# Find a polynomial $P_n$ such that the closed disk of radius $\frac{1}{n}$ is not contained in $P_n(\mathbb{D})$ [closed]

I try to solve the following problem:

Let $\mathbb D=\{z\in \mathbb C: |z| < 1\}$. For an integer $n\geq 1$, find a polynomial of degree $n$ $$P_n(z)=z+a_2z^2+\cdots +a_nz^n$$ such that the closed disk of radius $\frac{1}{n}$ is not contained in $P_n(\mathbb{D})$.

My attempt:

To solve this, I want to directly find a polynomial such that $P_n(\mathbb{D})$ is strictly contained in a circle of radius $\frac{1}{n}$. And if $|z|<1$, we have $$|P_n(z)|=|z+a_2z^2+\cdots +a_nz^n|\leq |z|+|a_2z^2|+\cdots +|a_nz^n|<1+a_2+\cdots a_n$$ Though it looks hopeless to get rid of the term ''$1+\cdots$'' so it cannot be less than $\frac{1}{n}$ by some coarse estimates...

I also try to consider the polynomial $$f(z)=z-\left(1-\frac{1}{n}\right)z^n$$

why I consider this is its image has some symmetric property (preserved after rotation by $\frac{\pi}{n-1}$). Due to this, we can reduce all disk to some disk ''near'' the real line.

Note that we need a normalized polynomial, namely, $f(0)=0$ and $f'(0)=1$. In that sense, some classic tricks, such as scaling and translation, are not allowed anymore.

## closed as unclear what you're asking by zhw., Nosrati, Claude Leibovici, Jack, user223391 Dec 4 '17 at 19:57

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• What does that even mean? Why do you expect us to know what these symobls mean? – zhw. Nov 30 '17 at 1:03
• @zhw. I am sorry, which symbols are ambiguous? – Aolong Li Nov 30 '17 at 1:05
• It means the closed disk should not be contained in the image of the open disk. – Robert Israel Nov 30 '17 at 2:22
• @stressed-out one of the disk is closed and the other is open. Just like what Robert said. – Aolong Li Nov 30 '17 at 2:30
• @stressed-out Notations differ. $\mathbb D$ or $D$ for the open unit disk is pretty common in complex analysis. See e.g. Wikipedia. – Robert Israel Nov 30 '17 at 3:21

Try $P_n(z) = \frac{1 - (1-z)^n}{n}$. Note that $P_n(z) = 1/n$ only for $z=1$ (which is not in the open unit disk).
• Thanks a lot! But I think your hint only can indicate that it cannot contain a closed disk whose center is $O$. But what if the circles centered at some other positions? – Aolong Li Nov 30 '17 at 22:55
• According to Landau's theorem, there is a positive constant $L$ such that $P_n(\mathbb D)$ always contains a disk of radius $L$. – Robert Israel Dec 1 '17 at 0:00