早教吧作业答案频道 -->数学-->
复数z满足z(2-i)=1+7i,则复数z的共轭复数为()A.-1-3iB.-1+3iC.1+3iD.1-3i
题目详情
复数z满足z(2-i)=1+7i,则复数z的共轭复数为( )
A. -1-3i
B. -1+3i
C. 1+3i
D. 1-3i
▼优质解答
答案和解析
∵z(2-i)=1+7i,
∴z=
=
=
=-1+3i,
∴
=-1-3i.
故选:A.
∴z=
| 1+7i |
| 2-i |
| (1+7i)(2+i) |
| (2-i)(2+i) |
| -5+15i |
| 5 |
∴
. |
| z |
故选:A.
看了复数z满足z(2-i)=1+7...的网友还看了以下:
已知复数z=a+3i(a∈R)在复平面内对应的点位于第二象限,且|z|=2,则复数z等于()A.- 2020-08-01 …
已知复数z=a+3i在复平面内对应的点位于第二象限,且|z|=2,则复数z等于()A.-1+3iB 2020-08-01 …
若复数z=3-2i,则z的共轭复数.z()A.-3+2iB.-3-2iC.-2+3iD.3+2i 2020-08-02 …
若复数z=3+6i,z的共轭复数为.z,则i•.z=()A.6+3iB.6-3iC.-6+3iD. 2020-08-02 …
复数z满足z(3i-4)=25(i是虚数单位),则z的共轭复数.z=()A.4+3iB.4-3iC 2020-08-02 …
把复数z的共轭复数记作,i为虚数单位.若z=1+i,则(1+z)·=[]A.3﹣iB.3+iC.1 2020-08-02 …
复数z满足z(2+i)=3-6i(i为虚数单位),则复数z的虚部为()A.3B.-3C.3iD.-3 2020-10-30 …
已知复数z满足z•i+z1-i=3+4i(i是虚数单位),则z=()A.3+iB.4-3iC.2-3 2020-11-01 …
若复数z满足z(4-3i)=(3+4i)2(i为虚数单位),则z=()A.4+3iB.4-3iC.- 2020-11-01 …
复数z满足z(2-i)=1+7i,则复数z的共轭复数为()A.-1-3iB.-1+3iC.1+3iD 2020-11-01 …