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I've been having some trouble with this complex question (not my best topic), and I was wondering if I could get any hints or explainations on how to do it.

Prove that all the roots of the equation $$z^n\cos(n\alpha)+z^{n-1}\cos((n-1)\alpha)+z^{n-2}\cos((n-2)\alpha)+\cdots+z\cos(\alpha)=1,$$ where $\alpha$ is real, lie outside the circle $|z|=\dfrac 12$.

Any help would be greatly appreciated

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Suppose that there exists a root $z_0$ of the equation such that $|z_0| \le \frac{1}{2}$. Now we have that $$1 = |z_0^n\cos(n\alpha) + z_0^{n-1}\cos((n-1)\alpha) + \dots + z_0\cos(\alpha)|\le$$ $$|z_0^n\cos(n\alpha)| + |z_0^{n-1}\cos((n-1)\alpha)| + \dots + |z_0\cos(\alpha)|\le|z_0^n| + |z_0^{n-1}| + \dots + |z_0|\le$$ $$\frac{1}{2^n} + \frac{1}{2^{n-1}} + \dots + \frac{1}{2} = 1 - \frac{1}{2^n}< 1 $$ Contradiction.

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  • $\begingroup$ So pretty much you show that it is impossible for any root to have modulus $\le \frac 12$ ? Also can you explain how you got from the modulus to the fractions please? $\endgroup$
    – user102120
    Nov 11, 2013 at 8:13
  • $\begingroup$ I proved that it doesn't exist root $z_0$ for which $|z_0| \le \frac{1}{2}$. This is logically equivalent to: for every root $z$, $|z| \gt \frac{1}{2}$ $\endgroup$
    – sve
    Nov 11, 2013 at 8:17
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    $\begingroup$ For example $|z_0^ncos(n\alpha)|$ (we use the fact that |ab|=|a||b|) =$|z_0^n||cos(n\alpha)|$ (we use the fact $|cos(n\alpha)|\le 1$) $\le$ $|z_0^n|$ = $|z_0|^n$ (we use the supposition) $\le \frac{1}{2^n}$ $\endgroup$
    – sve
    Nov 11, 2013 at 8:28
  • $\begingroup$ thank you for the explaination $\endgroup$
    – user102120
    Nov 11, 2013 at 9:01

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