Rudin theorem 3.44

  • Thread starter jecharla
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  • #1
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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?
 

Answers and Replies

  • #2
606
1
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]
 

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