Show that tanh(z/2) = (sinhx+isiny)/(coshx+cosy) Arfken book

  • Thread starter Felipe Lincoln
  • Start date
  • Tags
    Book
In summary: However, this definition does not include the imaginary part, while the one you started with does. In general, the imaginary part appears only in the numerator, while the left side of the equation that you start with will have a complex term in the denominator. There is a trick to getting rid of a complex term in the denominator. Hopefully this is helpful.
  • #1
Felipe Lincoln
Gold Member
99
11

Homework Statement


Show that ##\tanh(z/2) = \dfrac{\sinh x+i\sin y}{\cosh x+\cos y}##

Homework Equations


Not clear what I'm supposed to use

The Attempt at a Solution


Tried using the relation of sin and cosine with ##e^{iz}##, and also worked with the expanded ##\sin(x+iy)## and cosine as well but I'm still stuck.
 
Physics news on Phys.org
  • #3
The answer has an imaginary part only in the numerator, while the left side of the equation that you start with will have a complex term in the denominator. There is a trick to getting rid of a complex term in the denominator. Hopefully this is helpful.
 
  • #4
Felipe Lincoln said:

Homework Statement


Show that ##\tanh(z/2) = \dfrac{\sinh x+i\sin y}{\cosh x+\cos y}##

Homework Equations


Not clear what I'm supposed to use

The Attempt at a Solution


Tried using the relation of sin and cosine with ##e^{iz}##, and also worked with the expanded ##\sin(x+iy)## and cosine as well but I'm still stuck.

Please read the "Guidelines".

You will see that your statements that you do not know what you are supposed to use, and that you are stuck (but with no explanation) are not considered as a proper attempt, and are thus against the PF rules. You need to show your work, and in enough detail that helpers can tell how to give appropriate hints.
 
  • #5
start from the definition of ##\tanh(u)=\frac{e^{2u}-1}{e^{2u}+1}## plug in ##u=z/2## and then plug in ##z=x+iy##, then use ##e^{iy}=\cos y+i\sin y## and then do a trick that @Charles Link mention to get rid of the imaginary part in the denominator, that is you ll have to multiply both the numerator and the denominator, by the complex conjugate of the denominator. Then you ll have to work out the algebra. Might have to use some trigonometry identities as well..
 
Last edited:
  • Like
Likes Felipe Lincoln and Charles Link
  • #6
Delta² said:
start from the definition of ##\tanh(u)=\frac{e^{2u}-1}{e^{2u}+1}## plug in ##u=z/2## and then plug in ##z=x+iy##, then use ##e^{iy}=\cos y+i\sin y## and then do a trick that @Charles Link mention to get rid of the imaginary part in the denominator, that is you ll have to multiply both the numerator and the denominator, by the complex conjugate of the denominator. Then you ll have to work out the algebra. Might have to use some trigonometry identities as well..
A more intuitive starting point is
##\tanh u=\frac{\sinh u}{\cosh u}=\frac{(e^u-e^{-u})/2}{(e^u+e^{-u})/2}.##
Then, multiply by ##1=\frac{e^u}{e^u}## to get your "definition".
 
  • #7
robphy said:
A more intuitive starting point is
##\tanh u=\frac{\sinh u}{\cosh u}=\frac{(e^u-e^{-u})/2}{(e^u+e^{-u})/2}.##
Then, multiply by ##1=\frac{e^u}{e^u}## to get your "definition".
Ultimately , you are right, the strict definition of ##\tanh u## is as you say .
 

Related to Show that tanh(z/2) = (sinhx+isiny)/(coshx+cosy) Arfken book

1. What is the definition of tanh(z/2)?

Tanh(z/2) is a hyperbolic tangent function, which is defined as (e^z - e^-z) / (e^z + e^-z), where z is a complex number.

2. How can you prove that tanh(z/2) = (sinhx+isiny)/(coshx+cosy)?

This equation can be proved by using the definitions of sinh, cosh, and tanh, along with some basic properties of complex numbers.

3. What is the significance of the expression tanh(z/2) in complex analysis?

In complex analysis, tanh(z/2) is commonly used to map points from the complex plane onto the unit disc. It is also useful in solving certain differential equations and in evaluating integrals.

4. How is the equation tanh(z/2) = (sinhx+isiny)/(coshx+cosy) related to the Möbius transformation?

The Möbius transformation is a special case of the more general conformal mapping, and tanh(z/2) is one of the simplest examples of a conformal mapping. The relationship between the two is that the Möbius transformation can be obtained by composing two tanh(z/2) mappings.

5. Are there any practical applications of the equation tanh(z/2) = (sinhx+isiny)/(coshx+cosy)?

Yes, this equation has applications in various fields such as physics, engineering, and mathematics. It is commonly used in solving problems involving heat conduction, fluid flow, and electromagnetic fields.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
44
Views
4K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
486
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
914
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top