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

2 Answers
Feb 13, 2016

Answer:

Solution is #-5/3+(2/3)i#

Explanation:

To solve #(−2−5i)/(3i)#

multiply numerator and denominator by #i# (note #i^2=-1#)

and we get

#(i*(−2−5i))/(i*(3i)# or #(-2i-5i^2)/(3i^2)#

or #(-2i-5(-1))/(3*-1)# i.e. #(5-2i)/-3#, which is equivalent to

#-5/3+(2/3)i#

Feb 13, 2016

Answer:

#2/3 i - 5/3 #

Explanation:

The denominator of the fraction is required to be real. To achieve this multiply the numerator and denominator by 3i.

#(-2 - 5i )/(3i) xx( 3i)/(3i )#

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

[Note: # i^2 = (sqrt(-1))^2 = -1 #]

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