How do you write the following quotient in standard form #(2+i)/(2-i)#?

2 Answers
May 21, 2017

Answer:

My preferred way to divide complex numbers is to multiply numerator and denominator by the complex conjugate of the denominator and then simplify.

Explanation:

Given: #(2+i)/(2-i)#

Multiply numerator and denominator by the complex conjugate of the denominator:

#(2+i)/(2-i)(2+i)/(2+i)#

The denominator becomes the difference of two squares:

#((2+i)(2+i))/(4-i^2)#

Use the F.O.I.L method to multiply the the numerator:

#(4+ 2i+2i+i^2)/(4-i^2)#

Replace #i^2# with #-1#:

#(4+ 2i+2i-1)/(4-(-1))#

Combine like terms:

#(3+4i)/5#

Convert do #a+bi# form:

#3/5+4/5i#

May 21, 2017

Answer:

#3/5+4/5i#

Explanation:

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

#"before dividing the complex numbers we require the"#
#"denominator to be real"#

#"this is achieved by multiplying the numerator/denominator"#
#"by the "color(blue)"complex conjugate "" of the denominator"#

#"Note " (a+bi)(color(red)(a-bi))=a^2+b^2larr" a real number"#

#"the conjugate has the same values of a and b with the"#
#"opposite sign"#

#"the conjugate of " (2-i)" is " (2+i)#

#rArr(2+i)/(2-i)xx((2+i))/((2+i))=((2+i)(2+i))/((2-i)(2+i))#

#"expanding numerator/denominator gives"#

#(4+4i+i^2)/(4-i^2)=(3+4i)/5#

#rArr(2+i)/(2-i)=3/5+4/5ilarrcolor(red)" in standard form"#