# How do you simplify \frac { 4- 3i } { 5+ 2i } assuming that i is an imaginary number?

Aug 31, 2017

$\frac{14}{29} - \frac{23}{29} i$

#### Explanation:

$\text{we require to "color(blue)"rationalise the denominator}$

$\text{this is achieved by multiplying the numerator/denominator}$
$\text{by the "color(blue)"complex conjugate}$ of the denominator.

$\text{multiply numerator/denominator by } \left(5 - 2 i\right)$

•color(white)(x)i^2=(sqrt(-1))^2=-1

$\Rightarrow \frac{4 - 3 i}{5 + 2 i} \times \frac{5 - 2 i}{5 - 2 i}$

$\text{expand the factors using FOIL}$

$= \frac{20 - 23 i + 6 {i}^{2}}{25 - 4 {i}^{2}}$

$= \frac{14 - 23 i}{29}$

$= \frac{14}{29} - \frac{23}{29} i \leftarrow \textcolor{red}{\text{ in standard form}}$

Aug 31, 2017

Multiply by the complex conjugate of the denominator to get $\frac{14 - 23 i}{29}$

#### Explanation:

We use "conjugates" to get messy things like square roots and imaginary numbers out of expressions. A conjugate is the same two-term expression with a different sign in the middle. Generally, we like to get those "messy terms" out of the denominators of fractions. Use the complex conjugate of $5 + 2 i$, which is $5 - 2 i$. Set it up like this:

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

Multiply the numerators and multiply the denominators to get

$\frac{\left(4 - 3 i\right) \left(5 - 2 i\right)}{\left(5 + 2 i\right) \left(5 - 2 i\right)} = \frac{20 - 15 i - 8 i + 6 {i}^{2}}{25 - 4 {i}^{2}} = \frac{20 - 23 i - 6}{25 - 4 {i}^{2}}$
Since ${i}^{2} = - 1 , - 4 {i}^{2} = - 4 \left(- 1\right) = 4$

When you add and subtract numbers with imaginary parts, just treat $i$ like a regular variable:

$\frac{20 - 23 i - 6}{25 - 4 {i}^{2}} = \frac{14 - 23 i}{29}$

...and you're done!