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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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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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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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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