factorize:
a2 + b2
=(a+bi)(a-bi)
但我有點不明。
此網頁中提到 i2=-1
http://hk.knowledge.yahoo.com/question/?qid=7006100204439
但expand 番 (a+bi)(a-bi)
=a(a-bi) + bi(a-bi)
=a2 - abi + abi - (bi)2
=a2 - (bi)2
但我唔明點解最後 - (bi)2= +b2
請各位幫忙，解答我。
a^2 + b^2 factorization
2007-03-09 4:01 am
回答 (5)
2007-03-09 4:17 am
✔ 最佳答案
沒有什麼矛盾(a+bi)(a-bi)
=a²-abi+abi-(bi)²
=a²-(bi)²
由於i²=-1
-(bi)²=-(b)²(i)²=-b²(-1)=b²
所以a²-(bi)²=a²+b²
2007-03-10 3:36 am
點解最後 - (bi)^2= +b^2
- (bi)^2
= - (b^2 i^2)
= - [b^2(-1)]
=+b^2
- (bi)^2
= - (b^2 i^2)
= - [b^2(-1)]
=+b^2
2007-03-09 5:03 am
關於 - (bi)^2= b^2 :
因為 - (bi)^2= - ( (b)^2 ) ( (i)^2 ) = - (b^2) (-1) = b^2
呢個問題涉及到complex no. system,
除o左要知道 i^2 = -1 之外仲要識d complex no. o既property
例如(a+bi)(a-bi)=a(a-bi) + bi(a-bi)就已經係complex no. o既definition之一
p.s. a^2+b^2=(a+b)^2-2ab 都係常用o既工具之一，不過呢個係completing square, 唔係factorization
因為 - (bi)^2= - ( (b)^2 ) ( (i)^2 ) = - (b^2) (-1) = b^2
呢個問題涉及到complex no. system,
除o左要知道 i^2 = -1 之外仲要識d complex no. o既property
例如(a+bi)(a-bi)=a(a-bi) + bi(a-bi)就已經係complex no. o既definition之一
p.s. a^2+b^2=(a+b)^2-2ab 都係常用o既工具之一，不過呢個係completing square, 唔係factorization
2007-03-09 4:17 am
(bi) ^ 2 = (b ^ 2)(i ^ 2) = (b ^ 2) (-1) = -b ^ 2
Therefore, -(bi) ^ 2 = -(-b ^ 2) = +b ^ 2
Therefore, -(bi) ^ 2 = -(-b ^ 2) = +b ^ 2
收錄日期: 2021-04-12 22:14:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070308000051KK03542