Show there is a point c that satisfy the inequality let $p(z)=z^n+a_{n-1}z^{n-1}+...+a_0$, be a polynomial of degree $n ≥ 1$. Show that there is a point $c$, $|c| = 1$ where $|p(c)| ≥ 1$.
Any hints on how to proceed
 A: Hint: Assume the contrary and estimate the absolute value of
$$
 1 = \frac{p^{(n)}(0)}{n!}
$$
using Cauchy's integral formula. 
A: Define $$f_1(z)=z^n,\quad f_2(z)=f_1(z)-f(z)=-a_{n-1}z^{n-1}-\cdots-a_0$$The sign is chosen because it makes the geometric argument below more intuitive.
Now let's evaluate $f(e^{it})$ for $0\leq t\leq 2\pi$. $f_1(z)$ just goes around the unit circle $n$ times. That means that if $|f(e^{it})|=|f_1(e^{it})-f_2(e^{it})|<1$ for all $t$, then $f_2(e^{it})$ must "keep up" with $f_1(e^{it})$. Specifically, $f_2(e^{it})$ can never be $0$, and it must also have winding number $n$ around $0$. But this is impossible, since its degree is $n-1$.
A: Orthogonality of the exponentials implies
$$\frac{1}{2\pi}\int_0^{2\pi} |p(e^{it})|^2\, dt = \frac{1}{2\pi}\int_0^{2\pi} |e^{int} + a_{n-1}e^{i(n-1)t} + \cdots +a_0|^2\, dt$$ $$ = 1^2+ |a_{n-1}|^2 + \cdots + |a_0|^2 \ge 1.$$
Since $|p(e^{it})|^2$ is continuous and the above is an average, there must exist a $t$ for which $|p(e^{it})|^2\ge 1,$ which implies $|p(e^{it})|\ge 1.$
