设z1=4-3i,z2=1+2i,则z1/z2在复平面对应的点所在的象限为?
问题描述:
设z1=4-3i,z2=1+2i,则z1/z2在复平面对应的点所在的象限为?
答
z1=4-3i,z2=1+2i,
z1/z2=(4-3i)/(1+2i)
=(4-3i)(1-2i)/(1+2i)(1-2i)
=(-2-11i)/5
=-2/5-11/5i
即点(-2/5,-11/5)在第三象限.