Evaluate the following absolute square of a complex number. Assume $a$ and $b$ are real. Express your answer in terms of a hyperbolic function. ${\left\vert\frac{(a+bi)^2e^b-(a-bi)^2e^{-b}}{4abie^{-ia}}\right\vert}^2$.
My attempt:
$$\begin{align} {\vert z \vert}^2&=
{\left\vert\frac{(a+bi)^2e^b-(a-bi)^2e^{-b}}{4abie^{-ia}}\right\vert}^2 \\\\&=
z\overline z \\\\ &=
\frac{(a+bi)^2e^b-(a-bi)^2e^{-b}}{4abie^{-ia}}
\cdot\frac{(a-bi)^2e^b-(a+bi)^2e^{-b}}{-4abie^{ia}},\\\\ &\text{then let $v=a+bi$}\\\\&=
\frac{(v\overline v)^2e^{2b}-v^4-\overline v^4+(v\overline v)^2e^{-2b}}{(4ab)^2}\\\\ &=
\frac{(a^2+b^2)^2e^{2b}-v^4-\overline v^4+(a^2+b^2)^2e^{-2b}}{(4ab)^2}\\\\ &=
\frac{(a^2+b^2)^2(e^{2b}+e^{-2b})-(v^4+\overline v^4)}{(4ab)^2}\\\\ &=
1+\frac{(a^2+b^2)^2(e^{2b}+e^{-2b})-(2a^4+4a^2b^2+2b^4)}{16a^2b^2}\\\\ &=
1+\frac{(a^2+b^2)^2\cosh 2b-(a^2+b^2)}{8a^2b^2}\\\\ &=
1+\frac{(a^2+b^2)^2(1+\sinh^2b+\sinh^2b-1)}{8a^2b^2}\\\\ &=
1+\frac{(a^2+b^2)^2\sinh^2b}{(2ab)^2}\\\\
\end{align} $$
I sense I am on a correct path, but can't seem to grasp how to simplify equation even more to achieve an answer given in a textbook. Any clues? Solved.