How do you divide #(3i)/(1+i) + 2/(2+3i) #?
2 Answers
Jan 28, 2016
Explanation:
Jan 28, 2016
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
#=(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#