The problem I'm working on first asks to show that if $p(z)$ is a polynomial, then for some $\epsilon > 0$, there exists $z_0$ with $|z_0| = 1$ such that $$ |p(z_0) - 1/z_0| \geq \epsilon $$ I've already done this by considering the contour integral of $p(z) - 1/z$ on the unit circle oriented counterclockwise centered at $0$. I found that such $z_0$ exists whenever $0 < \epsilon \leq 1$.
The second part of the problem asks to find the "best possible $\epsilon$" and for that $\epsilon$, find an approximating polynomial. I believe that best $\epsilon$ is $1$, but I'm not completely sure. I'm also not sure how to find that approximating polynomial. I've tried to compute $|p(z) - 1/z|$ by hand but the summations and products quickly become cumbersome.
As you observed, no matter what $p$ is, the contour integral of $p(z)-1/z$ around the circle is $2\pi i$, and the length of the path is $2\pi$, so there is a point $z_0$ on the circle where $|p(z_0)-1/z_0|$ is at least $1$.
Now to make the maximum of a function with a fixed integral as small as possible, you want to make it constant (assuming this doesn't violate any other constraint). In other words you want $|p(z)-1/z| \equiv 1$. Can you do that? If so, what $p$ does the job?