Answer:
[tex]\large\boxed{\dfrac{2+3i}{i(4-5i)}+\dfrac{2}{i}=\dfrac{22}{41}-\dfrac{75}{41}i}[/tex]
Step-by-step explanation:
[tex]i=\sqrt{-1}\to i^2=-1[/tex]
[tex]\dfrac{2+3i}{i(4-5i)}+\dfrac{2}{i}\qquad\text{use distributive property}\\\\=\dfrac{2+3i}{4i-5i^2}+\dfrac{2}{i}=\dfrac{2+3i}{4i-5(-1)}+\dfrac{2}{i}=\dfrac{2+3i}{4i+5}+\dfrac{2}{i}\\\\\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{2+3i}{4i+5}\cdot\dfrac{4i-5}{4i-5}+\dfrac{2}{i}\cdot\dfrac{i}{i}=\dfrac{(2+3i)(4i-5)}{(4i)^2-5^2}+\dfrac{2i}{i^2}\\\\\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=\dfrac{(2)(4i)+(2)(-5)+(3i)(4i)+(3i)(-5)}{16i^2-25}+\dfrac{2i}{-1}[/tex]
[tex]=\dfrac{8i-10+12i^2-15i}{16(-1)-25}-2i=\dfrac{(-10-12)+(8i-15i)}{-16-25}-2i\\\\=\dfrac{-22-7i}{-31}-2i=\dfrac{-22}{-41}+\dfrac{-7i}{-41}-2i=\dfrac{22}{31}+\dfrac{7}{41}i-\dfrac{82}{41}i\\\\=\dfrac{22}{41}-\dfrac{75}{41}i[/tex]
