Proving the Complex Hyperbolic Property Using Trigonometric Construction

Click For Summary
SUMMARY

The discussion focuses on proving the complex hyperbolic property, specifically showing that sinh(z + i2π) = sinh(z) using the definition of sinh(z) as (e^z - e^(-z))/2. The user attempts to manipulate the equation by defining z' = x + i(2π + y) but recognizes that this leads to sinh(z') instead of sinh(z). The solution involves applying the laws of exponents and leveraging trigonometric identities to reach the conclusion.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically sinh
  • Familiarity with complex numbers and their properties
  • Knowledge of Euler's formula, e^(iθ) = cos(θ) + i sin(θ)
  • Basic proficiency in manipulating exponential expressions
NEXT STEPS
  • Study the properties of hyperbolic functions in complex analysis
  • Learn about Euler's formula and its applications in trigonometric identities
  • Explore the laws of exponents in the context of complex numbers
  • Practice problems involving hyperbolic functions and complex variables
USEFUL FOR

Mathematics students, particularly those studying complex analysis or hyperbolic functions, as well as educators looking for examples of trigonometric constructions in proofs.

nichkis
Messages
2
Reaction score
0

Homework Statement



Show: sinh(z + i2(pi)) = sinh(z) using sinh(z) = (ez - e-z)/2

Homework Equations


The Attempt at a Solution



So far I have (ex + i(2∏+y) - e-(x+i(2∏+y))/2.

Need help proceeding from here. My thoughts were to define a z' = x + i(2∏+y) but I don't think that I can then say that it is equal to sinh(z) but rather sinh(z').
 
Physics news on Phys.org
What is [itex]e^{2 \pi i}[/itex]? Then use the laws of exponents.
 
Last edited:
Using the trigonometric construction I was able to get to the answer. Thanks a bunch.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K