Show that $P(z)=a_kZ^k$ has $n$ zeros in $B(0,1)$.

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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 $\Bbb D$.