How do you simplify #((5+i)(4-3i)) / ((2-5i)(1-i))#?

1 Answer
Feb 15, 2016

Answer:

#4/29 + 97/29i #

Explanation:

making use of the following facts related to complex numbers.

• If a + bi , is a complex number then a - bi , is it's conjugate.

#•[ i^2 = (sqrt-1)^2 = -1 ]#

first step in this question is to distribute the brackets using FOIL ( or any method that you use ).

numerator : #(5+i)(4-3i) = 20-15i+4i-3i^2 =20-11i +3 = 23-11i#

denominator :#(2-5i)(1-i) = 2-2i-5i+5i^2=2-7i-5#
=-3-7i

hence #(23-11i)/(-3-7i) #
Require the denominator to be real so multiply numerator and denominator by the conjugate of (-3-7i) , which is (-3+7i)

#rArr( (23-11i)(-3+7i))/((-3-7i)(-3+7i))#

# = (-69+161i+33i-77i^2)/(9-21i+21i-49i^2) = (-69+194i+77)/(9+49)#

#=(8+194i)/58 = 4/29 + 97/29i #