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gerryliyana

  • 2 years ago

hi guys..., i have a complex trigonometry question.

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  1. gerryliyana
    • 2 years ago
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    Take a look at the picture below and try to work out how to prove it.

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  2. gerryliyana
    • 2 years ago
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    i've tried ..., using all identity above.., i have \[|\cos z|^{2} = |\cos (x+iy)|^{2} = \cos (x) \cos (iy) - \sin (x) \sin (iy)\] because of: \(\cos (iy) = \cosh(y) \) and \(\sin (iy) = i \sinh (y)\), then i have \[|\cos (x+iy)|^{2} = \left( \cos (x)\cosh(y) - i \sin(x)\sinh(y) \right)^{2}\] \[|\cos (x+iy)|^{2} = \cos^{2}(x)\cosh^{2}(y) - \sin^{2}(x) \sinh^{2}(y) - 2isin^{2}(x)\sinh^{2}(y)\]

  3. gerryliyana
    • 2 years ago
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    and then...., what should i do ?? have you idea, guys ?? @UnkleRhaukus @sirm3d @hartnn

  4. hartnn
    • 2 years ago
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    the first thing coming to my mind is to write sin^2x = (1-cos^2x) and simplify

  5. hartnn
    • 2 years ago
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    also, i think you missed the |...| shouldn't it be \(|\cos z|^{2} = |\cos (x+iy)|^{2} = |\cos (x) \cos (iy) - \sin (x) \sin (iy)|^2 \\= | \left( \cos (x)\cosh(y) - i \sin(x)\sinh(y) \right)|^{2}= [\cos (x)\cosh(y)]^2 +[ \sin(x)\sinh(y)]^2\) because |a+ib| = \(\sqrt{a^2+b^2}\)

  6. hartnn
    • 2 years ago
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    the point is, since you are squaring the magnitude, there should be no imaginary part....

  7. hartnn
    • 2 years ago
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    so you should've got, \(|\cos z|^{2} = [\cos (x)\cosh(y)]^2 +[ \sin(x)\sinh(y)]^2\) did you understand how ? now write sin^2x = 1- cos^2 x

  8. gerryliyana
    • 2 years ago
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    i'm little bit confused, why \[|(\cos(x)\cosh(y) - i \sin(x)\sinh(y))|^{2} = [\cos(x)\cosh(y)]^{2} + [\sin(x)\sinh(y)]^{2}\]

  9. hartnn
    • 2 years ago
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    \(|a-ib|^2=[\sqrt{a^2+b^2}]^2\) a=cos x coshy , b= sinx sinhy

  10. hartnn
    • 2 years ago
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    i think afterthat the required result is easy to get

  11. gerryliyana
    • 2 years ago
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    ah ok.., i'll try., :),, thank you

  12. hartnn
    • 2 years ago
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    ask if doubts, and having difficulty with any other part of the question ?

  13. gerryliyana
    • 2 years ago
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    hmm.., would u kindly help me with another questions ?? still talk about complex variable.,

  14. hartnn
    • 2 years ago
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    another Q ?? taylor series one ?

  15. gerryliyana
    • 2 years ago
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    yes..., taylor series

  16. gerryliyana
    • 2 years ago
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    find the taylor series expansion of f (z) = z exp (2z) about z = -1, and Determine the region of convergence.

  17. gerryliyana
    • 2 years ago
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    have idea @Roya ??

  18. gerryliyana
    • 2 years ago
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    yes., for for trigonometry, i got it now.., thank you so much @hartnn

  19. Roya
    • 2 years ago
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    accept my apology dear @gerryliyana i was so busy . i see @hartnn solved it. tnx

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