Complex analysis - integral independent of path

Click For Summary
SUMMARY

The integral of the function \(\cos\frac{z}{2}\) along any curve \(C\) from \(0\) to \(\pi + 2i\) is independent of the path taken due to the analyticity of \(\cos(z/2)\). This allows for the simplification of the integral to evaluating the endpoints directly. The final evaluation involves calculating \(\left|2\sin\frac{z}{2}\right|^{z=\pi+2i}_{z=0}\), leading to the expression \(2\sin(\pi + 2i)\) as the solution.

PREREQUISITES
  • Understanding of complex analysis and analytic functions
  • Familiarity with contour integrals and path independence
  • Knowledge of the sine function in complex analysis
  • Basic skills in evaluating integrals in the complex plane
NEXT STEPS
  • Study the properties of analytic functions in complex analysis
  • Learn about contour integration techniques
  • Explore the evaluation of complex integrals using the residue theorem
  • Investigate the behavior of trigonometric functions in the complex plane
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on complex analysis, as well as anyone interested in evaluating complex integrals and understanding path independence in integrals.

player1_1_1
Messages
112
Reaction score
0

Homework Statement


integral: \int\limits_C\cos\frac{z}{2}\mbox{d}z where C is any curve from 0 to \pi+2i

The Attempt at a Solution


can i do this like in real analysis when counting work between two points, just count this integral and put given data in?
 
Physics news on Phys.org
Yes, since [math]cos(z/2)[/math] is an analytic function, the integral is independent of the path, so you can just go ahead and integrate, the set z= \pi+ 2i and 0.
 
and then: \ldots=\left|2\sin\frac{z}{2}\right|^{z=\pi+2i}_{z=0}=2\sin(\pi+2i) yeah?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K
Replies
32
Views
4K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
9
Views
2K
Replies
3
Views
2K
Replies
3
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K