# How do you write the following quotient in standard form 1/(4-5i)^2?

Nov 13, 2017

$- \frac{9}{1681} + \frac{40}{1681} i$

#### Explanation:

$\text{expand the denominator using FOIL method}$

•color(white)(x)i^2=(sqrt(-1))^2=-1

$\Rightarrow {\left(4 - 5 i\right)}^{2} = 16 - 40 i + 25 {i}^{2} = - 9 - 40 i$

$\Rightarrow \frac{1}{4 - 5 i} ^ 2 = \frac{1}{- 9 - 40 i}$

$\text{to rationalise the denominator multiply }$
$\text{numerator/denominator}$

$\text{by the "color(blue)"conjugate " "of } - 9 - 40 i$

$\text{the conjugate of "-9-40i" is } - 9 \textcolor{red}{+} 40 i$

$\Rightarrow \frac{- 9 + 40 i}{\left(- 9 - 40 i\right) \left(- 9 + 40 i\right)}$

$= \frac{- 9 + 40 i}{1681}$

$= - \frac{9}{1681} + \frac{40}{1681} i \leftarrow \textcolor{b l u e}{\text{in standard form}}$