[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=

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[(1-i)/2^1/2]^2=a+bi则a^2-b^2=[(1-i)/2^1/2]^2=a+bi则a^2-b^2=[(1-i)/2^1/2]^2=a+bi则a^2-b^2=[(1-i)/2^1/2]

[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=
[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=
[(1-i)/2^1/2]^2=a+bi
则a^2-b^2=

[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=