Question #49a01

Apr 4, 2017

Given: $\frac{5 + 2 i}{3 + 2 i}$

Please notice that the denominator is $3 + 2 i$, this tells us to we multiply by 1 in the form of $\frac{3 - 2 i}{3 - 2 i}$:

$\frac{5 + 2 i}{3 + 2 i} \frac{3 - 2 i}{3 - 2 i}$

The reason why we chose to do this is, because it make the denominator fit the pattern $\left(x + y\right) \left(x - y\right) = {x}^{2} - {y}^{2}$.

Please look at what happens to the denominator:

$\frac{\left(5 + 2 i\right) \left(3 - 2 i\right)}{\left(3 + 2 i\right) \left(3 - 2 i\right)} = \frac{\left(5 + 2 i\right) \left(3 - 2 i\right)}{{3}^{2} - {\left(2 i\right)}^{2}}$

See how the denominator fits the pattern?

Square the terms in the denominator:

$\frac{\left(5 + 2 i\right) \left(3 - 2 i\right)}{9 - 4 {i}^{2}}$

Use the property ${i}^{2} = - 1$

$\frac{\left(5 + 2 i\right) \left(3 - 2 i\right)}{9 - 4 \left(- 1\right)}$

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

$\frac{\left(5 + 2 i\right) \left(3 - 2 i\right)}{9 + 4}$

$\frac{\left(5 + 2 i\right) \left(3 - 2 i\right)}{13}$

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

$\frac{\left(15 - 10 i + 6 i - 4 {i}^{2}\right)}{13}$

Do the same thing with ${i}^{2} = - 1$ in the numerator:

$\frac{\left(15 - 10 i + 6 i - 4 \left(- 1\right)\right)}{13}$

Change the sign:

$\frac{15 - 10 i + 6 i + 4}{13}$

Combine like terms:

$\frac{19 - 4 i}{13}$

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

$\frac{19}{13} - \frac{4}{13} i$