How do you simplify #(-2-5i)/(3i)#?

1 Answer
Sep 20, 2016

Answer:

#-5/3+2/3i#

Explanation:

We require the denominator to be a real number. This can be achieved by multiplying the numerator/denominator by 3i.

#(-2-5i)/(3i)xx(3i)/(3i)=(-6i-15i^2)/(9i^2)#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))#

#rArr(-6i-15i^2)/(9i^2)=(15-6i)/(-9)#

and dividing.

#=15/(-9)+(-6)/(-9)i=-5/3+2/3i#