Using Cauchy's integral formula, what is $\int_\gamma \frac{p'(z)}{p(z)}dz$ Let $p(z)$ be a polynomial with degree $n$ and $R > 0$ such that $|z_0| < R$ for all zeroes $z_0$. Using Cauchy's integral formula, show that
$$\int_{|z| = R} \frac{p'(z)}{p(z)}dz = 2 \pi i n$$
I am not asking for a full solution to this problem, but more or less for a hint.
 A: Hint:
$$\frac{p'(z)}{p(z)} = \frac{d}{dz}\log p(z) = \frac{d}{dz}\left[\log A+\sum_{k=1}^n\log(z-z_k)\right] =\sum_{k=1}^n\frac{1}{z-z_k}$$
A: Let $p(z)$ be a polynomial of order $N$ with $p(z)=\sum_{n=0}^N a_nR^ne^{in\phi}$, where $a_N\ne 0$.  Then, we can write
$$\begin{align}
\oint_{|z|=R}\frac{p'(z)}{p(z)}\,dz&=\int_0^{2\pi} \frac{\sum_{n=0}^{N-1} (n+1)a_{n+1}R^{n}e^{in\phi}}{\sum_{n=0}^N a_nR^ne^{in\phi}}\,iRe^{i\phi}\,d\phi\\\\
&=i\int_0^{2\pi} \frac{\sum_{n=1}^{N} na_{n}R^{n}e^{in\phi}}{\sum_{n=0}^N a_nR^ne^{in\phi}}\,d\phi\tag1
\end{align}$$
Since all zeroes of $p(z)$ are contained in the circle $|z|=R$, Cauchy's Integral Theorem guarantees that we can take $R$ arbitrarily large.  Letting $R\to \infty$ in $(1)$ reveals
$$\begin{align}
\oint_{|z|=R}\frac{p'(z)}{p(z)}\,dz&=\lim_{R\to\infty}\oint_{|z|=R}\frac{p'(z)}{p(z)}\,dz\\\\
&=i\int_0^{2\pi} \lim_{R\to\infty} \left(\frac{\sum_{n=1}^Nna_nR^{n-N}e^{in\phi}}{\sum_{n=0}^N a_nR^{n-N}e^{in\phi}}\right)\,d\phi\\\\
&=i\int_0^{2\pi}\frac{Na_Ne^{iN\phi}}{a_Ne^{iN\phi}}\,d\phi\\\\
&=2\pi iN
\end{align}$$
as was to be shown!
