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

2 Answers
Dec 20, 2016

Answer:

#2/5+1/5i#

Explanation:

To simplify the fraction we require to have a rational denominator.

This is achieved by multiplying the numerator/denominator by the #color(blue)"complex conjugate"# of the complex number on the denominator.

#"The conjugate of " 3-4i" is " 3+4i#

#rArr(2-i)/(3-4i)=((2-i)(3+4i))/((3-4i)(3+4i))#

distributing the numerator/denominator using the FOIL method.

#=(6+8i-3i-4i^2)/(9+12i-12i-16i^2)#

#color(orange)"Reminder: " i^2=(sqrt(-1))^2=-1#

#=(6+5i+4)/(9+16)=(10+5i)/25larr" rational denominator"#

#"Expressing in " color(blue)"standard form"#

#=10/25+5/25i=2/5+1/5i#

Dec 20, 2016

Answer:

#(2+i)/5#.

Explanation:

Multiply by the complex conjugate of the denominator, in this case #3+4i#.

#(2-i)/(3-4i)=(2-i)/(3-4i)*(3+4i)/(3+4i)=((2-i)(3+4i))/((3-4i)(3+4i))#.

Numerator: #(2-i)(3+4i)=6+8i-3i-4i^2=6+5i+4=10+5i#.

Denominator: #(3-4i)(3+4i)=9+12i-12i-16i^2=9+16=25#.

So we have #((2-i)(3+4i))/((3-4i)(3+4i))=(10+5i)/25=(2+i)/5#.