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How to simplify the following polynomial?

$$ \begin{align} (t - \sqrt{3} \; e^{ \frac{\pi}{3} i }) (t - \sqrt{3} \; e^{ -\frac{\pi}{3} i }) &= t^2 - \sqrt{3} \; e^{ \frac{\pi}{3} i } \; t - \sqrt{3} \; e^{ -\frac{\pi}{3} i } \; t + 3\\ &= t^2 - \sqrt{3} \; t \; ( e^{ \frac{\pi}{3} i } + e^{ -\frac{\pi}{3} i } ) + 3 \end{align} $$

I know the result is

$$ t^2 - \sqrt{3} \; t + 3 $$

but I can't see how to simplify the two complex numbers.

Is it a general rule, that if I've got a polynomial with $(t-c)(t-\bar c)$ where $c,\bar c \in \mathbb{C}$ and $\bar c$ is the complex complement of $c$, the resulting polnomial is $$(t-c)(t-\bar{c})=t^2-ct-\bar{c}t+c\bar{c}=t^2-(c+\bar{c})t+|c|^2,$$


$\mathbb{C}[t]$ means that the coefficients are in $\mathbb{C}$? Why we usually write $\mathbb{K}[t]$?

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You're almost there. Write out $e^{\frac{\pi}{3}i}+e^{-\frac{\pi}{3}i}$ in trigonometric form... – Potato Feb 3 '12 at 22:02
up vote 0 down vote accepted

$\bf Hint:$ $e^{\theta i}=cos\theta+isin\theta$.

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Yes. I think there's no way to »see« it without using the Euler identity to transform the numbers to trigonometric form? – meinzlein Feb 3 '12 at 22:12

We have that

$$e^{ \frac{\pi}{3} i } + e^{ -\frac{\pi}{3} i }=(\cos(\tfrac{\pi}{3})+i\sin(\tfrac{\pi}{3}))+(\cos(-\tfrac{\pi}{3})+i\sin(-\tfrac{\pi}{3}))=$$ $$(\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2})+(\tfrac{1}{2}-i\tfrac{\sqrt{3}}{2})=\tfrac{1}{2}+\tfrac{1}{2}=1$$

Your equation $$(t-c)(t-\bar{c})=t^2 - |c|t + |c|^2$$ is not quite correct; expanding out the left side, $$(t-c)(t-\bar{c})=t^2-ct-\bar{c}t+c\bar{c}=t^2-(c+\bar{c})t+|c|^2,$$ but it is not generally true that $c+\bar{c}=c\bar{c}$.

To answer your other question: in general, if $R$ is a ring (an abstract kind of "number system" where we can add and multiply), then $R[x]$ denotes the collection of polynomials in the variable $x$ having coefficients in $R$, i.e. $$R[x]=\{a_0+a_1x+\cdots+a_nx^n\mid a_i\in R\}.$$ Thus, $\mathbb{C}[t]$ denotes the polynomials in the variable $t$ with complex coefficients. A good way to think of it is as "adding in" the new element $x$ to the original number system $R$. This is also its meaning when talking about, for example, $\mathbb{Z}[\sqrt{2}]$.

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Thank you! So there's no way to »see« it when the numbers are written in polar form? Is there a general rule as mentioned in the original post? – meinzlein Feb 3 '12 at 22:09
@meinzlein: There is, but the general rule in your post was not quite right. I've explained the correct version. – Zev Chonoles Feb 3 '12 at 22:21
Thank you for your explanation. So we »add in« $\sqrt{2}$ when talking about $\mathbb{Z}[\sqrt{2}]$? But usually we don't exactly declare what e. g. variable $t$ should be: $\mathbb{Z}[t]$ means that we »add in« $t$ in $\mathbb{Z}$; but why don't we declare what $t$ should be like (real number, matrix, ...)? When talking about $\mathbb{Z}[\sqrt{2}]$ it's clear, that we »add in« the real number $\sqrt{2}$. – meinzlein Feb 3 '12 at 22:59

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