Evaluating the Complex Limit: Proving Existence and Value

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SUMMARY

The limit \(\lim_{z \to 1} \frac{\log{z}}{z-1}\) exists and equals 1, as established through the application of L'Hôpital's Rule and the properties of the logarithm. Specifically, the limit can be transformed to \(\lim_{z \to 0} \frac{\log(1+z)}{z}\), which confirms the result. The discussion emphasizes the importance of the principal branch of the logarithm and the convergence criteria for the series expansion of the logarithm when \(|z| < 1\).

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  • Understanding of complex analysis, particularly limits involving complex functions.
  • Familiarity with L'Hôpital's Rule for evaluating indeterminate forms.
  • Knowledge of the principal branch of the logarithm in complex analysis.
  • Concept of series expansion and convergence criteria for logarithmic functions.
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  • Study the application of L'Hôpital's Rule in complex analysis contexts.
  • Explore the properties and applications of the principal branch of the logarithm.
  • Learn about series expansions of complex functions, focusing on logarithmic series.
  • Investigate convergence criteria for complex series and their implications in limit evaluations.
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Students and professionals in mathematics, particularly those focusing on complex analysis, limit evaluation, and logarithmic functions.

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[SOLVED] complex limit

Homework Statement


Evaluate the complex limit if it exists:

\lim_{z \to 1} \frac{\log{z}}{z-1}

where log denotes the principal branch of the logarithm.

Homework Equations


The Attempt at a Solution


I am pretty sure it exists and equals 1, because that is what it equals when I approach with specific sequences. But how can I prove that?
 
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\lim_{z \to 1} \frac{\log{z}}{z-1} = \lim_{z\to 0} \frac{\log (1+z)}{z}.

Are you allowed to use the property that the series for that log term converges to that log term iff |z| < 1?
 

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