题目内容
已知z1,z2为复数,(3+i)z1为实数,z2=
且|z2|=5
,则z2=______.
| z1 |
| 2+i |
| 2 |
设z1=a+bi,
∵(3+i)z1为实数,
∴(3+i)(a+bi)=3a-b+(a+3b)i
∴a+3b=0,
∴z1=a+bi=-3b+bi
∵z2=
=
=
=
=-b+bi
∵|z2|=5
,
∴
=5
,
∴b=±5,
∴z2=±(5-5i)
故答案为:±(5-5i)
∵(3+i)z1为实数,
∴(3+i)(a+bi)=3a-b+(a+3b)i
∴a+3b=0,
∴z1=a+bi=-3b+bi
∵z2=
| z1 |
| 2+i |
| -3b+bi |
| 2+i |
| (-3b+bi)(2-i) |
| (2+i)(2-i) |
| -5b+5bi |
| 5 |
∵|z2|=5
| 2 |
∴
| 2b2 |
| 2 |
∴b=±5,
∴z2=±(5-5i)
故答案为:±(5-5i)
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