Bounds for the number of Imaginary (Nonreal Complex) roots

Turning the two real-root bounds into a non-real-root bound. Let $m$ be the number of sign changes in $P(x)$ (so at most $m$ positive roots) and $k$ the number of sign changes in $P(-x)$ (so at most $k$ negative roots). Since a degree-$n$ equation has exactly $n$ roots counted with multiplicity (§3.3.2.1), and at most $m+k$ of them can be real,

Worked illustration (Example 3.30 — proving a lower bound on imaginary roots, with no solving). $P(x)=9x^9+2x^5-x^4-7x^2+2$ (degree $9$): signs are $+,+,-,-,+$: 2 sign changes, so at most $2$ positive roots. $P(-x)=-9x^9-2x^5-x^4-7x^2+2$: signs $-,-,-,-,+$: 1 sign change, so at most $1$ negative root. Clearly $0$ is not a root (constant term $2\ne0$), so the maximum possible number of real roots is $2+1=3$. Since the equation has $9$ roots in all (Fundamental Theorem of Algebra), at least $9-3=6$ roots must be non-real (imaginary) — proved entirely from the two sign-change counts, with no solving at all.

Worked illustration (Example 3.31(i) — all roots non-real, in one line). $x^{2018}+1947x^{1950}+15x^8+26x^6+2019$: every coefficient is positive, so $P(x)$ has $0$ sign changes (no positive roots) and $P(-x)$ (all even powers, unchanged) also has $0$ sign changes (no negative roots). Since $0$ is clearly not a root either (constant term $\ne0$), every one of the $2018$ roots is non-real. …