About the general method to compute the number of zeros of complex polynomials

Question

I read the following questions accidentally:

For any fixed real number t, what is the number of zeros of polynomial $p(z) = z^5+2z^3+z^2+z-t$. (in this case, let's say exclude those duplicated roots, which will make this question trivial.)

Then another question is, to the above polynomial, how to calculate its root numbers in a specific area, for example, the left side of the complex plane (Real z < 0?)

Are there any suitable tools in, for example, contour analysis, to solve this general question?

Answer 1

To find the amount of roots without counting the multiplicity, we just need to filter out the duplicate roots.

Suppose the $\alpha$ is a root with multiplicity $m$, then $(x-\alpha)^m\mid f(x)$. Writing $f(x)=(x-\alpha)^mP(x)$ with $x-\alpha\not \mid P(x)$. Then by calculus, we know: $$(x-\alpha)^{m-1}\mid f'(x)=(x-\alpha)^{m-1}((x-\alpha)P'(x)+mP(x))$$ Since $(x-\alpha)\not\mid P(x)$, then $(x-\alpha)^{m}\not\mid f'(x)$. This shows that for each root $\alpha$, $f'(x)$ has exactly one less the multiplicity, then $\frac {f(x)}{\gcd(f(x),f'(x))}$ contains exactly the roots with multiplicity $1$. So the amount of unique roots is $\deg(f)-\deg(\gcd (f,f'))$

Answer 2

The number of zeros (or number of roots) of a polynomial, including multiplicities is the degree of the polynomial. Here the degree is five (the term with the highest power of $z$ is $z^5$).

So $p(z)$ has five zeros in the complex plane.