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

  • #1
Felipe Lincoln
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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.
 

Answers and Replies

  • #3
Charles Link
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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
Ray Vickson
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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
Delta2
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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..
 
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  • #6
robphy
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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
Delta2
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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 .
 

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