Let $n>k$ and $p(z)=z^n+az^k$. Then I've got to show the following:

$p(\mathbb{D}) \subset \mathbb{D} \Rightarrow a=0$, where $\mathbb{D}=\{z \in \mathbb{C} | |z|<1\}$.

I've tried it per contraposition. That is if $a\neq 0$ then $p(\mathbb{D})$ is not in $ \mathbb{D} $. That is $|z^n+az^k|<|1+a|$, because $|z^n|$ and $|z^k|$ are both less than one in $\mathbb{D}$. Is that right?

  • i think your argument implies that $p(\mathbb{D}) \subseteq 1 + |a| \cdot \mathbb{D}$, which is not enough to imply $p(\mathbb{D}) \nsubseteq \mathbb{D}$ – JayTuma Sep 14 at 13:39
  • But when it's shown $p(\mathbb{D}) \subset 1 + |a|* \mathbb{D} $. Then for any $a \neq 0$ it follows directly that $p(\mathbb{D})$ is not in $\mathbb{D}$ – Thesinus Sep 14 at 13:52

Since $p$ is continuous on the whole complex plane, $p(\mathbb D) \subseteq \mathbb D$ also tells us that $p(\overline{\mathbb D}) \subseteq \overline{\mathbb D}$.

We have $$ |p(z)| = |z^k||z^{n-k} + a|. $$ Now if $a \neq 0$, we can consider $$ \left|p\left(\frac{\sqrt[n-k]{a}}{\left|\sqrt[n-k]{a}\right|}\right)\right| = \left|\left(\frac{\sqrt[n-k]{a}}{\left|\sqrt[n-k]{a}\right|}\right)^k\right|\left|\left(\frac{\sqrt[n-k]{a}}{\left|\sqrt[n-k]{a}\right|}\right)^{n-k} + a\right| = 1 \times \left|\frac{a}{|a|} + a\right| = 1 + |a| > 1. $$

  • Why do you put in $\sqrt[n-k]{a}$ in $p$? – Thesinus Sep 14 at 14:31

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