# How do you simplify (2-3i)(4+i)?

Feb 5, 2016

To simplify this expression, multiply top and bottom by the complex conjugate, $\left(4 - i\right)$. This yields $\frac{11 - 4 i}{17}$

#### Explanation:

To simplify an expression like this, we multiply top and bottom of the fraction (denominator and numerator) by the complex conjugate of the bottom (numerator). For a complex expression $\left(a + b i\right)$, the complex conjugate is $\left(a - b i\right)$. When we do the calculation, it will become clear why this works so well.

Note that $\frac{a - b i}{a - b i}$ is just the same as 1, so when we do this multiplication the result is the same number we started with.

$\frac{2 - 3 i}{4 + i} \cdot \frac{4 - i}{4 - i} = \frac{8 - 12 i - 2 i + 3 {i}^{2}}{16 + 4 i - 4 i - {i}^{2}}$

But since $i = \sqrt{-} 1$, ${i}^{2} = - 1$. Using this and collecting like terms:

$\frac{8 - 14 i - 3}{16 + 1} = \frac{11 - 4 i}{17}$