Consequence of Cauchy's Integral Formula

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    Formula Integral
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The discussion revolves around proving that if a function f is analytic inside and on the unit circle with a bounded magnitude |f(z)| ≤ M for |z| = 1, then it follows that |f(0)| ≤ M and |f'(0)| ≤ M. Participants express confusion about the implications of the bounded condition and seek clarification on how to apply Cauchy's Integral Formula. It's noted that the magnitude of f being bounded on the unit circle indicates that the function does not exceed M in that region. Additionally, the conversation touches on estimating |f^n(0)|, suggesting that further exploration of Cauchy's formulas is necessary. The thread emphasizes the importance of understanding the conditions of analyticity and boundedness in complex analysis.
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Homework Statement


Suppose f is analytic inside and on the unit circle |z|=1. Prove that if |f(z)|\leqM for |z|=1, then |f(0)|\leqM and |f'(0)|\leqM.

Finally, what estimate can you give for |f^n(0)|?

Homework Equations


This problem is in a section dealing with Cauchy's Integral Formula and it's generalized form.
http://en.wikipedia.org/wiki/Cauchy_integral_formula

The Attempt at a Solution


Im not quite sure how to get started here. Obviously Cauchy's formulas apply to our function f considering its analiticity. I guess I don't really understand the if part of our statement. IE if |f(z)|\leqM for |z|=1. The magnitude of the function f is bounded? I don't know what it means "for |z|=1." I'm missing something here. A nudge in the right direction would be appreciated. Thanks!
 
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The absolute value of the integral of a function is less or equal to the integral of the absolute value of the function times the length of the curve you are integrating over. The 'a' the Cauchy formula is zero. |z|=1. The function f is bounded on |z|=1 by M. That's a nudge.
 
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