# How do you write the expression (1 - i)^5 in the standard form a + bi?

Mar 16, 2016

${\left(1 - i\right)}^{5} = - 4 + 4 i$

#### Explanation:

The exponent in this case isn't too difficult to calculate directly using the binomial theorem. Proceeding with that, we have:

${\left(1 - i\right)}^{5} = {\sum}_{n = 0}^{5} \left(\begin{matrix}5 \\ n\end{matrix}\right) {1}^{n} {\left(- i\right)}^{5 - n} = {\sum}_{n = 0}^{5} \left(\begin{matrix}5 \\ n\end{matrix}\right) {\left(- i\right)}^{5 - n}$

$= {\left(- i\right)}^{5} + 5 {\left(- i\right)}^{4} + 10 {\left(- i\right)}^{3} + 10 {\left(- i\right)}^{2} + 5 {\left(- i\right)}^{1} + {\left(- i\right)}^{0}$

$= - i + 5 + 10 i - 10 - 5 i + 1$

$= - 4 + 4 i$