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So if I have a polynom with real coefficients and the solution $x+iy$, why is $x-iy$ always a solution too?

Let $z$ and $w$ be complex numbers, with $w^{\ast}$ = complex-conjugated of $w$, then $$(z-w)(z-w^{\ast}) = z^2 -2z\mathrm{Re}(w) + |w|^2$$ All coefficients are real, but how do I prove the other direction?

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Well, to get non-real roots to a real polynomial, you extend the reals by adding a formal symbol, which is normally called 'i', but I will call 'h' just to be different, with the constraint that h^2=-1. So then you get some root involving h.

But now we say, "really" the root of -1 is 'i', so setting i=h gives the root we just found. But then we ask, what if 'i' is really -h? Could there be any difference? Since (-h)^2 is also -1 by elementary algebra, there is no way to say that h is "really" i or -i. But if a non-real value were a root without its complex conjugate being a root, then there would be a contradiction, because we would have to have a way of saying which of i and -i is "really" h.

This is an extremely informal argument, but I think the crux of the matter is that i is only defined in this formal manner.

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If $f$ is the polynomial and $z$ is a root, then $f(z)=0$. But $f(\bar z)=\overline{f(z)}=\bar 0=0$, so $\bar z$ is also a root.

EDIT: (Why is $f(\bar z)=\overline{f(z)}$)
First, note that for every pair of complex numbers $w,z$ you have:

  • $\bar z\bar w=\overline{zw}$
  • $\bar z+\bar w=\overline{z+w}$

Now put $f(X)=\sum_{j=0}^n a_jX^j$. Then, $$f(\bar z)=\sum_{j=0}^n a_j\bar z^j=\sum_{j=0}^n a_j\overline{z^j}\stackrel*=\sum_{j=0}^n \overline{a_jz^j}=\overline{\sum_{j=0}^na_jz^j}=\overline{f(z)}$$

(*) Here is essential that $a_j\in\Bbb R$.

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    $\begingroup$ why is $f(\bar z)=\overline{f(z)}$ ? $\endgroup$ – Christian Nov 23 '14 at 10:16
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    $\begingroup$ @Christian: Because the coefficients of $f$ are real. And complex conjugates commutes with powers of $z$, i.e. $(\overline{z})^{n} = \overline{z^{n}}$ for every $n\in\mathbb{N}$. This follows from $\overline{zw}=\overline{z}\cdot \overline{w}$. $\endgroup$ – Prism Nov 23 '14 at 10:24
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Firstly, we show that if a complex number is a root of a polynomial equation, then its complex conjugate is also root of the polynomial equation. Consider $$f(x) = \sum_{i = 1}^{n} a_{i} x^{i} = 0,$$ in which $a_{i} \in \mathbb{R}$. Suppose some complex number $\xi$ is a root of $f(x) = 0$; that is, $$f(\xi) = \sum_{i = 1}^{n} a_{i} \xi^{i} = 0.$$ Now, let $\xi^{\ast}$ be the complex conjugate of $\xi$. By the properties of complex conjugation, $$ f(\xi^{\ast}) = \sum_{i = 1}^{n} a_{i} (\xi^{\ast})^{i} = \sum_{i = 1}^{n} a_{i} (\xi^{i})^{\ast} = \sum_{i = 1}^{n} (a_{i} \xi^{i})^{\ast} = \left(\sum_{i = 1}^{n} a_{i} \xi^{i}\right)^{\ast}. $$ Since $$\left(\sum_{i = 1}^{n} a_{i} \xi^{i}\right)^{\ast} = 0^{\ast},$$ it follows that $$ f(\xi^{\ast}) = \sum_{i = 1}^{n} a_{i} (\xi^{\ast})^{i} = 0^{\ast} = 0, $$ as required.

Since complex conjugate roots come in conjugate pairs, there are even number of them. It follows from this and the fundamental theorem of algebra that a polynomial of even degree has either even number of real roots or even number of complex conjugate roots, or both, whereas a polynomial of odd degree always has an odd number of real roots, the rest of them being complex conjugate pairs.

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  • $\begingroup$ nice answer, thank you! $\endgroup$ – Christian Nov 23 '14 at 10:25

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