i got different answers as my friends.
remember i^0=1 i^1=i i^2=-1 i^3=- i
i got 18-36i but i didn't simplify it completely
Find (6+3i)(2i^3-4i^5) in simplest a+bi form?
2009-01-06 5:12 am
回答 (2)
2009-01-06 7:02 am
✔ 最佳答案
( 6 + 3i )( 2i^3 - 4i^5 )i^3 = - i i^5 = i
( 6 + 3i )( -2i - 4i )
= ( 6 + 3i )( -6i )
= -36i + 18
= 18 - 36i
= 18 (1 - 2i )
2009-01-06 4:53 pm
(6 + 3i)(2i^3 - 4i^5)
= 6*2i^3 + 3i*2i^3 - 6*4i^5 - 3i*4i^5
= 12i^3 + 6i^4 - 24i^5 - 12i^6
= 12(-i) + 6(1) - 24(i) - 12(-1)
= -12i + 6 - 24i + 12
= 6 + 12 - 12i - 24i
= 18 - 36i
= 18(1 - 2i)
= 6*2i^3 + 3i*2i^3 - 6*4i^5 - 3i*4i^5
= 12i^3 + 6i^4 - 24i^5 - 12i^6
= 12(-i) + 6(1) - 24(i) - 12(-1)
= -12i + 6 - 24i + 12
= 6 + 12 - 12i - 24i
= 18 - 36i
= 18(1 - 2i)
參考: Imaginary Unit
http://en.wikipedia.org/wiki/Imaginary_unit
收錄日期: 2021-05-01 11:48:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090105211203AAISTsJ