Let z be a complex number with |z| = 1
Prove that
|z+1|+|z^(2k)+1|+|z^(2k+1)+1|>=2, for any positive integers k.
AL complex number
2010-09-11 7:46 pm
回答 (1)
2010-09-13 5:16 pm
✔ 最佳答案
|x|+|y| >= |x-y|, then |z^(2k)+1|+|z^(2k+1)+1| >= |z^(2k)*(z-1)| = |z-1||z+1|+|z^(2k)+1|+|z^(2k+1)+1|
>= |z+1| + |z-1|
>= |(z+1)-(z-1)| =2
2010-09-13 09:16:52 補充：
參考一下:
|x|+|y| >= |x-y|, then |z^(2k)+1|+|z^(2k+1)+1| >= |z^(2k)*(z-1)| = |z-1|
|z+1|+|z^(2k)+1|+|z^(2k+1)+1|
>= |z+1| + |z-1|
>= |(z+1)-(z-1)| =2
收錄日期: 2021-04-29 00:03:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100911000051KK00469