Question

(3i/√2-i)^2的虚部是多少

Answer 1

(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