# How do you simplify (15-8i)^2 and write the complex number in standard form?

Jul 28, 2016

$161 - 240 i$

#### Explanation:

Since
${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$,

then

${\left(15 - 8 i\right)}^{2} = {15}^{2} + 2 \left(15\right) \left(- 8 i\right) + {\left(- 8 i\right)}^{2}$

that is

$225 - 240 i + 64 {i}^{2}$

and, since ${i}^{2} = - 1$, it becomes:

$225 - 240 i - 64 = 161 - 240 i$ in standard form