Question 475227
{{{(3-2i-4i^3) / (i+2i^2-5i^3)}}}

Now {{{i^2 = -1}}} and {{{i^3 = -i}}}

==> {{{(3-2i + 4i)/(i - 2 + 5i) = (3+2i)/(-2 + 6i) = (3+2i)/(-2(1 - 3i))

= ((3+2i)/(-2(1 - 3i))) *((1+3i)/(1+3i)) 

= (3+9i+2i-6)/(-2(1+9))

= (-3 + 11i)/(-20)
}}}