained in $|z|<r$.

 A: Here's an integration path that should prove helpfull. Let $\gamma_{\varphi,R}$ be the path


*

*(Part $\gamma^1_{\varphi}$). Starts at $z_0=re^{i\varphi}$ and goes along the circle $|z|=r$ counter-clockwise until it reaches $z_1=re^{-i\varphi}$.

*(Part $\gamma^2_{\varphi,R}$) Then proceeds from $z_1$ parallel to the real axis until it reaches $z_2 = Re^{-i\theta}$

*(Part $\gamma^3_{\varphi,R}$) Continues by going along the circle $|z|=R$ clockwise until it reaches $z_3 = Re^{i\theta}$

*(Part $\gamma^4_{\varphi,R}$) Finally goes back to $z_0$ parallel to the real axis.
Then show that 


*

*$\oint_{\gamma_{\varphi,R}} \frac{1}{P(z)} \,dz = 0$. Hint: Does the integrand have any poles within the area enclosed by that path?

*$\int_{\gamma^1_{\varphi}}  \frac{1}{P(z)} \,dz \to \oint_{|z|=r} \frac{1}{P(z)} \,dz$ as $\varphi \to 0$. Use that the path $\gamma^1_{\varphi}$ converges to the path $|z|=r$.

*$\int_{\gamma^3_{\varphi,R}} \frac{1}{P(z)} \,dz \to 0$ as $R \to \infty$. You'll need that $P$ has degree two or higher for that.

*$\int_{\gamma^2_{\varphi,R}} \frac{1}{P(z)} + \int_{\gamma^4_{\varphi,R}} \frac{1}{P(z)}\,dz \to 0$ as $\varphi \to 0$. Use that $\gamma^1_{\varphi,R}$ and $\gamma^4_{\varphi,R}$ converge towards the same line $[r,R]$, but run through it in different directions.
Putting it all together, the situation is that if you let $\varphi \to 0$ and $R \to \infty$ then $$
  0 = \oint_{\gamma_{\varphi,R}} \frac{1}{P(z)} \,dz =
  \underbrace{\int_{\gamma^1_\varphi} \frac{1}{P(z)} \,dz}_{\to \oint_{|z|=r} \frac{1}{P(z)} \,dz} +
  \underbrace{\int_{\gamma^3_{\varphi,R}} \frac{1}{P(z)} \,dz}_{\to 0} +
  \underbrace{\int_{\gamma^2_{\varphi,R}} \frac{1}{P(z)} \,dz +
  \int_{\gamma^4_{\varphi,R}} \frac{1}{P(z)} \,dz}_{\to 0}
$$
from which you can then conclude that indeed $$
  \oint_{|z|=r} \frac{1}{P(z)} \,dz = 0
$$
