Z=(1-i)/√2 ,则Z^100+Z^50+1的值是?

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Z=(1-i)/√2 ,则Z^100+Z^50+1的值是?

z=cos(-π/4)+isin(-π/4)
所以原式=[cos(-100*π/4)+isin(-100*π/4)]+[cos(-50*π/4)+isin(-50*π/4)]+1
=cos(-25π)+isin(-25π)+cos(-25/2π)+isin(-25/2π)+1
=cosπ+isinπ+cos(-π/2)+isin(-π/2)+1
=-1-i+1
=-i

z²=(1-i)²/(√2)²=(1-2i-1)/2=-i
z⁴=(z²)²=(-i)²=-1
z^100+z^50+1
=(z⁴)^25 +[(z⁴)^12]×z²+1
=(-1)^25+[(-1)^12]x(-i) +1
=-1-i+1
=-...

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z²=(1-i)²/(√2)²=(1-2i-1)/2=-i
z⁴=(z²)²=(-i)²=-1
z^100+z^50+1
=(z⁴)^25 +[(z⁴)^12]×z²+1
=(-1)^25+[(-1)^12]x(-i) +1
=-1-i+1
=-i


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z²=(1-i)²/2=-i
z^4=-1
Z^100+Z^50+1
=(z^4)^25+(z^4)^12×z²+1
=(-1)^25+(-1)^12×(-i)+1
=-1-i+1
=-i