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.
A more intuitive starting point isDelta² 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..
Ultimately , you are right, the strict definition of ##\tanh u## is as you say .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".
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.
This equation can be proved by using the definitions of sinh, cosh, and tanh, along with some basic properties of complex numbers.
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.
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.
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.