Show that for any polynomial P(z) $max_{|z|=1}|\frac{1}{z^2} - P(z)| \geq 1$
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What have you tried so far? – Benedict W. J. Irwin Mar 20 '19 at 11:01
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2Assume that the maximum is $<1$. Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$. Apply the Argument principle together with this inequality to $1/z^2$ and $P$ to conclude that $1/z^2$ and $f$ have the same variation of the argument around $0$ along the loop $|z|=1$. But since this value is $-2$ for $1/z^2$, so should be for $P$. This implies that $P$ should have some poles inside $|z|<1$, which is a contradiction. – user647486 Mar 20 '19 at 11:37
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Essentially a duplicate of Show that for any polynomial $p(z)$ there is a $z$ with $|z|=1$ such that $|p(z)-1/z|\geq 1$.. – Martin R Mar 20 '19 at 12:18
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1Possible duplicate of Complex zeros of the polynomials $\sum_{k=0}^{n} z^k/k!$, inside balls – Lee David Chung Lin Mar 21 '19 at 00:56
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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.
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 $|z|<1$.
user647486
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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 $M \ge1.$
Fred
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