Question

How do you simplify `(5-i)/(2-i) - (3-7i)/(2-3i)` and write in a+bi form?

Answer 1

`8/65 + 64/65 i`

Explanation:

There are two different possibilities:

• either you combine the two fractions into one first and later write it in the `a + bi` form
• or you first write each of the fractions in `a + bi` form and later subtract the one from the other

Which approach to pick is a matter of personal preference and doesn't make a big difference.

Let me do the second approach if you don't mind.

1) Transform `(5-i)/(2-i)` into `a + bi` form.

To do so, you need to make use of the formula `(x+y)(x-y) = x^2 - y^2`.
In order to achieve such a situation, you need to multiply both the numerator and the denominator with `2+i`, the conjugate of `2-i`:

`(5-i)/(2-i) = ((5-i)(2+i))/((2-i)(2+i)) = (10 + 3 i - i^2)/(2^2 - i^2)`

... use `i^2 = -1` and simplify...

`= (10 + 3i + 1)/(4 + 1) = (11 + 3i)/5 = 11/5 + 3/5 i`

2) Transform `(3-7i)/(2-3i)` into `a + bi` form.

The approach is the same: multiply the numerator and the denominator with the conjugate of the denominator, namely `2+3i`, and simplify:

`(3-7i)/(2-3i) = ((3-7i)(2+3i))/((2-3i)(2+3i)) = (6 - 5 i - 21 i^2)/(2^2 - (3i)^2) = (6 - 5i + 21)/(4 + 9)`

`color(white)(xxxxx) = (27-5i)/13 = 27/13 - 5/13 i`

3) Subtract the two terms

`(5-i)/(2-i) - (3-7i)/(2-3i) = (11/5 + 3/5 i) -(27/13 - 5/13 i)`

`color(white)(xxxxxxxxxxx) = 11/5 - 27/ 13 + (3/5 + 5/13) i`

`color(white)(xxxxxxxxxxx) = 8/65 + 64/65 i`