Complex numbers is an algebraic extension of Reals so what real Polynomial has root x+iy

Question

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?

Answer 1

$z= (x+yi)$

$z-x = yi$

$(z-x)^2 = -y^2$

$z^2 -2xz + (x^2 + y^2)$ 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 $z^2 -2xz + (y^2 + x^2)$

Answer 2

Note that for any complex number $w$, $$(x-w)(x-\bar w) = x^2 -\bar w x-wx + w\bar w$$ $$=x^2 - (w+\bar w)x + w\bar w$$ $$=x^2 + (2\operatorname{Re}w)x + |w|^2$$ which has real coefficients.

Remember, a monic polynomial has real coefficients if and only if all complex roots occur in conjugate pairs.

Answer 3

If a polynomial in $p(t) \in \Bbb R[t]$ has $z = x + iy$ as a root, it also has $\bar z = x - iy$ as a root, for if

$p(t) = \displaystyle \sum_0^n p_k t^k \in \Bbb R[t], \tag 1$

then

$p_k \in \Bbb R, \; 0 \le k \le n; \tag 2$

hence with

$\displaystyle \sum_0^n p_k z^k = p(z) = 0, \tag 3$

we have

$p(\bar z) = \displaystyle \sum_0^n p_k \bar z^k = \sum_0^n \bar p_k \bar z^k =\overline{ \displaystyle \sum_0^n p_k z^k} = 0, \tag 4$

since

$\bar p_k = p_k \tag 5$

by virtue of (2).

Clearly the same argument shows that $x + iy$ is a root of $p(t)$ whenever $x - iy$ is; simply interchange $y \longleftrightarrow -y$ in (3), (4).

It follows then that any polynomial (1) such that (3) binds for some $z = x + iy$ with $y \ne 0$ must be of degree at least $2$, since

$\bar z = x - iy \ne x + iy = z \tag 6$

in this case.

Since the simplest possible real polynomial satisfied by $z$ is also satisfied by $\bar z$, it must be of degree at least $2$, so we are led to consider

$m(t) = (t - z)(t - \bar z) = t^2 - (z + \bar z)t + z\bar z; \tag 7$

we note that the coefficients of this quadratic are invariant under complex conugation:

$\overline{z + \bar z} = \bar z + \bar{\bar z} = \bar z + z, \tag 8$

$\overline{z \bar z} = \bar z \bar{\bar z} = \bar z z, \tag 9$

and hence are both real; indeed we have

$z + \bar z = (x + iy) + (x - iy) = 2x, \tag{10}$

$z \bar z = (x + iy)(x - iy) = x^2 - ixy + iyx - (iy)^2 = x^2 + y^2, \tag{11}$

whence

$m(t) = (t - z)(t - \bar z) = t^2 - 2x t + (x^2 + y^2) \in \Bbb R[t] \tag{12}$

is the simplest real polynomial satisfied by $z$ provided $y \ne 0$.

In the event that $y = 0$,

$z = x, \tag{13}$

and hence $z$ satisfies

$t - x \in \Bbb R[t], \tag{14}$

with

$\deg(t - x) = 1. \tag{15}$

There are of course many other real polynomials which have $z$ and $\bar z$ as roots; indeed for any

$f(t) \in \Bbb R[t] \tag{16}$

we may take

$p(t) = (t - z)(t - \bar z)f(t), \tag{17}$

and then

$p(z) = (z - z)(z - \bar z)f(z) = 0, \tag{18}$

$p(\bar z) = (\bar z - z)(\bar z - \bar z)f(z) = 0, \tag{19}$

showing that

$p(z) = p(\bar z) = 0 \tag{20}$

for any $p(t)$ of the form (17).

The content of the preceding paragraph may evidently be cast in ring-theoretic terms; we observe that (16)-(17) essentially affirm that every element of the proper principal ideal

$\langle (t = z)(t - \bar z) \rangle = ((t - z)(t - \bar z))\Bbb R[t]$ $= (t^2 - (z + \bar z)t + z \bar z)\Bbb R[t] \subsetneq \Bbb R[t] \tag{21}$

vanishes at $z$ and $\bar z$.

Finally, the picture may be completed if we can establish that every $p(t) \in \Bbb R[t]$ such that

$p(z) = 0 \tag{22}$

may be divided by $(t - z)(t - \bar z)$; this of course is a consequence of Euclidean division of polynomials which allows us to write

$p(t) = q(t)((t - z)(t - \bar z)) + r(t), \tag{23}$

with

$\deg r(t) \le 1; \tag{24}$

evaluating (23) at $z$ and $\bar z$ yields

$0 = p(z) = q(z)((z - z)(z - \bar z)) + r(z) = q(z)((0)(z - \bar z)) + r(z) = r(z), \tag{25}$

and

$0 = p(\bar z) = q(z)((\bar z- z)(\bar z - \bar z)) + r(z) = q(z)((\bar z - z)(0)) + r(\bar z) = r(\bar z); \tag{26}$

but a polynomial of degree $1$ cannot have the two distinct roots $z$ and $\bar z$; thus

$\deg r(t) = 0, \tag{27}$

from which we easily see

$r(t) = 0, \tag{28}$

which implies

$(t - z)(t - \bar z) \mid p(t). \tag{29}$