How do you use demoivre's theorem to simplify #(1+i)^5#?

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Answer 1 · 2016-08-08T22:31:19

#(i+1)^5=-4(i+1)#

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

where

#e^{i phi} = cos(phi)+i sin(phi)#

and

#phi = arctan(y/x)#

Applying to our example

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

#(i+1)^5 = (sqrt(2) e^{i pi/4})^5 = (sqrt(2))^5 e^{i (5pi)/4}#

but #e^{i (5pi)/4} = e^{i (4pi)/4} dot e^{i pi/4} = -e^{i pi/4}# and #(sqrt(2))^5 = 4 sqrt(2)#

finally

#(i+1)^5=-4(i+1)#