What is #(7+4i)/(6-8i)#?

1 Answer
Dec 8, 2015

#1/10+4/5i#

Explanation:

Multiply the expression by the complex conjugate.

#(7+4i)/(6-8i)((6+8i)/(6+8i))=(42+56i+24i+32i^2)/(36+48i-48i-64i^2)=(42+80i+32i^2)/(36-64i^2)#

Recall that since #i=sqrt(-1)#, we can say that #i^2=-1#.

#=(42+80i+32(-1))/(36-64(-1))=(10+80i)/100=1/10+4/5i#

Notice that the answer is written in the #a+bi# form of a complex number.