# When are complex conjugates of solutions not also solutions?

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?

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I think here "normal" is most likely "polynomials with real coefficients". – tabstop Jan 29 '14 at 15:39

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."

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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.

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I think you should reformulate your first sentence as it doesn't make sense. (the conjugate of what ?) – Thomas Produit Jan 29 '14 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 '14 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 '14 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 '14 at 16:11

Key Idea $\$ Conjugation $\rm\:x\mapsto \bar x\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products}\,$ and $\rm\:\color{#0a0}{fixes\ coefficients}\in\color{#0a0}{\Bbb R}.\:$ Thus by induction it commutes with 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.  Explicitly $$\begin{eqnarray} \rm \overline{f(w)}\: &=&\rm\ \ \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\ &=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\ \ by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in\! \Bbb C\\ &=&\rm\,\ \overline a_n\, \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\ \ by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in\! \Bbb C \\ &=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\ \ by\ \ \ \color{#0a0}{\overline a = a}\ \ \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb R}\\ &=&\rm\ f(\overline w)\\ \rm So\ \ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w),\, \ \ \text{i.e. \ \rm w root of \,\rm f\,\Rightarrow\,\bar w root of \,\rm f}\quad {\bf QED} \end{eqnarray}$$

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

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