(3i/√2-i)^2的虚部是多少
问题描述：
(3i/√2-i)^2的虚部是多少
答
(3i/√2-i)^2
=（-9）/(2-1-2√2i)
=-9/(1-2√2i)
=-9(1+2√2i)/[(1-2√2i)(1+2√2i)]
=-9(1+2√2i)/(1+8)
=-1-2√2i
虚部是-2√2