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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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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

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  • $\begingroup$ 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? $\endgroup$ – Aolong Li Nov 30 '17 at 22:55
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    $\begingroup$ According to Landau's theorem, there is a positive constant $L$ such that $P_n(\mathbb D)$ always contains a disk of radius $L$. $\endgroup$ – Robert Israel Dec 1 '17 at 0:00
  • $\begingroup$ No. Bloch’s Theorem said that we have a disk in the domain, not the image. Also, I think “closed” is not satisfied, which actually doesn’t matter. $\endgroup$ – Aolong Li Dec 1 '17 at 0:05
  • $\begingroup$ Actually Landau's theorem, although Conway calls it Bloch's theorem. $\endgroup$ – Robert Israel Dec 1 '17 at 0:07
  • $\begingroup$ Oh! I see! I misunderstood what it asks. $\endgroup$ – Aolong Li Dec 1 '17 at 0:15

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