Question #49a01

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Answer 1 · 2017-04-04T02:51:27

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:

$19/13-4/13i$