Suppose that $a_0 >a_1 >...>a_{2013} >0.$ Prove that $\sum_{n = 0}^{2013}a_nz^n \neq 0$ when $|z|<1$ Suppose that $a_0 >a_1 >...>a_{2013} >0.$  Prove that $\sum_{n = 0}^{2013}a_nz^n \neq 0$ when $|z|<1$
Not sure where to begin with this.  Any suggestions?  Thanks.
 A: Elementary, my dear Watson.

If $a_0\geq a_1\geq\cdots\geq a_n>0$,
  then $\sum_{k=0}^na_kz^k\neq 0$ for every $z\in\mathbb{C}$ with $|z|< 1$.

$\newcommand{\CC}{\mathbb{C}}
$Write $f(z):=\sum_{k=0}^na_kz^k$.
Suppose that $z\in\CC$, $|z|<1$.
Then
$$
(1-z)f(z) \:=\: a_0 - \sum_{k=1}^n(a_{k-1}-a_k)z^k - a_nz^{n+1}~,
$$
and hence
$$
\begin{aligned}
\bigl|(1-z)f(z)\bigr|
    ~&\:\geq\: a_0 \:-\, \sum_{k=1}^n(a_{k-1}-a_k)|z|^k \,-\: a_n|z|^{n+1} \\[-.5ex]
    ~&\:>\: a_0 \:-\, \sum_{k=1}^n(a_{k-1}-a_k) \;-\; a_n \\
    ~&\:=\: a_0 - (a_0-a_n) - a_n \\[1ex]
    ~&\:=\:0~,
\end{aligned}
$$
where the second inequality is strict because $a_n>0$ and $|z|^{n+1}<1$.
Now you're ready to prove a related assertion:

If $a_0>0$ and $a_0\geq a_1\geq a_2\geq \cdots \geq a_n \geq \cdots \geq 0$,
  then $f(z) := \sum_{n=0}^\infty a_nz^n$ converges everywhere on the open disk
  $|z|<1$, and has no zero on this disk.

You can follow the same line of reasoning as above, with a twist.
Let's push the envelope:

Let $a_0\geq a_1\geq \cdots \geq a_n>0$, $n\geq 1$,
  and set $f(z):=\sum_{k=0}^na_kz^k$ for $z\in\CC$.
  Suppose that $f(z_0)=0$ for some $z_0$ with $|z_0|\leq 1$.
  Then there exists a divisor $d>1$ of $n+1$ so that
  $a_k=a_{k-1}$ for every $k$, $1\leq k\leq n$, which is not a multiple of $d$,
  and $z_0=\exp(l\!\cdot\!2\pi i/d)$
  for some integer $l$, $0<l<d$.
  In particular, if $n+1$ is a prime, then $a_0=a_1=\cdots=a_n$
  and $z_0=\exp(l\!\cdot\!2\pi i/(n\!+\!1))$ for some integer $l$, $1\leq l\leq n$.

In order to prove this you will need, besides the trick of multiplying $f(z)$ by $1-z$,
the following

Lemma. $~$Let $z_1,\ldots,z_n\in\CC$ $\,$($n\geq 1$)$\,$ and $z=z_1+\cdots+z_n$.
  Then $|z|=|z_1|+\cdots+|z_n|$ if and only if
  there exist nonnegative real numbers $\lambda_1$, $\ldots$, $\lambda_n$ whose sum is $1$
  such that $z_k=\lambda_k z$ for $1\leq k\leq n$.

Lemma says that the line segment from $0$ to $z$
is the only shortest polynomial line from $0$ to $z$.
Can I hear you protesting
that stating more problems is no way of answering a question?
I am doing it for a reason:
since I served you only a simple trick on a platter,
I feel I owe it to you to add to your plain fare
a dollop or two of something tastier to chew on.
