Question on Rudin Theorem 3.44 Inequality

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The discussion centers on Rudin's Theorem 3.44, specifically the inequality \(\left|\frac{1-z^{m+1}}{1-z}\right| \leq \frac{2}{|1-z|}\) for complex numbers \(z\) where \(|z| < 1\). The user queries the validity of the assumption that \(|z^{m+1}| < 1\) given \(|z| < 1\). The conclusion drawn is that since \(|z| < 1\), it follows that \(|z|^k < 1\) for all natural numbers \(k\), confirming the inequality holds true. The discussion emphasizes the importance of understanding the properties of complex numbers in this context.

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  • Understanding of complex numbers and their properties
  • Familiarity with Rudin's Theorem 3.44
  • Knowledge of inequalities in mathematical proofs
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  • Review the proof of Rudin's Theorem 3.44 in "Principles of Mathematical Analysis" by Walter Rudin
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jecharla
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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?
 
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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]
 

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