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

Gold Member

## 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.

fresh_42
Mentor
2021 Award
Homework Helper
Gold Member
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.

Ray Vickson
Homework Helper
Dearly Missed

## 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.

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.

Delta2
Homework Helper
Gold Member
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:
robphy
Homework Helper
Gold Member
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".

Delta2
Homework Helper
Gold Member
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 .