Is the Integral Zero for Closed Paths in Complex Analysis?

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SUMMARY

The integral of a function over closed paths in complex analysis is zero if the function is analytic within the enclosed region. The discussion highlights that if the modulus of a function, denoted as ##\left|f(z)\right|##, is less than or equal to zero, it must be equal to zero since it cannot be negative. The key takeaway is that the integral is zero due to the properties of analytic functions and the application of the Cauchy-Goursat theorem, which states that integrals of analytic functions over closed paths yield zero.

PREREQUISITES
  • Understanding of complex analysis concepts, particularly analytic functions.
  • Familiarity with the Cauchy-Goursat theorem.
  • Knowledge of complex function properties, including modulus and inequalities.
  • Basic skills in evaluating integrals in the complex plane.
NEXT STEPS
  • Study the Cauchy-Goursat theorem in detail.
  • Explore the implications of analytic functions on contour integrals.
  • Learn about the properties of complex functions and their moduli.
  • Investigate examples of closed path integrals in complex analysis.
USEFUL FOR

Students and professionals in mathematics, particularly those focused on complex analysis, as well as educators seeking to clarify the concept of integrals over closed paths.

Gwinterz
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Hey, I have been stuck on this question for a while:

WClcaAN.jpg

I have tried to follow the hint, but I am not sure where to go next to get the result.



Have I started correctly? I am not sure how to show that the integral is zero.

If I can show it is less than zero, I also don't see how that shows it is always zero.

Thanks in advance for any help.
 

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The condition in the premise is on ##\left|f(z)\right|##, not ##f(z)##. If you have in fact shown that ##\left|f(z)\right| \leq 0##, then it must be equal to 0 since it can't be negative.
 
Gwinterz said:
Hey, I have been stuck on this question for a while:

View attachment 232707
I have tried to follow the hint, but I am not sure where to go next to get the result.



Have I started correctly? I am not sure how to show that the integral is zero.

If I can show it is less than zero, I also don't see how that shows it is always zero.

Thanks in advance for any help.

Something cannot be less than zero and equal to zero at the same time. However, since you have non-strict inequalities "##\leq##" there is a chance you can show the thing is ##\leq 0##. Then (being a norm in the complex plane) it must also be ##\geq 0##, hence must ##= 0.##
 
Thanks guys that makes sense.

Is the integral equal to zero because its a closed path?
 

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