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