Modulus of a complex number with hyperbolic functions

[roam]

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.

 

[kuruman]

Starting from (1), I first multiplied the numerator and denominator by the complex conjugate of the denominator ...

I think your life would be simpler if you multiplied $r$ by its complex conjugate to get $|r|^2$ directly.

 

[haruspex]

Kuruman is right, but to answer your question

But this is not the correct answer given in (2).

It is the same. You just need to find and eliminate the common factor.