Question

Question #49a01

Answer 1

Given: `(5+2i)/(3+2i)`

Please notice that the denominator is `3+2i`, this tells us to we multiply by 1 in the form of `(3-2i)/(3-2i)`:

`(5+2i)/(3+2i)(3-2i)/(3-2i)`

The reason why we chose to do this is, because it make the denominator fit the pattern `(x+y)(x-y) = x^2-y^2`.

Please look at what happens to the denominator:

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

See how the denominator fits the pattern?

Square the terms in the denominator:

`((5+2i)(3-2i))/(9-4i^2)`

Use the property `i^2 = -1`

`((5+2i)(3-2i))/(9-4(-1))`

That changes the minus sign in the denominator to a plus sign:

`((5+2i)(3-2i))/(9+4)`

Perform the addition:

`((5+2i)(3-2i))/13`

Use the F.O.I.L method on the numerator:

`((15-10i+6i-4i^2))/13`

Do the same thing with `i^2= -1` in the numerator:

`((15-10i+6i-4(-1)))/13`

Change the sign:

`(15-10i+6i+4)/13`

Combine like terms:

`(19-4i)/13`

You can leave it like this or you can divide each term by 13:

`19/13-4/13i`