Let $f(z)=z^4-2z^3+z^2-12z+20$. Then evaluate the integral by using the argument principle $$\oint_C \frac{zf'(z)}{f(z)} \,ds$$ Where $C$ is the circle $|z|=5$.
What I've tried: I tried using the residue theorem but this is a cumbersome method and calculations get too long. Also the book wants us to solve it by the argument theorem and the problem is that the integral is not in the formal form described in the theorem. What's the correct way to solve this problem?
The answer in the book is $4\pi i.$
An integral of the form
$$\int_C \frac{zf'(z)}{f(z)}\,dz,$$
where $C$ is the positively oriented boundary curve of a domain $V$ and $f$ is holomorphic in $V$, continuous on $\overline{V}$ and has no zeros on $C$, is $2\pi i$ times the sum of the zeros of $f$ in $V$, counted according to multiplicity:
Let $\zeta$ be a zero of order $k$ of $f$. Then we have $f(z) = (z-\zeta)^k\cdot h(z)$, with a holomorphic $h$ that satisfies $h(\zeta) \neq 0$. Thus
$$\frac{f'(z)}{f(z)} = \frac{k(z-\zeta)^{k-1} h(z) + (z-\zeta)^k h'(z)}{(z-\zeta)^k h(z)} = \frac{k}{z-\zeta} + \frac{h'(z)}{h(z)},$$
and
$$\frac{zf'(z)}{f(z)} = \frac{k\zeta}{z-\zeta} + k + z\frac{h'(z)}{h(z)},$$
so the residue of $\dfrac{zf'(z)}{f(z)}$ in $\zeta$ is $k\zeta$.
There is a simple relation between the sum of the zeros of a polynomial (counting multiplicity) and its coefficients. Here, all zeros of $f$ lie inside the disk bounded by $C$.