# How can this inequality imply the polynomial has no roots?

Given a complex polynomial $p(z)=z^n+a_1z^{n-1}+\ldots+a_n$, and a large real number $r=|z|$, how can the inequality $|z^n|>|a_1z^{n-1}+\ldots+a_n|$ imply that the polynomial $p_t(z)=z^n+t(a_1z^{n-1}+\ldots+a_n)$ has no roots on the circle $|z|=r$ when $0\leq t\leq1$?

I try to reason as follows:

I tried to work backward, if it has no roots then $p_t(z)>0\implies z^n+t(a_1z^{n-1}+\ldots+a_n)>0\implies z^n>-t(a_1z^{n-1}+\ldots+a_n)$, but I still failed to see the connection.

Could somebody please lend a help? Thanks.

We will show something a stronger statement: for $t\in [0,1]$, the polynomial $p_t(z)$ has all its roots inside the disk $|z|<r$.
By Rouché's theorem, if $\gamma$ is a simple curve in the complex plane and if $P$ and $Q$ are two polynomials such that $$|P(z)| > |Q(z) − P(z)|\quad \mbox{for every z\in\gamma},$$ then $P$ and $Q$ have the same number of zeros (counting multiplicity) in the interior of $\gamma$.
Now let $\gamma$ be the circle $|z|=r$, $P(z)=z^n$ and $Q(z)=p_t(z)$. Then $Q(t)-P(t)=t(a_1z^{n-1}+\ldots+a_n)$ and for $t\in [0,1]$, $$|P(z)|=|z^n|>|a_1z^{n-1}+\ldots+a_n|\geq t|a_1z^{n-1}+\ldots+a_n|=|Q(t)-P(t)|.$$ Hence $z^n$ and $p_t(z)$ has the same number of roots in $|z|<r$, that is the polynomial $p_t(z)$ has $n$ roots inside the disk $|z|<r$. Since the degree of $p_t(z)$ is $n$, we conclude that it has no roots in $|z|\geq r$.