# Homework Help: Modulus of a complex number with hyperbolic functions

1. Oct 21, 2017

### roam

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? Or do I need to use some hyperbolic identities to make simplifications?

Any help is greatly appreciated.

2. Oct 21, 2017

### kuruman

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

3. Oct 22, 2017