复数(1+i)2−1+3i的虚部为-12-12．

Question

复数((1+i)2)/(−1+√(3)i)的虚部为

-(1)/(2)

-(1)/(2)

．

Answer 1

复数((1+i)2)/(−1+√(3)i)=(2i)/(−1+√(3)i)=(2i(−1−√(3)i))/((−1+√(3)i)(−1−√(3)i))=(2√(3)−2i)/(4)=(√(3))/(2)−(1)/(2)i．
所以复数的虚部为：-(1)/(2)．
故答案为：-(1)/(2)．