# How do you perform the operation and write the result in standard form given (4+5i)^2?

Aug 22, 2016

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

#### Explanation:

Whenever multiplication or division operations of complex numbers one need to always remember that ${i}^{2} = - 1$.

As such ${\left(4 + 5 i\right)}^{2}$

= 4^2+2×4×5i+(5i)^2

= 16+40i+25×(-1)

= $16 + 40 i - 25$

= $- 9 + 40 i$