How do you divide #(3i)/(1+i) + 2/(2+3i) #?

2 Answers
Jan 28, 2016

Answer:

#27/26i+47/26#

Explanation:

#(3i)/(1+i)+2/(2+3i)#
#=(3i(1-i))/((1+i)(1-i))+(2(2-3i))/((2+3i)(2-3i)#
#=(3i-3i^2)/(1-i^2)+(4-6i)/(4-9i^2)#
#=(3i-3(-1))/(1-(-1))+(4-6i)/(4-9(-1))# [as, #i^2=-1# ]
#=(3i+3)/(1+1)+(4-6i)/(4+9)#
#=(3i+3)/2+(4-6i)/13#
#=3/2i+3/2+4/13-6/13i#
#=27/26i+47/26#

Jan 28, 2016

Answer:

#47/26+27/26i#

Explanation:

Get a common denominator.

#=(3i(2+3i))/((1+i)(2+3i))+(2(1+i))/((1+i)(2+3i))#

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

#=(2+8i+9i^2)/(2+5i+3i^2)#

This can be simplified since #i^2=-1#.

#=(2+8i-9)/(2+5i-3)#

#=(-7+8i)/(-1+5i)#

Now, multiply by the complex conjugate of the denominator.

#=(-7+8i)/(-1+5i)((-1-5i)/(-1-5i))#

#=(7+27i-40i^2)/(1-25i^2)#

#=(7+27i+40)/(1+25)#

#=(47+27i)/26#

#=47/26+27/26i#