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

1 Answer
Aug 12, 2016

Answer:

#3/5+4/5i#

Explanation:

To simplify this fraction, we require the denominator to be real.

To achieve this multiply both the numerator and denominator by the #color(blue)"complex conjugate"# of the denominator.

Given z = a ± bi then the conjugate is #color(red)(bar(z))=color(red)(a)∓color(red)(b)i#

Note that the real part remains unchanged while the sign of the imaginary part is reversed.

Also #color(red)(|bar(ul(color(white)(a/a)color(black)((a+bi)(a-bi)=a^2)color(white)(a/a)|)))" the result is real"#
where a and b are real.

#rArr3-4i" has conjugate " 3+4i#

multiply numerator/denominator by (3+4i)

#rArr(5(3+4i))/((3-4i)(3+4i))=(15+20i)/25=3/5+4/5i#