题目内容
z为复数,(1+i)z为纯虚数,若|
|=
,则z=______.
| z |
| 2+i |
| 2 | ||
|
设z=a+bi(a,b∈R),∴(1+i)z=a-b+(a+b)i,
∵(1+i)z为纯虚数,∴
,即a=b≠0,
∵
=
=
=
,且|
|=
,
∴
+
=
,解得a=±
,∴z=±
(1+i),
故答案为:±
(1+i).
∵(1+i)z为纯虚数,∴
|
∵
| z |
| 2+i |
| a+ai |
| 2+i |
| a(1+i)(2-i) |
| (2+i)(2-i) |
| a(3+i) |
| 5 |
| z |
| 2+i |
| 2 | ||
|
∴
| 9a2 |
| 25 |
| a2 |
| 25 |
| 4 |
| 5 |
| 2 |
| 2 |
故答案为:±
| 2 |
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