[tex]\sqrt{-1}=i\to i^2=-1\\\\therefore\ we\ have:\\\\\dfrac{i}{(3+8i)-(2+5i)}=\dfrac{i}{3+8i-2-5i}=\dfrac{i}{1+3i}\\\\=\dfrac{i}{1+3i}\cdot\dfrac{1-3i}{1-3i}=\dfrac{i(1-3i)}{1^2-(3i)^2}=\dfrac{i-3i^2}{1-9i^2}=\dfrac{i-3(-1)}{1-9(-1)}\\\\=\dfrac{i+3}{1+9}=\dfrac{3+i}{10}=0.3+0.1i[/tex]
Used:[tex](a+b)(a-b)=a^2-b^2[/tex]
