TITLE: Is this theorem already in existence?
QUESTION [10 upvotes]: I came up with this theorem (shown below) a few months ago, and I haven't been able to find anything like it on the web.  This theorem will give you the quadratic expression that results from $(x-(a+bi))(x-(a-bi))$.  This theorem is useful for finding all roots of an expression when given a complex root (because complex roots come in conjugate pairs, remember?).  Can anyone tell me if this is old news, or if I have discovered something?
Some context:
I'm the student not the teacher. My teacher (which, BTW, she's my Mom, since I'm homeschooled) is trying to get me to re-prove every time.
Basic Theorem (which if it doesn't have a name, I'll call it Smith's Product of Complex Conjugate Roots Theorem.  A tad long, don't you think?):
$(x-(a+bi))(x-(a-bi)) = x^2-2ax+a^2+b^2$
Proof 1:
$(x-(a+bi))(x-(a-bi))$
$(x-a-bi)(x-a+bi)$
$x^2-ax+bxi-ax+a^2-abi-bxi+abi-b^2i^2$
$x^2-2ax+a^2-b^2i^2$
$x^2-2ax+a^2+b^2$
Proof 2:
$(x-(a+bi))(x-(a-bi))$
$(x-a-bi)(x-a+bi)$
$((x-a)-bi)((x-a)+bi)$
$(x-a)^2-b^2i^2$
$(x^2-2ax+a^2)+b^2$
$x^2-2ax+a^2+b^2$

REPLY [4 votes]: You get to the result much faster if you set $z=a+bi$, so $\bar{z}=a-bi$, the complex conjugate. Then
$$
(x-(a+bi))(x-(a-bi))=(x-z)(x-\bar{z})=x^2-(z+\bar{z})x+z\bar{z}
$$
Since $z+\bar{z}=2a$ and $z\bar{z}=|z|^2=a^2+b^2$, the final form of the expression is
$$
x^2-2ax+a^2+b^2
$$
You might note also that if you have a polynomial $x^2-2ax+q$ with negative discriminant, so $a^2-q<0$, you can set $b=\sqrt{q-a^2}$ and thus $q=a^2+b^2$. Then the polynomial is
$$
(x-z)(x-\bar{z})
$$
where $z=a+bi$.