Question on Rudin Theorem 3.44 Inequality

  • Context: Graduate 
  • Thread starter Thread starter jecharla
  • Start date Start date
  • Tags Tags
    Theorem
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
jecharla
Messages
24
Reaction score
0
I have a question about the last inequality Rudin uses in his proof of this theorem. Given that |z| < 1 he gets the inequality

|(1-z^(m+1)) / (1-z)| <= 2 / (1-z)

I think he is using the fact that |z| = 1, so

|(1-z^(m+1)) / (1-z)| <= (1 + |z^(m+1)|) / |1-z|

So i am guessing that

|z^(m+1)| < 1 since |z| < 1

But I don't know why this would be true?
 
Physics news on Phys.org
jecharla said:
I have a question about the last inequality Rudin uses in his proof of this theorem. Given that |z| < 1 he gets the inequality

|(1-z^(m+1)) / (1-z)| <= 2 / (1-z)

I think he is using the fact that |z| = 1, so

|(1-z^(m+1)) / (1-z)| <= (1 + |z^(m+1)|) / |1-z|

So i am guessing that

|z^(m+1)| < 1 since |z| < 1

But I don't know why this would be true?


Please, do USE Latex to write mathematics in this site!

[tex]\left|\frac{1-z^{m+1}}{1-z}\right|\leq \frac{|1|+|z|^{m+1}}{|1-z|}\leq\frac{1+1}{|1-z|}= \frac{2}{|1-z|}[/tex]

DonAntonio

Ps. Of course, [itex]\,|z|<1\Longrightarrow |z|^k<1\,\,\,\forall\,k\in\Bbb N[/itex]