ｉのｉ乗について

ｉのｉのｉ乗について扱ってきたが、ここでは

ｉ=cos(π/2+2nπ)+isin(π/2+2nπ)

ｉ^1/3=cos(π/6+2nπ/3)+isin(π/6+2nπ/3)

n=0のとき

ｉ^1/3=cos(π/6)+isin(π/6)＝1/2(√3+i)

n=1のとき

ｉ^1/3=cos(5π/6)+isin(5π/6)＝1/2(-√3+i)

n=2のとき

ｉ^1/3=cos(3π/2)+isin(3π/2)＝-i

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±１の3乗方根も求めてみたい。

1=cos(2nπ)+isin(2nπ)

-1=cos(π+2nπ)+isin(π+2nπ)

1^1/3=cos(2nπ/3)+isin(2nπ/3)

(-1)^1/3=cos(π/3+2nπ/3)+isin(π/3+2nπ/3)

n=0のとき

1^1/3=cos(0)+isin(0)＝1

(-1)^1/3=cos(π/3)+isin(π/3)=1/2(1+√3i)

n=1のとき

1^1/3=cos(2π/3)+isin(2π/3)＝1/2(-1+√3i)＝ω

(-1)^1/3=cos(π)+isin(π)＝-1

n=2のとき

1^1/3=cos(4π/3)+isin(4π/3)＝1/2(-1-√3i)＝ω^2

(-1)^1/3=cos(5π/3)+isin(5π/3)＝1/2(1-√3i)

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