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

1 Answer
sente
May 19, 2016

Divide by the norm of 1+i1+i and apply De Moivre's theorem to find that

(1+i)^20=-1024(1+i)20=−1024

Explanation:

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

(cos(theta)+isin(theta))^n = cos(ntheta)+isin(ntheta)(cos(θ)+isin(θ))n=cos(nθ)+isin(nθ)

However, as 1+i1+i does not lie on the unit circle, we cannot use the formula directly. To fix this, we can divide 1+i1+i by its norm sqrt(2)√2 and factor out the necessary constant:

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

=(sqrt(2))^20(sqrt(2)/2+sqrt(2)/2i)^20=(√2)20(√22+√22i)20

=1024(cos(pi/4)+isin(pi/4))^20=1024(cos(π4)+isin(π4))20

=1024(cos(20*pi/4)+isin(20*pi/4))=1024(cos(20⋅π4)+isin(20⋅π4))

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

=1024(-1+i*0)=1024(−1+i⋅0)

=-1024=−1024

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