Question

How do you simplify `(6+3i)/(2-i)`?

Answer 1

`9/5 +12/5 i`

Explanation:

Require to rationalise the denominator of the fraction. To do this we multiply numerator/denominator by the`color(blue)" complex conjugate ""of the denominator"`

If `color(blue)" a± bi ""is a complex number"`

then `color(red) "a ∓ bi ""is it's conjugate"`

Note that the 'real part'(a) remains unchanged, while the sign of the 'imaginary part' changes.

Note also : `color(blue)"(a + bi)"color(red)"(a-bi)"=a^2+b^2" a real number"`

using the fact `[ i^2 =(sqrt(-1))^2 = -1 ]`

here the conjugate of 2 - i is 2 + i

`rArr (6+3i)/(2-i) = ((6+3i)(2+i))/((2-i)(2+i))=(12+12i-3)/(4+1)`

`=(9+12i)/5 = 9/5 + 12/5 i`