Assuming $a, b \in \mathbb{R}$ and $b >0$, is there a way we can simplify the real part of this expression? $$ i ab \left(e^{-i \sqrt{-a^4 -i a b}}-e^{i \sqrt{-a^4 +i a b}}\right) $$
In case it matters, it was derived using the $(-\infty, 0)$ branch cut.
Assuming $a,b\in\mathbb{R}$:
$$\Re\left(iab\left(e^{-i\sqrt{-a^4-iab}}-e^{i\sqrt{-a^4+iab}}\right)\right)=$$ $$\Im\left(ab\left(e^{i\sqrt{-a^4+iab}}-e^{-i\sqrt{-a^4+iab}}\right)\right)=$$ $$\Im\left(ab\cdot e^{i\sqrt{-a^4+iab}}\right)-\Im\left(ab\cdot e^{-i\sqrt{-a^4+iab}}\right)=$$ $$-\Im\left(ab\left(e^{-i\sqrt{-a^4-iab}}-e^{i\sqrt{-a^4+iab}}\right)\right)=$$ $$2\Re\left(ab\cdot\sin\left(\sqrt{-a^4+iab}\right)\right)=$$
$$2ab\sin\left(\sqrt[4]{a^8+a^2b^2}\cdot\cos\left(\frac{1}{2}\arg\left(-a^4+iab\right)\right)\right)\cdot\cosh\left(\sqrt[4]{a^8+a^2b^2}\cdot\sin\left(\frac{1}{2}\arg\left(-a^4+iab\right)\right)\right)$$
