How do you simplify #(4-4i)/(5+3i)#?

1 Answer
Nov 21, 2015

Answer:

#2/9-8/9i#

Explanation:

Use the conjugate of the denominator. (Recall that the conjugate of #a+b# is #a-b#.)

We can multiply the expression by #(5-3i)/(5-3i)#.

#(4-4i)/(5+3i)((5-3i)/(5-3i))=(20-20i-12i+12i^2)/(25cancel(-15i+15i)-9i^2)=(20-32i+12i^2)/(25-9i^2)#

Recall that #i=sqrt1#, so #i^2=-1#.

#(20-32i+12(-1))/(25-9(-1))=(20-12-32i)/(25+9)=(8-32i)/36=2/9-8/9i#

Note that the answer is written in the complex number format #a+bi#.