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I am trying to show that the noted inequality holds without using the Euclidean Algorithm. This appears in Complex Analysis by Newman and Bak on the top of page 63 (remark 1, my paraphrase):

If $\alpha$ is a zero of an $n^{th}$ degree polynomial $P_n$, then $P_n(z)=(z-\alpha)P_{n-1}$, where $P_{n-1}$ is a polynomial of degree $n-1$. This can be seen by noting that

$$|\frac{P_n(z)}{z-\alpha}|\le A+B|z|^{n-1}$$

and hence is equal to an $(n-1)^{st}$ degree polynomial by the Extended Liouville Theorem.

I am having trouble showing the noted inequality. Furthermore, even if the inequality holds, doesn't the Extended Liouville Theorem merely say that the degree of the polynomial is at most $n-1$?

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I suggest a completely different approach:

From the geometric sum formula (which I assume you are familiar with) $$ (z^k-\alpha^k)=(z-\alpha)(z^{k-1}+z^{k-2}\alpha+\cdots+z\alpha^{k-2}+\alpha^{k-1})\tag{1}. $$

Since $P$ is a polynomial of degree $n$, it is on the form $$ P(z)=a_nz^n+\cdots+a_1z+a_0. $$ In particular, if $\alpha$ is a zero of $P$, then $$ \begin{aligned} P(z)&=P(z)-P(\alpha)\\ &=a_n(z^n-\alpha^n)+\cdots+a_k(z^k-\alpha^k)+\cdots+ a_1(z-\alpha)+a_0-a_0 \end{aligned} $$ Here, we can factor $z-\alpha$ from each term according to $(1)$. From the first term, we get the highest order term, $$ a_n(z^n-\alpha^n)=(z-\alpha)(a_n z^{n-1}+\cdots). $$ Thus, we condlude that $$ P(z)=(z-\alpha)\tilde{P}(z), $$ where $\tilde{P}$ is a polynomial of degree $n-1$ (the coefficient in front of $z^{n-1}$ is $a_n$).

Now, I guess you can proceed proving the inequality (if you really want to).

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  • $\begingroup$ Thank you! This was a much clearer way of seeing things. $\endgroup$ – Greyson Jun 6 '15 at 22:58

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