<s>[INST] Complex numbers is an algebraic extension of Reals so what real Polynomial has root x+iy If I have an arbitrary complex number $x+iy$ then how do I work out a polynomial in the ring of polynomials over $\mathbb{R}$ for which it is a root?
 [/INST] $z= (x+yi)$
$z-x = yi$
$(z-x)^2 = -y^2$
$z^2 -2xz + (x^2 + y^2)$ is a polynomial that has $x+yi$ as a root.
..... or .....
$P(z) = az^2 + bz + c$ then the root of $P$ is $\frac {-b\pm\sqrt{b^2-4ac}}{2a}$ so we let $2a=1$ need $x=-b$ and $b^2-4ac=b^2+2c= -y^2$
So let $b=-x$, $a=\frac 12$ then $c=\frac {y^2 + x^2}2$.
So $x+yi$ is root to $\frac 12z^2 -xz +\frac {y^2 + x^2}2$ or to $z^2 -2xz + (y^2 + x^2)$
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