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

Nov 12, 2016

${\left(2 + 3 i\right)}^{2} + {\left(2 - 3 i\right)}^{2} = - 10$

#### Explanation:

${\left(2 + 3 i\right)}^{2} + {\left(2 - 3 i\right)}^{2} = \left\{{2}^{2} + 2 \left(2\right) \left(3 i\right) + {\left(3 i\right)}^{2}\right\} + \left\{{2}^{2} - 2 \left(2\right) \left(3 i\right) + {\left(3 i\right)}^{2}\right\}$
$\therefore {\left(2 + 3 i\right)}^{2} + {\left(2 - 3 i\right)}^{2} = \left\{4 + 12 i + 9 {i}^{2}\right\} + \left\{4 - 12 i + 9 {i}^{2}\right\}$
$\therefore {\left(2 + 3 i\right)}^{2} + {\left(2 - 3 i\right)}^{2} = 8 + 18 {i}^{2}$

Now, ${i}^{2} = - 1$, so

$\therefore {\left(2 + 3 i\right)}^{2} + {\left(2 - 3 i\right)}^{2} = 8 - 18$
$\therefore {\left(2 + 3 i\right)}^{2} + {\left(2 - 3 i\right)}^{2} = - 10$