What is ${(1 + i)}^{6}$ $(1+i)^6$ ?

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Answer 1 · 2017-10-14T17:56:40

${(1 + i)}^{6} = - 8 i$ $(1+i)^6 = -8i$

${(1 + i)}^{2} = 1 + 2 i + i^{2} = 2 i$ $(1+i)^2 = 1+2i+i^2 = 2i$

${(2 i)}^{3} = 2^{3} \cdot i^{2} \cdot i = - 8 i$ $(2i)^3 = 2^3*i^2*i = -8i$

So:

${(1 + i)}^{6} = {({(1 + i)}^{2})}^{3} = - 8 i$ $(1+i)^6 = ((1+i)^2)^3 = -8i$

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