I want to do the following:
If p is a non-constant polynomial and R>0 such that all roots of p lie inside $B_R(0)$, compute $\int_{B_R(0)} \frac{p’(z)}{p(z)}dz$
I believe that the answer is $2\pi i n$, where n is the number of roots of the polynomial. I got this answer using Residue Theorem.
However, I would like to evaluate this integral without using this theorem. Does anyone know of such a way to evaluate this integral?
My second thought was, since the degree of the numerator is strictly less than the degree of the denominator, to use partial fraction decomposition and assume that the zeroes were distinct. Then the integral could be split up into n many integrals over the ball centered around each zero with sufficiently small radius to assume they don’t intersect. Then, each of these n mant integrals has value $2\pi i$, so the integral is $2\pi i n$.
The fundamental theorem of algebra implies that the considered $n$-degree polynomial can be rewritten as $p(z) = a(z-z_1)\cdots(z-z_n)$, where $a$ is a complex coefficient and $z_1,\ldots,z_n$ its roots. Then, the integrand takes the following form : $$ \frac{p'(z)}{p(z)} = \frac{\mathrm{d}}{\mathrm{d}z} \ln p(z) = \frac{\mathrm{d}}{\mathrm{d}z} \left(\ln a + \sum_{k=1}^n \ln(z-z_k)\right) = \sum_{k=1}^n \frac{1}{z-z_k}, $$ hence
$$ \oint_C \frac{p'(z)}{p(z)} \mathrm{d}z = \sum_{k=1}^n \oint_C \frac{\mathrm{d}z}{z-z_k} = 2\pi in $$ by the residue formula.