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If the equation $x^4 + ax^3 + bx^2 + cx + 1 = 0$ ($a,b,c$ are real numbers) has no real roots and if at least one root is of modulus one, then what is the relation between $a,b$ and $c$?

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3 Answers 3

Hint: first prove that all the roots have modulus $1$. Then consider the reciprocal polynomial.

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Could you please explain with more details? –  Nimit May 6 '13 at 7:00
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As the roots come in pairs of complex conjugates and the constant part is their product, we see that all roots have modulus $1$.

As a consequence some more estimates can be found such as $-2<a<2$.

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What about the polynomial $x^4-3x^3+4.25x^2-3x+1$? –  Jared May 6 '13 at 14:35

Factoring $x^4+ax^3+bx^2+cx+1=(x^2+dx+1)(x^2+ex+1)$ shows that $a=c=d+e$ and $b=2+de$. To require no real roots we have the inequalities $-2<d,e<2$. Hence all possible triplets $(a,b,c)$ have the form:

$$(a,b,c)=(d+e,2+de,d+e)$$

for some $(d,e)\in(-2,2)\times (-2,2)$.

Intuitively, $d$ is twice the real part of one pair of complex roots, $z,\bar{z}$. Since $|z|=1$, and $z\ne\bar{z}$, it follows that $-1<\operatorname{Re}z<1$, which gives us our inequalities for $d$. Similarly for $e$.

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