Proving |z|<1 and n is a positive integer: Complex Analysis Proof

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SUMMARY

The discussion focuses on proving the inequality \(\left|\frac{1-z^n}{1-z}\right|\le n\) for complex numbers \(z\) where \(|z|<1\) and \(n\) is a positive integer. The proof involves differentiating the function \(\frac{1-z^n}{1-z}\) to find its maximum, leading to the conclusion that \(\left|z^{n-1}\right|\le 1\) under the given conditions. An alternative proof using the geometric series expansion \(\left|\sum_{k=0}^{n-1}z^k\right|\) is also presented, confirming the inequality holds true.

PREREQUISITES
  • Complex analysis fundamentals
  • Understanding of differentiation in complex functions
  • Knowledge of geometric series
  • Familiarity with absolute values in complex numbers
NEXT STEPS
  • Study the properties of complex functions and their differentiability
  • Learn about the geometric series and its convergence criteria
  • Explore advanced topics in complex analysis, such as Cauchy's integral theorem
  • Investigate other inequalities in complex analysis, such as the Maximum Modulus Principle
USEFUL FOR

Students and educators in mathematics, particularly those studying complex analysis, as well as anyone interested in proving inequalities involving complex numbers.

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Homework Statement


Given |z|<1 and n a positive integer prove that
[tex] \left|\frac{1-z^n}{1-z}\right|\le n[/tex]

The Attempt at a Solution


I try to find the maximum of the function by differentiation

[tex] \frac{d}{dz}\frac{1-z^n}{1-z}=\frac{-nz^{n-1}*(1-z)+(1-z^n)}{(1-z)^2}=0\Rightarrow (1-z^n)=nz^{n-1}*(1-z)[/tex]

I then plug this in

[tex] \left|\frac{nz^{n-1}*(1-z)}{1-z}\right|=n\left|z^{n-1}\right|\le n[/tex]

I guess this works. Does someone have another way to prove it?
 
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Divide the expression inside absolute values and see what you have - use the bound on |z|.
 
Ah that is clever

[tex] \left|\frac{1-z^n}{1-z}\right|=\left|\sum_{k=0}^{n-1}z^k\right|\le \sum_{k=0}^{n-1}|z|^k\le n[/tex]
 

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