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Very difficult complex no question

  1. Nov 27, 2005 #1
    very difficult complex no question plzzzz help

    Hi Guys!
    well! im facing problems in the following question:
    [tex]\tan{x+iy} = \alpha + i\beta[/tex]
    prove that
    [tex]\alpha^2 + \beta^2 = \frac{\cosh{(h^2)(y)}-(\cos{x}^2)}{\cosh{(h^2)(y)}-(\sin{x}^2)}[/tex]

    what im not getting here is this that you have to prove that

    [tex]\alpha^2 + \beta^2[/tex]

    how to transform this question to give [tex]\alpha^2 + \beta^2[/tex]

    now i can make it like

    [tex]\alpha^2 - \beta^2[/tex]

    but because of that i , im unable to transform it into

    [tex]\alpha^2 + \beta^2[/tex]

    plz help me. I have just one day remaining in my exams.


    Yeah! if you are unable to read the above code im attaching an attachment as well.

    Thanks and Good Bye

    Attached Files:

  2. jcsd
  3. Nov 27, 2005 #2


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    Homework Helper

    HINT: [itex](\alpha + i \beta)(\alpha - i \beta) = \alpha^2 + \beta^2[/itex]
  4. Nov 28, 2005 #3
    Im still unable to solve this
    im so near to the required answer but couldnt solve. Can you tell me if there is any special thing to keep in mind while solving this question.

    Thanks in advance
  5. Nov 28, 2005 #4
    see if im correct here

    Hi guys!

    thanks for your help.

    Well! i want to show you how much i have done this:

    i have written this


    so when i solve this thing

    im getting an answer like this

    [tex]\tan{(x+iy)}=\frac{\sin{(x+iy)}}{\cos{(x+iy)}}= \frac{\sin{2x}+i\sinh{2y}}{\cos{2x}+\cosh{2y}}=\alpha + i\beta[/tex]

    now according to Mr. Tide I should write

    [itex](\alpha + i \beta)(\alpha - i \beta) = \alpha^2 + \beta^2[/itex]

    it means that

    [tex]\alpha - i\beta= \frac{\sin{2x}-i\sinh{2y}}{\cos{2x}+\cosh{2y}} [/tex]

    and than i can do it like

    [itex](\alpha + i \beta)(\alpha - i \beta) = \alpha^2 + \beta^2[/itex]

    but im unable to solve from this point.

    what should i do?

    plz help me

    i have just one day left now and on wednesday i have my exams.

    Thanks in advance
    Last edited: Nov 28, 2005
  6. Nov 28, 2005 #5
    Have you tried writing the same thing in terms of e's, that might work.
    [tex]cosh(x) = \frac{1}{2}(e^x + e^(-x))[/tex]
  7. Nov 28, 2005 #6


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    Homework Helper

    HINT: Multipley [itex]\tan (x+iy) \tan (x-iy)[/itex] and use [itex]\sin z = (e^{iz} - e^{-iz})/(2i)[/itex] and similarly for the cosine.
  8. Aug 7, 2008 #7
    Re: very difficult complex no question plzzzz help

    hi shaikh
    i think d q s of coll level
    it includes circular and hyperbolic fncns
  9. Aug 7, 2008 #8
    Re: very difficult complex no question plzzzz help

    see it nd tell if u need soln
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