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#