# How do you simplify (7+5i)(7-5i) and write the complex number in standard form?

Feb 12, 2017

$\left(7 + 5 i\right) \left(7 - 5 i\right) = 74 + i 0$

#### Explanation:

Using the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

$\left(7 + 5 i\right) \left(7 - 5 i\right)$

= ${7}^{2} - {\left(5 i\right)}^{2}$

= $49 - 25 {i}^{2}$ - but ${i}^{2} = - 1$

Hence this is $49 - 25 \times \left(- 1\right) = 49 + 25 = 74$

and in standard form, this can be written as $74 + i 0$