How to find no. of complex root of any polynomial?

Question

For $f(x)=x^5+x^3-2x+1$, we know there should be $5$ roots . We can also find real roots of this polynomial.

Can we conclude that:
No of complex roots = No of total roots - No of real roots .

Do we have any rule for this ?

I am keeping above polynomial short for best understanding , so I am asking for a concept that applies to higher degree polynomial . also I am interested only in no of complex roots ( not in magnitude of roots ) .

Answer 1

There are a number of tests for the number of positive (real) roots of polynomials, like Descartes' rule of signs. See e.g. Jameson "Counting Zeros of Generalized Polynomials: Descartes' Rule of Signs and Laguerre's Extensions", Mathematical Gazette 90:518 (June 2006), pp. 223-234. This (and other, more fine, criteria) have been the objective of much research on polynomials. Being able to determine the number of positive roots of $p(x)$ gives you the number of negative roots by considering $p(-x)$, and by the fundamental theorem of algebra the number of complex roots (which you know come in conjugate pairs if the coefficients are real).

Answer 2

Yes, we can conclude. For a polynomial with real coefficients the complex roots (which are not real) appear in conjugate pairs i.e., if $a+ib$ is a root then $a-ib$ is also a root. Since degree of polynomial is five therefore it has at least 1 real root. Thus number of complex root(s) (which are not real) is either 0, 2 or 4.

Answer 3

Descartes rule of signs gives partial information. The rule says the number of real positive roots is less than or equal to the number of sign changes, in this case $x^5+x^3−2x+1$ two sign changes. We can find a bound on the number of negative roots: $(-x)^5+(-x)^3−2(-x)+1=-x^5-x^3+2x+1$ here just one sign change. So altogether at most $3$ real roots. We know therefore that there is at least $2$ complex roots, conjugate of course.