Complex Analysis: Solving Logarithmic Equations and Limits

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SUMMARY

The discussion focuses on solving logarithmic equations and limits in complex analysis, specifically evaluating ln(i), ln(-i), and i^i. The values derived are ln(i) = iπ/2, ln(-i) = -iπ/2, and i^i = exp(-π/2). Additionally, the limit of ln(zo + e) - ln(zo + i*e) as e approaches 0 is determined to be 0 for both zo = -1 - i and zo = 1 + i, highlighting the continuity of the logarithmic function away from its branch cut on the negative real axis.

PREREQUISITES
  • Complex number theory
  • Understanding of logarithmic functions in the complex plane
  • Knowledge of limits and continuity in calculus
  • Familiarity with the exponential function and its properties
NEXT STEPS
  • Study the properties of complex logarithms and their branch cuts
  • Learn about the continuity of complex functions and their limits
  • Explore the implications of the exponential function in complex analysis
  • Investigate further examples of limits involving complex variables
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on complex analysis, as well as educators preparing materials on logarithmic equations and limits in the complex domain.

Dassinia
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Hello,
I'm solving the problems given in previous exams, and there's this question:

Homework Statement



a/ Give the value of ln(i), ln(-i) and i^i
b/ If zo=-1-i , what is the value of

lim [ ln(zo+e)-ln(zo+i*e) ] when e-> 0
Same question with zo=1+i


Homework Equations





The Attempt at a Solution


a. ln(i)=ln|i|+i*arg(i)=i*pi/2
ln(-i)=-i*pi/2
i^i=exp(-pi/2)
b. I found that when zo=-1-i and zo=1+i the limit is 0, I don't know if I'm missing something, this question is so easy that it seems suspect

Thanks !
 
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There is a cut line where the limits will differ. Since exp is a many to one function in the complex plane we have to make a restriction on it's domain when we invert. But that cut is typically made on the negative real ray so only there do you have discontinuity in the Ln function. But away from there it is a continuous function so limit=value.

Easy \ne Simple, sometimes the easy answer is easy because you understand some very non-simple concepts. Sometimes a simple answer isn't easy too.
 

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