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What is #(1+i)^6# ?

1 Answer

Oct 14, 2017

#(1+i)^6 = -8i#

Explanation:

#(1+i)^2 = 1+2i+i^2 = 2i#

#(2i)^3 = 2^3*i^2*i = -8i#

So:

#(1+i)^6 = ((1+i)^2)^3 = -8i#

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