I am told that a nonzero polynomial $p$ in $P_3(\mathbb{R})$ cannot both have 1 and $i$ as roots.
Solutions to quadratic equations (which are the highest order polynomials in $P_3(\mathbb{R})$), can be found with the well-known formula:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
How can I show that 1 and $i$ cannot both be roots at the same time for a $p\in P_3(\mathbb{R})$?
Best Answer
We can certainly follow your idea to show that no quadratic in $P_3(\Bbb R)$ can have both of them as roots.
On the one hand, if $$1=\frac{-b+\sqrt{b^2-4ac}}{2a},$$ then $$2a=-b+\sqrt{b^2-4ac},$$ so $$b+2a=\sqrt{b^2-4ac},$$ and so $$i=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-b-(b+2a)}{2a}=\frac{-2a-2b}{2a}=-1-\frac{b}{a},$$ which is clearly false, since $a$ and $b$ are real. On the other, if $$1=\frac{-b-\sqrt{b^2-4ac}}{2a},$$ then $$2a=-b-\sqrt{b^2-4ac},$$ so $$b+2a=-\sqrt{b^2-4ac},$$ so $$-b-2a=\sqrt{b^2-4ac},$$ and so $$i=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-b+(-b-2a)}{2a}=\frac{-2a-2b}{2a}=-1-\frac{b}{a},$$ which is clearly false again.
At that point, we'd want to justify that linear polynomials can't have them both as roots, which is straightforward.
A simpler way to deal with it, though, is to recall that $c$ is a root of a polynomial $p(x)$ if and only if $(x-c)$ is a factor of $p(x).$ Thus, in order to have $p(1)=0$ and $p(i)=0$ for a non-zero polynomial $p(x),$ we would have $$p(x)=q(x)(x-1)(x-i)$$ for some non-zero polynomial $q(x).$ Since $p$ has degree at most two, then $q$ must be a non-zero constant polynomial, meaning that $$p(x)=a(x-1)(x-i)=a\left(x^2-x-ix+i\right)=ax^2-a(1+i)x+ai$$ for some $a\ne 0.$ Since the coefficients of $p$ are real, then $a$ is real, but since $a\ne 0,$ then $ai$ can't be real, so no such polynomial $p$ can exist.
One further way we can go is to suppose there exist some $a,b,c\in\Bbb R$ such that $p(x)=ax^2+bx+c$ has both $1$ and $i$ as zeros. Put another way, $$0=p(1)=a+b+c$$ and $$0=p(i)=-a+bi+c.$$ Since $0=-a+bi+c,$ then $$a-c=bi,$$ but the right-hand side is imaginary and the left-hand side is real, so we must have $b=0,$ so our equations reduce to $$0=a+c$$ and $$0=-a+c.$$ Solving the second equation for $c$ gives us $c=a,$ so the first equation becomes $0=2a.$ But then $a=0,$ so $c=a=0,$ and so $p(x)=0x^2+0x+0=0.$ Thus, no non-zero element of $P_3(\Bbb R)$ has $1$ and $i$ as roots.