Question

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

Answer 1

`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`

Answer 2

`(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`.