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Modulus of a complex number with hyperbolic functions

  1. Oct 21, 2017 #1
    1. The problem statement, all variables and given/known data
    For the expression

    $$r = \frac{i\kappa\sinh(\alpha L)}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)} \tag{1}$$

    Where ##\alpha=\sqrt{\kappa^{2}-\delta^{2}}##, I want to show that:

    $$\left|r\right|^{2} = \left|\frac{i\kappa\sinh(\alpha L)}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)}\right|^{2} = \boxed{\frac{\kappa^{2}\sinh^{2}(\alpha L)}{\alpha^{2}\cosh^{2}(\alpha L)+\delta^{2}\sinh^{2}(\alpha L)}} \tag{2}$$

    2. Relevant equations
    For a complex number ##z## with complex conjugate ##\bar{z}##:

    $$\left|z\right|^{2}=z\bar{z} \tag{3}$$

    3. The attempt at a solution

    Starting from (1), I first multiplied the numerator and denominator by the complex conjugate of the denominator to get it in the form ##\underline{a+bi}##:

    $$\frac{i\kappa\sinh(\alpha L)}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)}.\frac{\alpha\cosh(\alpha L)+i\delta\sinh(\alpha L)}{\alpha\cosh(\alpha L)+i\delta\sinh(\alpha L)}$$

    $$=\frac{i\kappa\alpha\sinh(\alpha L)\cosh(\alpha L)-\kappa\delta\sinh^{2}(\alpha L)}{\alpha^{2}\cosh^{2}(\alpha L)+\delta^{2}\sinh^{2}(\alpha L)}.$$

    Then I multiplied the expression by its own complex conjugate to find ##\left|r\right|^{2}##:

    $$\frac{-\kappa\delta\sinh^{2}(\alpha L)+i\kappa\alpha\sinh(\alpha L)\cosh(\alpha L)}{\alpha^{2}\cosh^{2}(\alpha L)+\delta^{2}\sinh^{2}(\alpha L)}.\frac{\left(-\kappa\delta\sinh^{2}(\alpha L)-i\kappa\alpha\sinh(\alpha L)\cosh(\alpha L)\right)}{\alpha^{2}\cosh^{2}(\alpha L)+\delta^{2}\sinh^{2}(\alpha L)}$$

    $$=\frac{\kappa^{2}\delta^{2}\sinh^{4}(\alpha L)+\kappa^{2}\alpha^{2}\sinh^{2}(\alpha L)\cosh^{2}(\alpha L)}{\alpha^{4}\cosh^{4}(\alpha L)+2\alpha^{2}\delta^{2}\cosh^{2}(\alpha L)\sinh^{2}(\alpha L)+\delta^{4}\sinh^{4}(\alpha L)}$$

    But this is not the correct answer given in (2). Am I doing something wrong? :confused: Or do I need to use some hyperbolic identities to make simplifications?

    Any help is greatly appreciated.
     
  2. jcsd
  3. Oct 21, 2017 #2

    kuruman

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    I think your life would be simpler if you multiplied ##r## by its complex conjugate to get ##|r|^2## directly.
     
  4. Oct 22, 2017 #3

    haruspex

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    Kuruman is right, but to answer your question
    It is the same. You just need to find and eliminate the common factor.
     
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