Respuesta :

[tex]\bf recall~that\qquad i^2=-1\\\\ -------------------------------\\\\ \cfrac{-5+i}{2i}\cdot \cfrac{i}{i}\implies \cfrac{i(-5+i)}{2i^2}\implies \cfrac{-5i+i^2}{2(-1)}\implies \cfrac{-5i+(-1)}{-2} \\\\\\ \cfrac{-5i-1}{-2}\implies \cfrac{1+5i}{2}[/tex]