Question

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

Answer 1

`(1-i)^31 = 32768+32768i`

Explanation:

Method using trigonometric form

`1-i = sqrt(2)(cos(-pi/4)+isin(-pi/4))`

So by de Moivre's rule:

`(1-i)^31 = sqrt(2)^31(cos(31(-pi/4))+isin(31(-pi/4)))`

`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)))`

`color(white)((1-i)^31) = 2^15(1+i)`

`color(white)((1-i)^31) = 32768+32768i`

Method using direct multiplication

`(1-i)^32 = ((1-i)^2)^16`

`color(white)((1-i)^32) = (1-2i+i^2)^16`

`color(white)((1-i)^32) = (-2i)^16`

`color(white)((1-i)^32) = ((-2i)^2)^8`

`color(white)((1-i)^32) = (4i^2)^8`

`color(white)((1-i)^32) = (-4)^8`

`color(white)((1-i)^32) = ((-4)^2)^4`

`color(white)((1-i)^32) = 16^4`

`color(white)((1-i)^32) = 2^16`

`color(white)((1-i)^32) = 65536`

So:

`(1-i)^31 = (1-i)^32/(1-i)`

`color(white)((1-i)^31) = (65536(1+i))/((1-i)(1+i))`

`color(white)((1-i)^31) = (65536(1+i))/(1^2-i^2)`

`color(white)((1-i)^31) = (65536(1+i))/2`

`color(white)((1-i)^31) = 32768+32768i`