How do you divide #(-4+10i)/(3+4i)#?

1 Answer
Aug 25, 2016

Answer:

#(-4+10i)/(3+4i)=28/25+46/25 i#

Explanation:

To divide #-4+10i# by #3+4i#, one should multiply numerator and denominator by complex conjugate of denominator #3+4i# i.e. #3-4i#. Also remember that #i^2=-1#.

#(-4+10i)/(3+4i)#

= #((-4+10i)×(3-4i))/((3+4i)×(3-4i))#

= #(-4)×3+(-4)(-4i)+10i×3+10i×((-4i))/(3×3+3×(-4i)+4i×3+4i×(-4i))#

= #(-12+16i+30i-40i^2)/(9-12i+12i-16i^2)#

= #(-12+46i-40×(-1))/(9-16×(-1))#

= #(-12+46i+40)/(9+16)#

= #(28+46i)/25#

= #28/25+46/25 i#