How do you use DeMoivre's Theorem to find #(1+i)^20# in standard form?

1 Answer
sente
May 19, 2016

Divide by the norm of #1+i# and apply De Moivre's theorem to find that

#(1+i)^20=-1024#

Explanation:

De Moivre's formula, which can be derived from Euler's formula that #e^(itheta)=cos(theta)+isin(theta)#, states that

#(cos(theta)+isin(theta))^n = cos(ntheta)+isin(ntheta)#

However, as #1+i# does not lie on the unit circle, we cannot use the formula directly. To fix this, we can divide #1+i# by its norm #sqrt(2)# and factor out the necessary constant:

#(1+i)^20 = (sqrt(2)(1/sqrt(2)+1/sqrt(2)i))^20#

#=(sqrt(2))^20(sqrt(2)/2+sqrt(2)/2i)^20#

#=1024(cos(pi/4)+isin(pi/4))^20#

#=1024(cos(20*pi/4)+isin(20*pi/4))#

#=1024(cos(5pi)+isin(5pi))#

#=1024(-1+i*0)#

#=-1024#

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