# How do you simplify (1-i)^31 ?

Oct 7, 2017

${\left(1 - i\right)}^{31} = 32768 + 32768 i$

#### Explanation:

Method using trigonometric form

$1 - i = \sqrt{2} \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

So by de Moivre's rule:

${\left(1 - i\right)}^{31} = {\sqrt{2}}^{31} \left(\cos \left(31 \left(- \frac{\pi}{4}\right)\right) + i \sin \left(31 \left(- \frac{\pi}{4}\right)\right)\right)$

color(white)((1-i)^31) = 2^(31/2)(cos(-8pi+pi/4)+isin(-8pi+pi/4)))

color(white)((1-i)^31) = 2^(31/2)(cos(pi/4)+isin(pi/4)))

$\textcolor{w h i t e}{{\left(1 - i\right)}^{31}} = {2}^{15} \left(1 + i\right)$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{31}} = 32768 + 32768 i$

Method using direct multiplication

${\left(1 - i\right)}^{32} = {\left({\left(1 - i\right)}^{2}\right)}^{16}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {\left(1 - 2 i + {i}^{2}\right)}^{16}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {\left(- 2 i\right)}^{16}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {\left({\left(- 2 i\right)}^{2}\right)}^{8}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {\left(4 {i}^{2}\right)}^{8}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {\left(- 4\right)}^{8}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {\left({\left(- 4\right)}^{2}\right)}^{4}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {16}^{4}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = {2}^{16}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{32}} = 65536$

So:

${\left(1 - i\right)}^{31} = {\left(1 - i\right)}^{32} / \left(1 - i\right)$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{31}} = \frac{65536 \left(1 + i\right)}{\left(1 - i\right) \left(1 + i\right)}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{31}} = \frac{65536 \left(1 + i\right)}{{1}^{2} - {i}^{2}}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{31}} = \frac{65536 \left(1 + i\right)}{2}$

$\textcolor{w h i t e}{{\left(1 - i\right)}^{31}} = 32768 + 32768 i$