Question

3+5i/-2+3i perform operation

Answer 1

`(9)/(13)-(19)/(13)i`

Explanation:

Remember
`i^2=-1`

You are given the fraction
`(3+5i)/(-2+3i)`

First, multiply the numerator and the denominator by -2-3i:

`(3+5i)/(-2+3i)=((3+5i)(-2-3i))/((-2+3i)(-2-3i))=((3+5i)(-2-3i))/((-2)^2-(3i)^2)=((3+5i)(-2-3i))/(4-9i^2)=((3+5i)(-2-3i))/(4+9)`

This gives you 13 in denominator, now multiply two complex numbers in numerator:

`(3+5i)(-2-3i)=-6-9i-10i-15i^2=-6-19i+15=9-19i`

Thus, the initial fraction is

`(9-19i)/(13)=(9)/(13)-(19)/(13)i`