复数z满足(1+2i)z=4+ai(a∈R,i是虚数单位),若复数z的实部与虚部相等,则a等于(  )A.12B.4C.

2025-04-01 16:56:36
推荐回答(1个)
回答1:

由(1+2i)z=4+ai,得:
z=

4+ai
1+2i
(4+ai)(1?2i)
(1+2i)(1?2i)
4+2a+(a?8)i
5
=
4+2a
5
+
a?8
5
i

∵复数z的实部与虚部相等,
∴4+2a=a-8,解得:a=-12.
故选:D.