Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 463: 55

$-\dfrac{5+4i}{2}$

Using $i^2=-1$, the expression $ \dfrac{4-5i}{2i} $ simplifies to \begin{array}{l} \dfrac{4-5i}{2i}\cdot\dfrac{2i}{2i} \\\\= \dfrac{8i-10i^2}{4i^2} \\\\= \dfrac{8i-10(-1)}{4(-1)} \\\\= \dfrac{10+8i}{-4} \\\\= -\dfrac{5+4i}{2} .\end{array}