Question

4

Ramon was asked to write the quotient 3 - i

in the

a + bi form. He began this way:

(3-i)

(3-i)

4(3-i)

(3² + (²)

112 - 41)

9 - 1

(12 - 4i)

8

11

(3-i)

2

Find Ramon's error and correct it (algebraically or in

words) so that he arrives at the accurate final solution.

Answer 1

`(4)/(3 - i) = (6)/(5) + (2)/(5)i`

Explanation:

Given

`(4)/(3 - i)`

See attachment for question

Required

Correct Ramon's error

Start by rationalizing the expression:

`(4)/(3 - i) = (4)/(3 - i)*(3 + i)/(3 + i)`

`(4)/(3 - i) = (4(3 + i))/((3 - i)(3 + i))`

Expand

`(4)/(3 - i) = (12 + 4i)/(3^2 - i^2)`

`(4)/(3 - i) = (12 + 4i)/(9 - i^2)`

In complex numbers:

`i^2 = -1`

So:

`(4)/(3 - i) = (12 + 4i)/(9 - (-1))`

`(4)/(3 - i) = (12 + 4i)/(9 +1)`

`(4)/(3 - i) = (12 + 4i)/(10)`

Split fraction

`(4)/(3 - i) = (12)/(10) + (4i)/(10)`

`(4)/(3 - i) = (6)/(5) + (2)/(5)i`