Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the following polynomial of x $$x^2+(a+b\,g)x+c\,g^2=0$$ where $a\in \mathbb{R}$, $b\in \mathbb{R}$, and $c\in \mathbb{R}$, whereas g can be complex. $c\neq 0$.

$c$ is a free variable and can be chosen to be ANY function of $a$ and $b$, for example $c=a+b$, $c=ab$, or $c=a^2$.

When $a$ is zero and $c=b^2$, we can factor this polynomial as $$(x+(\frac{1}{2}+\frac{\sqrt{3}}{2}i)b\,g)(x+(\frac{1}{2}-\frac{\sqrt{3}}{2}i)b\,g)=0$$ What is the factorization $(x+x_1)(x+x_2)=0$ when $a\neq 0$ and $b\neq0$, $c\neq 0$?

share|improve this question
2  
The quadratic formula works the same for polynomials with either real or complex coefficients. –  user7530 Jan 22 '13 at 18:21
    
I know you can use the solution of the quadratic equations to obtain $x_1$ and $x_2$, but I was wondering if there is a clean representation like the example I gave above for a=0. –  sara Jan 22 '13 at 18:28

1 Answer 1

up vote 2 down vote accepted

Just like in the real-coefficient case, the roots of a quadratic complex polynomial are given by the quadratic formula. In your case,

\begin{align*}&\quad x^2 + (a+bg)x + cg^2\\ &= \left(x + \frac{a+bg}{2} + \frac{\sqrt{a^2+2abg + (b^2-4c)g^2}}{2}\right)\left(x + \frac{a+bg}{2} - \frac{\sqrt{a^2+2abg + (b^2-4c)g^2}}{2}\right), \end{align*} where $\sqrt{z}$ is the principal square root.

There is no significantly simpler formulation unless some special relationship exists between $a,b,c$ and $g$.

EDIT: I assume you're looking for special values of $a,b,c$ so that the discriminant "factors nicely":

$$\sqrt{a^2+2abg + (b^2-4c)g^2} = dg+e.$$ Squaring both sides $$a^2 + 2abg + (b^2-4c)g^2 = e^2 + 2deg + d^2g^2$$ and equating coefficients of $g$ gives $a=\pm e$, from which it follows that either $a=0$ or $b=\pm d\Rightarrow c =0$. There are no other "simple cases."

share|improve this answer
    
"There is no significantly simpler formulation unless some special relationship exists between a,b,c.", I am exactly looking for that special relation so I can have a simple factorization. Even special choices like a=1, b=2, c=6 etc is fine. –  sara Jan 22 '13 at 18:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.