Is this theorem already in existence?

Solution 1:

While it should be encouraged that you reach your own results and ways of thinking, you should be more careful about the importance (?) of the results you achieve, other than to yourself.

Your question can have many interpretations, really: an excited student that tries to reach its own conclusions, an arrogant student, an independent student etc.

But, responding to your question: No, there is no name for this. This is a trivial consequence of basic properties of multiplication and complex numbers.

Solution 2:

Your proof looks correct. However, this is highly unlikely to be a real theorem, because it is just simple binomial multiplication. This is similar to proving a theorem like "Addition Theorem" which states that 1+5=6. (Note I didn't use 1+1=2 because Peano and such.)

Solution 3:

It is a well-known fact (even in High School) that if $u$ and $v$ are the roots of a second degree, monic polynomial, then the polynomial is $$X^2-sX+p$$ where $s$ and $p$ are, respectively, $u+v$ and $uv$. Your "theorem" is a trivial consequence of this.

Solution 4:

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 $z=a+bi$.