1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cos and Cosign Equation help

  1. May 6, 2005 #1
    Hello, can someone please help me out with the following question?

    Express [tex]\sinh \left( {x + yi} \right)[/tex] in terms of [tex]\sinh (x)[/tex], [tex]\cosh(x)[/tex], [tex]\cos(y)[/tex] and [tex]\sin(y)[/tex]. Hence find all solutions [tex]z \in C[/tex] of the equation [tex]\sinh(z) = i[/tex].

    I have little idea as to how to do this question. Here is my working.

    [tex]\sinh \left( {x + yi} \right) = \sinh \left( x \right)\cosh \left( {iy} \right) + \cosh \left( x \right)\sinh \left( {iy} \right)[/tex]

    [tex] = \sinh \left( x \right)\left( {\frac{{e^{iy} + e^{ - iy} }}{2}} \right) + \cosh \left( x \right)\left( {\frac{{e^{iy} - e^{- iy} }}{2}} \right)[/tex]

    [tex]{\rm = sinh}\left( {\rm x} \right)\left( {\frac{{e^{iy} + e^{iy} }}{2}} \right) + i\cosh \left( x \right)\left( {\frac{{e^{iy} - e^{ - iy} }}{{2i}}} \right)[/tex]

    [tex] = \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right)[/tex]

    So [tex]\sinh \left( z \right) = i \Rightarrow \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right) = i[/tex]

    Equating real and imaginary parts:

    [tex]\sinh \left( x \right)\cos \left( y \right) = 0...(1)[/tex]

    [tex]\cosh \left( x \right)\sin \left( y \right) = 1...(2)[/tex]

    At this point I am unsure of how to proceed. The equations do not look solvable, directly anyway. So I decided to start off with equation 1 and see where that would lead. Here is what I have done.

    [tex]\sinh \left( x \right)\cos \left( y \right) = 0[/tex]

    [tex] \Rightarrow \cos (y) = 0[/tex] or [tex]\sinh (x) = 0[/tex]

    [tex]\frac{{e^x - e^{ - x} }}{2} = 0 \Rightarrow x = 0[/tex] and [tex]y = \frac{{n\pi }}{2}[/tex] where n is an odd ineger.

    Now since equations (1) and (2) need to be satisfied simultaneously then x is necessarily equal to zero and equation (2) reduces to [tex]\sin \left( y \right) = 1[/tex] which has solutions: [tex]y = \frac{\pi }{2} + 2k\pi[/tex] where k is an integer.

    I need y to satisfy both equations (1) and (2) so I take y to be the 'least general' answer so that [tex]y = \frac{\pi }{2} + 2k\pi[/tex].

    So I end up with [tex]\sinh \left( z \right) = i \Leftrightarrow z = \left( {\frac{\pi }{2} + 2k\pi } \right)i[/tex]. I'm not sure if my method is right. I pretty much got stuck at the simultaneous equations part. Can someone please hep me out with this question?
     
  2. jcsd
  3. May 6, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It looks okay to me...

    You may have tried to check whether

    [tex] \sinh\left(\frac{i\pi}{2}+2ki\pi\right)=i [/tex]


    Daniel.
     
  4. May 6, 2005 #3
    Thanks for your response dextercioby. I have checked that my solution works for k = 0 but not for any other values.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cos and Cosign Equation help
  1. Cos inverse help (Replies: 8)

  2. Help with Cos and Sin (Replies: 3)

  3. Sin/Cos In Equation (Replies: 3)

Loading...