If $P$ is a polynomial, then there exists $z_0$ such that $|z_0|=1$ and $|P(z_0)-\frac{1}{z_0}|\geq 1$.

Question

If $P$ is a polynomial, then there exists $z_0$ such that $|z_0|=1$ and $|P(z_0)-\frac{1}{z_0}|\geq 1$.

Assume P is a polynomial of degree $m$.

Assume for all $|z|=1$, we have $|p(z)-\frac{1}{z}|<1$, so $|zp(z)-1|<|z|=1$, so $zp(z)-1$ has $m$ roots outside the disc $D(0,1)$, but I cannot go further. I appreciate any help.

Answer 1

Hint Use the Maximum Modulus Principle to the function $f(z)=zP(z)-1$ on the domain $\overline{D}=\{z| |z|\leq 1\}$. The maximum of $|f(z)|$ is attained on $z_0$ such that $|z_0|=1$ and $$ |f(z_0)|\geq |f(0)|. $$

Answer 2

Or, using Cauchy's integral formula, because $f(z)=zP(z)-1$ is holomorphic: $$f(0)=\frac{1}{2 \pi i} \int\limits_{|z|=1}\frac{f(z)}{z}dz$$ then $$\color{red}{1}=|f(0)|=\left|\frac{1}{2 \pi i} \int\limits_{|z|=1}\frac{f(z)}{z}dz\right|\leq \frac{1}{2 \pi}\int\limits_{|z|=1} \left|\frac{f(z)}{z}\right||dz|\color{red}{<} \frac{1}{2 \pi}\int\limits_{|z|=1} |dz|=\color{red}{1}$$ contradiction.