MHB Help with Hyperbolic Functions

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To solve the problem involving hyperbolic functions, the function w=f(z)=coth(z/2) can be expressed in exponential form. By using the identity \(\coth(u)=(e^u + e^{-u})/(e^u - e^{-u})\), the function simplifies to \(f(z)=(e^{z/2} + e^{-z/2})/(e^{z/2} - e^{-z/2})\). Multiplying the numerator and denominator by \(e^{z/2}\) yields \(f(z)=(e^{z} + 1)/(e^{z} - 1)\). Substituting \(C=e^{z}\) confirms that w=f(z)=h(C)=(C+1)/(C-1). This demonstrates the relationship between the hyperbolic function and the exponential function.
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Hi guys, I just need some help with a complex variables question
I need to "Let w=f(z)=coth(z/2). Show that w=f(z)=h(C) = (C+1)/(C-1) where C=g(z)=e^z"Thanks guys, any help will be super appreciated and best answer given!
CB
 
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I need to "Let w=f(z)=coth(z/2). Show that w=f(z)=h(C) = (C+1)/(C-1) where C=g(z)=e^z"
Write \(\coth(z/2)\) in exponential form using:

\[\coth(u)=( e^u + e^{-u} )/( e^u - e^{-u} )\]

Then:

\[f(z)=( e^{z/2} + e^{-z/2} )/( e^{z/2} - e^{-z/2} )\]

now multiply top and bottom by \(e^{z/2}\) to get:

\[f(z)=( e^{z} + 1 )/( e^{z} - 1 )\]

so when you substitute \(C=e^{z}\) you are done.

CB
 
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