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.

I've heard that for "normal" equations (e.g. $3x^2-2x=0$), if $(a+bi)$ is a solution then $(a-bi)$ will be a solution as well.

This is because, if we define $i$ in terms of $i^2=-1$ then we might as well define $i^\prime=-i$. Since ${i^\prime}^2=-1$ we find $(a+bi)$ has the same algebraic behaviour as $(a+bi^\prime)$.

So what are non-"normal" equations? When are conjugates not also solutions?

share|improve this question
2  
I think here "normal" is most likely "polynomials with real coefficients". –  tabstop Jan 29 at 15:39

3 Answers 3

up vote 2 down vote accepted

An example of a polynomial with non-real coefficients that has two non-conjugate solutions is $$ (x - i) (x - 1) = 0 $$ whose roots are clearly $i$ and $1$. Written out, it looks like $$ x^2 - (1+i) x + i = 0. $$

As others have said, I think that the person using the word "normal" meant "equation with real-number coefficients."

share|improve this answer

If the polynomial has real coefficients, and there is a nonreal root, then its conjugate is also a root. Otherwise, there would be at least one nonreal coefficient.

share|improve this answer
    
I think you should reformulate your first sentence as it doesn't make sense. (the conjugate of what ?) –  Thomas Produit Jan 29 at 15:47
    
Thanks @ThomasProduit, I'm afraid auto-correct changed what I typed. It was supposed to say "a nonreal root". Corrected. –  MPW Jan 29 at 15:59
    
Now it's OK but to improve your answer you should prove it or give some hints to do it –  Thomas Produit Jan 29 at 16:01
    
I didn't see that a proof was required. But it follows simply from the fact that $\overline{p(z)} = p(\bar{z})$ if the coefficients are real. –  MPW Jan 29 at 16:11

Key Idea $\ $ Conjugation $\rm\:x\mapsto \bar x\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products}.\:$ Also it $\rm\:\color{#0a0}{fixes\ coefficients}\in\color{#0a0}{\Bbb R}.\:$ Therefore, by induction, it preserves polynomials $\rm\ \overline{f(w)} = f(\overline w),\ \ f(x)\in\color{#0a0}{\Bbb R}[x],\ $ having all $\,\rm\color{#0a0}{real}$ coefficients, since such polynomials are compositions of said basic operations. $ $ More explicitly $$ \begin{eqnarray} \rm \overline{f(w)}\: &=&\rm\ \ \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\ &=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in \Bbb C\\ &=&\rm\,\ \overline a_n\, \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in \Bbb C \\ &=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \ \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb R}\\ &=&\rm\ f(\overline w)\\ \rm Hence\ \ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w)\quad {\bf QED} \end{eqnarray}$$

Generally this fails if $f$ has non-real coefficients, e.g. $\,\bar w\,$ is a root of $\,x-w\,$ iff $\,\bar w = w,\,$ i.e. $\,w\in \Bbb R.$

Remark $\ $ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $0$-ary operations), they also preserve the "polynomial" terms composed of these basic operations. Said equivalently, hom's commute with polynomials.

share|improve this answer

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.