How do you simplify #(2+i)/(1+2i)#?

2 Answers
Jan 2, 2018

Answer:

#4/5-3/5i#

Explanation:

#"multiply the numerator/denominator by the"#
#color(blue)"complex conjugate"" of the denominator"#

#"the complex conjugate of "1+2i" is "1color(red)(-)2i#

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

#"expand factors on numerator/denominator using FOIL"#

#=(2-3i-2i^2)/(1-4i^2)#

#[i^2=(sqrt(-1))^2=-1]#

#=(4-3i)/5=4/5-3/5ilarrcolor(blue)"in standard form"#

Jan 2, 2018

Answer:

#4/5-(3i)/5#

Explanation:

To divide complex numbers we first remove the complex number from the denominator, by multiplying by the complex conjugate of the denominator:

This is #1-2i#

The product of a complex number and its conjugate is always a real number.

#:.#

#((1-2i)(2+i))/(( 1-2i)(1+2i))=((1-2i)(2+i))/5=(4-3i)/5=4/5-(3i)/5#