Reduce $\left\lvert \frac{(a+ib)^2 e^{ia+b}-(a-ib)^2 e^{ia-b}}{4abi} \right\rvert ^2$, expressing the result with hyperbolic functions

Question

I have to reduce:

$$\left\lvert \frac{(a+ib)^2 e^{ia+b}-(a-ib)^2 e^{ia-b}}{4abi} \right\rvert ^2$$

$a$ and $b$ have to be real, and I have to express the answer in hyperbolic functions.

My guess is that I have to use the next formulas: $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$

but I also see that the denominator is two complex numbers in polar form.

Edit: I rewrote $(a+ib)^2$ and $(a-ib)^2$ in polar form $$\left\lvert \frac{(a^2+b^2)e^{2iϕ} e^{ia+b}-(a^2+b^2)e^{-2iϕ} e^{ia-b}}{4abi} \right\rvert ^2$$

Answer 1

Here is literally what you have to do.

(1) Simplify by noting that the absolute of some of the terms are unity,

$$ \begin{align} \left\lvert \frac{(a+ib)^2 e^{ia+b}-(a-ib)^2 e^{ia-b}}{4abi} \right\rvert ^2 &=\left\lvert \frac{e^{ia}}{4abi}\right\rvert ^2~\left\lvert (a+ib)^2 e^{b}-(a-ib)^2 e^{-b}\right\rvert ^2\\ &=\frac{1}{16a^2b^2}~\left\lvert (a+ib)^2 e^{b}-(a-ib)^2 e^{-b}\right\rvert ^2\\ \end{align} $$

(2) expand and collect the real and imaginary parts

$$ \begin{align} \left\lvert (a+ib)^2 e^{ia+b}-(a-ib)^2 e^{ia-b}\right\rvert ^2 &=\left\lvert (a^2-b^2) (e^{b}-e^{-b})+2abi(e^{b}+e^{-b}) \right\rvert ^2\\ &=\left\lvert (a^2-b^2) 2\sinh b+2abi 2\cosh b \right\rvert ^2\\ &=\sqrt{((a^2-b^2) 2\sinh b)^2+(2ab 2\cosh b)^2}^2\\ &=(a^2-b^2)^2 4\sinh^2 b+16a^2b^2\cosh^2 b \end{align} $$

So that finally,

$$\left\lvert \frac{(a+ib)^2 e^{ia+b}-(a-ib)^2 e^{ia-b}}{4abi} \right\rvert ^2=\frac{(a^2-b^2)^2 \sinh^2 b}{4a^2b^2}+\cosh^2 b$$

I have verified this solution numerically for arbitrary (random) values of $a$ and $b$.

Answer 2

A couple of hints:

1- $|e^{i \theta}|^2=1$.

2- $a+ib = r e^{i \phi}$, and $a-ib = r e^{-i \phi}$.

Using these hints, solving the question should be easy.