复数(2i/1+i)^2=?

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复数(2i/1+i)^2=?
复数(2i/1+i)^2=?

复数(2i/1+i)^2=?
= (2i)²/(i+1)²
=-4/(1+2i-1)
=-4/2i
=-2i/i²
=2i

(2i/1+i)^2=-4/(2i)=2i

(2i/1+i)^2= [ 2i*(1-i) / (1+1) ] ^2 = [ i*(1-i) ] ^2 = (1+i)^2 = 1-1+2i = 2i

复数(2i/1+i)^2=? 复数 i-2/1+2i= 复数2-2i/1+i=? 复数(1-i)^2/1+i= 复数1+i/2i-1= 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i/(2+i)= 复数i/2+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(i-2)/(i+2)=? 复数1+2i/2-i等于? 复数1+2i/2-i等于? 复数(4+3i)(2+i)=