# Proof real coefficients complex analysis

Show that if the polynomial $p(z)$ has real coefficients, it can be expressed as a product of linear and quadratic factors, each having real coefficients.

I am not sure how to prove this. From what I know thus far is if there exists $p(z)$, where $z = x +ib$ has real coefficients, but does that mean this $\Re(z)$ or that $x,b \in \mathbb{R}$ or both? And if it can be expressed as a product of linear and quadratic factors, would that necessarily mean that it is a polynomial?

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Hint: If $z=x+iy$ is a root, can you show that $\bar{z}=x-iy$ is also a root? – Brett Frankel Jan 25 '13 at 4:50
I don't know if it helps, but remember that if a polynomial has real coefficients, then if $a+ib$ is a solution, so is $a-ib$ (i.e., the conjgate). Then we see that $[z-(a+bi)][z-(a-bi)]=z^2-2za+a^2+b^2$. – anon271828 Jan 25 '13 at 4:51
@BrettFrankel Actually I cannot, roots are right after this section I am learning now. – Q.matin Jan 25 '13 at 4:52
@anon271828 It does help, thanks! – Q.matin Jan 25 '13 at 4:53

Start with the equation $$0=a_0+a_1z+a_2z^2+a_3z^3+\cdots=a_0+\sum_{k=1}^{n}a_kz^k.$$ Now, "take the conjugate" of both sides: $$0=a_0+a_1\bar{z}+a_2\bar{z}^2+a_3\bar{z}^3+\cdots=a_0+\sum_{k=1}^{n}a_k\bar{z}^k$$ This means that, whenever $z$ is a solution, $\bar{z}$ is also a solution. By the fundamental theorem of algebra, you know that you can split $P(z)$ into a product of linear factors of the from $(z-r_k)$. But if two roots are $r$ and $\bar{r}$, then the two factors $(z-r)$ and $(z-\bar{r})$ nicely multiply together as $$(z-r)(z-\bar{r})=z^2-(r+\bar{r})z+r\bar{r}.$$ Try to convince yourself that this is a quadratic real polynomial by knowing that $r\bar{r}=|r|^2$ and $r+\bar{r}=2\Re(r)$ -- and you are done!

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Thanks a lot!!! – Q.matin Jan 25 '13 at 5:43

When a polynomial has real coefficients, all complex roots are conjugate pairs. Say two complex roots are $a+i b$ and $a-i b$; they are roots of the quadratic $y = x^2-2 a x+a^2+b^2$.

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Not sure exactly how this follows the question? You think you can explain a bit further? – Q.matin Jan 25 '13 at 4:54
The case with complex roots means that the polynomial has a factor of a quadratic with real coefficients. For a real, there is a linear factor (i.e. $x-c$). Over all possible roots, a general polynomial has linear and quadratic factors, each of which has real coefficients. – Ron Gordon Jan 25 '13 at 4:57
Thanks a lot , rlgordonma! – Q.matin Jan 25 '13 at 5:44

Step 1: Show that every odd degree polynomial with real coefficients has a real root. Try the intermediate value theorem.

Step 2: If $z \in \mathbb C \setminus \mathbb R$ is a root of $p(x)$ then show that $\bar{z}$ is as well. Then deduce that $(x-z)(x-\bar{z})$ has real coefficients and divides $p(x)$.

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Thanks for the hint! – Q.matin Jan 25 '13 at 5:43