How do you use demoivre's theorem to simplify #(-1+i)^10#?

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Answer 1 · 2016-08-09T16:12:48

#(i-1)^10=-32i#

#x+i y = sqrt(x^2+y^2) e^{i phi}#

where

#e^{i phi} = cos(phi)+i sin(phi)# (de Moivre's identity)

and

#phi = arctan(y/x)#

Applying to our example

#-1+i = sqrt(2)e^{-i pi/4# then

#(i-1)^{10} = (sqrt(2) e^{-i pi/4})^{10} = (sqrt(2))^10 e^{-i (10pi)/4}#

but #e^{-i (10pi)/4} = e^{-i (8pi)/4} dot e^{-i pi/2} = -e^{i pi/4}# and #(sqrt(2))^10 =32#

finally

#(i-1)^10=-32i#