complex polynomials problem with a weird inequality Show that if $P(z) = a_nz^n+a_{n-1}z^{n-1}+...+a_1z + a_0$
is a polynomial of degree n then there is $R_0>0$ such that for $|z| > R_0$ the polynomial can be bound from below as 
$|P(z)| > \frac{1}{2}|a_n||z|^n$
but what if you have the polynomial $P(z) = a_1z+a_0$ where $a_1 = -100$ and $a_0 = 100$ and we fill in $z = 1$ then we have the left side equal to $0$ and the right side equal to 50 so it does not hold. How does that work? 
I also do not understand what the $R_0$ adds to the problem. Do i use it somewhere in the proof because i do not see its relevance otherwise than stating that absolute value is bigger than 0 also.
 A: The loose idea is that the lower bound applies when $|z|$ is "large enough".
The role of $R_0$ in the problem, which you asked about - "I also do not understand what the $R_0$ adds to the problem" - is to give a precise measure of what the vague phrase "large enough" means, in this context.
As has been pointed out in the comments, you needn't worry about what happens when $R_0$ is "small". You're perfectly at liberty to make $R_0$ as "large" as you like.
As your example illustrates, it is sometimes necessary to take $R_0 \geqslant 1$.
I just want to point out that if you make the assumption $R_0 \geqslant 1$ in the general case, it is easy to work out an exact value for $R_0$ that is good enough to do the job.
You don't have to do this. The problem doesn't ask you to name a particular value of $R_0$. As in Marios Gretsas's answer - in my opinion it is the "right" answer, but I wanted to add an observation - it enough to apply general theorems about limits to show that some value of $R_0$ exists.
I think that's what you should take away from this question, but at the risk of merely distracting you, here is a concrete way to make $|z|$ "large" enough:
Because $P$ is of degree $n,$ we have $a_n \ne 0$.
Let $A = |a_{n-1}| + \cdots + |a_1| + |a_0| \geqslant 0$.
For all $z \ne 0,$ write:
\begin{align*}
\frac{P(z)}{a_nz^n} & = \frac{a_nz^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0}{a_nz^n} \\
& = 1 + \frac{a_{n-1}}{a_nz} + \cdots + \frac{a_1}{a_nz^{n-1}} + \frac{a_0}{a_nz^n}.
\end{align*}
If $R_0 \geqslant 1$, then for all $z$ such that $|z| > R_0,$ we have $|z|^m > R_0^m \geqslant R_0 \quad (m = 1, 2, \ldots),$ therefore:
\begin{align*}
\frac{|P(z)|}{|a_n||z|^n} = \left\lvert\frac{P(z)}{a_nz^n}\right\rvert
& \geqslant 1 - \frac{|a_{n-1}|}{|a_n||z|} - \cdots - \frac{|a_1|}{|a_n||z|^{n-1}} - \frac{a_0}{|a_n||z|^n} \\
& \geqslant 1 - \frac{A}{|a_n|R_0} \\
& > \frac12 \quad \text{if } R_0 > \frac{2A}{|a_n|}.
\end{align*}
So it is good enough to define:
$$
R_0 = 1 + \frac{2A}{|a_n|}.
$$
In particular, for your example ($n = 1$), you could take:
$$
R_0 = 1 + \frac{2|a_0|}{|a_1|}.
$$
A: We have that $\lim_{z \to \infty}\frac{|p(z)|}{|a_n||z|^n}=1$
So for $\epsilon=\frac{1}{2}$ exists $R_0>0$ such that $$\frac{|p(z)|}{|a_n||z|^n}>1-\epsilon=\frac{1}{2},\forall z:|z|>R_0$$
