Question

8-6i / 3i


3i / 4+2i


I = imaginary numbers

Answer 1

I assume you want it 'simplified' and that you forgot the brackets

(8-6i)/(3i)
= (8-6i)/(3i) * i/i
= (8i - 6i^2)/(3i^2)
= (6 + 8i)/-3
or
=( -6 - 8i)/3

hint for the second...
multiply top and bottom by (4-2i)/(4-2i)