Complex analysis, P(z)

Question

Show that for any polynomial P(z) $max_{|z|=1}|\frac{1}{z^2} - P(z)| \geq 1$

Answer 1

Assume that the maximum is $<1$.

Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$.

Neither $1/z^2$ nor $P$ has poles on $|z|=1$ and by the inequality $P$ doesn't vanish on $|z|=1$, nor does $1/z^2$. This allows us to use the Argument principle to compute that $-2$ is the winding number of the image of the unit circle $|z|=1$ by $1/z^2$ around $0$.

Likewise, $0$ is the winding number of the image of $C$ by $P(z)$ around $0$.

Since $|1/z^2-P(z)|<1=|1/z^2|$, we have that $1/z^2$ is not a negative multiple of $P(z)$, for $|z|=1$. Therefore,

$$H(t,z)=t/z^2+(1-t)P(z)$$

provides a homotopy between those two curves that never touches $0$. Therefore, those two curves ought to have the same winding number with respect to $0$.

Contradiction.

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Or you could apply Rouche's theorem to $1$ and $z^2P(z)$, since $$|z^2P(z)-1|<|z^{2}|=1$$ for $|z|=1$. The theorem implies that $1$ and $z^2P(z)$ have the same number of zeros inside $|z|<1$. But $1$ never vanishes and $z^2P(z)$ has two zeros in $|z|<1$.

Answer 2

Consider the entire function $f(z)=1-z^2P(z)$. For $|z|=1$ we have

$$|\frac{1}{z^2} - P(z)| =|f(z)|.$$

Hence we have to show that $m:=\max \{|f(z)|: |z|=1\} \ge 1.$

If $M:=\max \{|f(z)|: |z| \le 1\}$, then the maximum - principle says:

$$M=m.$$

From $|f(0)|=1$, we get $M \ge1.$