Minimum number of imaginary roots based on number of terms missing from polynomial.

The comprehension in my book states the fact mention in the image. I am interested about how do I prove it. I don't see any approach to prove that. I thought to do something with Descartes rules of signs and relate it with this but I fail to find connection with this.

Solution 1:

(Too long for a comment.)

• A polynomial $a(x)$ of degree $n$ has at most $n+1$ terms, so $\sigma_a \le n$ sign changes. The associated $a(-x)$ has the same degree $n$, so it has $\tau_a \le n$ sign changes. It can be easily shown that $\sigma_a+\tau_a \le n$.

• A polynomial of degree $n$ with with $p$ missing terms can be written as $c(x) = x^{k+p}\,a(x) + b(x)$ where $\deg a = n-k-p\,$, $\deg b = k-1$.

• Then $\sigma_c=\sigma_a+\gamma+\sigma_b$ where $\gamma=1$ if there is a sign change between the constant term of $a(x)$ and the leading term of $b(x)$, or $0$ otherwise.

If $p$ is odd, then $\tau_c=\tau_a+\gamma+\tau_b$ since the constant term of $a(x)$ and the leading term of $b(x)$ either both change sign or both stay the same. It follows that the maximum number of real roots is $\sigma_c+\tau_c=(\sigma_a+\tau_a)+(\sigma_b+\tau_b)+ 2 \gamma \le (n-k-p)+(k-1)+2 \gamma=n-p-1+2 \gamma$, and therefore there must be at least $n - \left(n-p-1+2 \gamma\right)=p+1-2\gamma$ non-real complex roots, which proves $(ii)$.

If $p$ is even, point $(i)$ follows via a similar argument.

[ EDIT ] $\;$ The intuition behind it is that Descartes' rule of signs, combined with the $x \mapsto -x$ substitution, provides an upper bound for the number of real roots of a polynomial. When the polynomial has large "gaps", there are not enough non-$0$ terms left to account for the number of sign changes required for all roots to be real, so there will necessarily exist some non-real complex roots. The rest is just a matter of counting them.