Quick question regarding Rudin's proof of the Fundamental Theorem of Algebra

Question

The first part of the proof needs some fixing, as mentioned here.

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I'm struggling to understand the inequality $$|Q(re^{i\theta })|\leq 1-r^k|b_k|-r^{k+1}|b_{k+1}|-\ldots -r^n|b_n|$$

I'm guessing it makes use of $$|1+r^ke^{i\theta k}b_k|=1-r^k|b_k|$$

yet I don't see where the other $r^i|b_i|$ come from.

Answer 1

We have that

$Q(re^{i\theta}) = 1 + r^ke^{ki\theta}(b_k + re^{i\theta}b_{k+1} + ... +r^{n-k}e^{(n-k)i\theta}b_n )$

so that

$|Q(re^{i\theta})| = |1 + r^ke^{ki\theta}(b_k + re^{i\theta}b_{k+1} + ... +r^{n-k}e^{(n-k)i\theta}b_n )| \\ \leq |1 + e^{ki\theta}r^kb_k | + |r^k(re^{i\theta}b_{k+1} + ... +r^{n-k}e^{(n-k)i\theta}b_n )| $

by the triangle inequality; but we have that $|1+r^ke^{ii\theta k}b_k|=1-r^k|b_k|$ which implies that

$|1 + e^{ki\theta}r^kb_k | + |r^k(re^{i\theta}b_{k+1} + ... +r^{n-k}e^{(n-k)i\theta}b_n )| = 1 - |r^kb_k | + |r^k(re^{i\theta}b_{k+1} + ... +r^{n-k}e^{(n-k)i\theta}b_n )| $

and using the triangle inequality and the fact $|e^{ii\theta}z| = |z|$ for any complex number we have that

$1 - |r^kb_k | + |r^k(re^{i\theta}b_{k+1} + ... +r^{n-k}e^{(n-k)i\theta}b_n )| \leq 1 - |r^kb_k | + |r^{k+1}b_{k+1}| + ... +|r^nb_n | \\ = 1-r^k(|b_k|-r|b_{k+1}|-\ldots -r^{n-k}|b_n|)$

which gives us the desired

$|Q(re^{i\theta })|\leq 1-r^k(|b_k|-r|b_{k+1}|-\ldots -r^{n-k}|b_n|)$