# How do you perform the operation and write the result in standard form given (2-3i)^2?

Mar 26, 2018

$- 5 - 12 i$

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{w h i t e}{x} {i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1$

$\text{note that } {\left(2 - 3 i\right)}^{2} = \left(2 - 3 i\right) \left(2 - 3 i\right)$

$\text{expand factors using FOIL}$

$= 4 - 12 i + 9 {i}^{2} = - 5 - 12 i \leftarrow \textcolor{red}{\text{in standard form}}$

Mar 26, 2018

$- 5 - 12 i$

#### Explanation:

First, multiply using the distributive property:

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

$\left(2 - 3 i\right) \left(2 - 3 i\right) = 4 - 6 i - 6 i + 9 {i}^{2}$

Then simplify;

$4 - 6 i - 6 i + 9 {i}^{2} = 4 - 12 i + 9 \left(- 1\right)$

$4 - 12 i + 9 \left(- 1\right) = 4 - 12 i - 9$

$4 - 12 i - 9 = - 5 - 12 i$