Maximum value of $|z^{-1}-p(z)|$ for any polynomial $p(z)$ Show that for any polynomial $p(z)$, $\max_{|z|=1}|z^{-1}-p(z)|\geq 1$. I've tried using $z=e^{i\theta}$ but haven't gotten anywhere. Don't really know where to start!
 A: Suppose not.  Then we have $$\left|\int_{|z|=1}\left(\frac1z-p(z)\right)\mathrm{d}z\right|<\int_{|z|=1}\mathrm{d}|z|=2\pi$$
On the other hand, we have by Cauchy's integral theorem and formula, $$\int_{|z|=1}\left(\frac1z-p(z)\right)\mathrm{d}z=2\pi i$$ contradiction.
The first statement needs a little bit of argument that the integrand is bounded away from $1$ on $|z|=1$ perhaps, but that's obvious by continuity.
A: $1-zp(z)$ is a holomorphic function in the unit disc that is $1$ at zero, hence its maximum modulus is bigger than $1$ on the unit circle. But for $|z|=1$, $|z^{-1}-p(z)|=|z||z^{-1}-p(z)|=|1-zp(z)|$, so we are done!
A: Here is a slightly different solution:
Suppose there is a polynomial $p_0(z)$ such that $\sup_{|z|=1}\big|\tfrac{1}{z}-p_0(z)\big|<1$
On $|z|=1$,
$$\big|\tfrac{1}{z}-p_0(z)\big|=|z|\big|\tfrac{1}{z}- p_0(z)\big|=|1-z p_0(z)|<1$$
Then, by Rouché's theorem, $f(z)\equiv1$ and $q(z)=zp(z)$ have the same number of roots inside the unit ball $B(0;1)$. The polynomial $q$ has at least one root ($z=0$) but $f$ has none. This leads to a contradiction and so, no such $p_0$ exists
