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Let $P(z)=\sum_{k=0}^n a_kz^k$ where $0＜a_1＜\cdots＜a_n$. Show that $P(z)$ has n zeros in unit disk $\Bbb D$. Moreover, show that $\sum_{k=0}^n a_kcoskt$ has 2n zeros in open interval $(0,2π)$.

I want to use Rouché theorem but I can't find a analytic function has n zero in $\Bbb D$.

Tristan Duquesne
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Hanyu.Chern
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    most likely a duplicate as both results are classic; for the first note use $Q(z)=(z-1)P(z)$ and apply the triangle inequality to show that $|Q(z)|>0, |z| \ge 1, z \ne 1$, for the second note that by the argument principle and first part $\arg P$ changes by $2n\pi$ on the unit circle and show that implies that $\Re P$ has $2n$ zeroes there – Conrad Jan 29 '22 at 15:32
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    The first question is answered here: https://math.stackexchange.com/q/188039/42969, the second question is answered here: https://math.stackexchange.com/q/2266985/42969. – Martin R Jan 29 '22 at 19:03

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