复数(1-i)^2/1+i=

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复数(1-i)^2/1+i=复数(1-i)^2/1+i=复数(1-i)^2/1+i=(1-i)^2/(1+i)=(1-2i-1)/(1+i)=-2i/(1+i)=-2i(1-i)/(1+1)=-i(1

复数(1-i)^2/1+i=
复数(1-i)^2/1+i=

复数(1-i)^2/1+i=
(1 - i)^2/(1 + i)
= (1 - 2i - 1)/(1 + i)
= -2i/(1 + i)
= -2i(1 - i)/(1 + 1)
= -i(1 - i)
= -1 - i

复数(1-i)^2/1+i= 复数1+i/2i-1= 复数(2i/1+i)^2=? 复数 i-2/1+2i= 复数2-2i/1+i=? 复数 (3-i/1+i)^2= 复数：5i/1-2i=? 复数1+3i/2+i= 复数-2+i/1+i= 复数1-i/i=? 计算复数z=(2-i)/(1-i)-i 复数z=1+2i,则复数z-i/z+i的虚部是复数z满足方程1-i/z+2i=i,则复数z等于复数1+2i/2-i等于? 复数1+2i/2-i等于? 复数(1+i）i= 复数(1-3i)/(3+i)=? 复数2i/1-i等于?