[(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]^2=a+bi 则a^2-b^2=](/uploads/image/z/12732073-25-3.jpg?t=%5B%EF%BC%881-i%EF%BC%89%EF%BC%8F2%5E1%EF%BC%8F2%5D%5E2%3Da%2Bbi+%E5%88%99a%5E2-b%5E2%3D)
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设a.b为实数,若复数1+2i/a+bi=1+i,则 a+bi =
(1+ai)(1-i)/(b+i)=2-i,则a+bi=?
[(1-i)/2^1/2]^2=a+bi 则a^2-b^2=
设z=a+bi,且a,b满足a(1+i)³+(2-5i)=bi-4,则z的共轭复数=
(2+i)/i=a+bi a+b=?
复数1-i/(1+i)^2=a+bi(a,b 是R)则b=?
证明:(a+bi)^2=(a+bi)(a-bi)
设a,b为实数,若复数(1+i)(a+bi)=1+2i,则a= b=
数学复数题计算a b为实数 复数-1+3i/a+bi=1+2i 则a+bi=?
设a,b∈R,(2+b)/(a-i)=0.5-i,则a+bi=
证:向量a*向量b=1/2(I向量a+向量bI)平方-I向量aI平方-I向量bI平方)
i是虚数单位,若(2+i)/(1+i)=a+bi,则a+bi的值