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