roam
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Homework Statement
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:
$$\left|r\right|^{2} = \left|\frac{i\kappa\sinh(\alpha L)}{\alpha\cosh(\alpha L)-i\delta\sinh(\alpha L)}\right|^{2} = \boxed{\frac{\kappa^{2}\sinh^{2}(\alpha L)}{\alpha^{2}\cosh^{2}(\alpha L)+\delta^{2}\sinh^{2}(\alpha L)}} \tag{2}$$
Homework Equations
For a complex number ##z## with complex conjugate ##\bar{z}##:
$$\left|z\right|^{2}=z\bar{z} \tag{3}$$
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?
Or do I need to use some hyperbolic identities to make simplifications?Any help is greatly appreciated.