Complex Conjugate Roots

Theorem (Complex Conjugate Roots). If $f(x)=0$ is an equation with real coefficients and $a+ib$ (with $b\neq0$) is a root, then its conjugate $a-ib$ is also a root, with the same multiplicity.

The proof divides $f(x)$ by the real quadratic $(x-a)^2+b^2$ -- the smallest-degree real polynomial having $a+ib$ as a root -- to get $f(x)=\big[(x-a)^2+b^2\big]Q(x)+Rx+S$ with $R,S$ real (since $f$ and the divisor both have real coefficients, ordinary polynomial long division cannot introduce non-real numbers). Substituting $x=a+ib$ makes the bracketed term vanish, leaving $0=R(a+ib)+S=(Ra+S)+iRb$; since $R,S,a,b$ are all real and $b\neq0$, both the real and imaginary parts must vanish separately, forcing $R=0$ and $S=0$. So the remainder is identically zero, meaning $(x-a)^2+b^2$ divides $f(x)$ exactly -- and this is exactly the statement that $a-ib$ is a root of $f$ too. …