f.4 數系-_-

其實我想知(9+6i)/(3i)呢度點解會計到2 -3i出黎既'?

-->首先, 我相信你應該知道 i^2 = -1 :p


( 9 + 6i ) / ( 3i )

= ( 9 + 6i ) / ( 3i ) X ( i / i ) <-- 目的係將分母3i 整走佢

= [ 9i + 6 (-1) ] / -3 <--- 因為 i^2 = -1

= 2 - 3i

跟住 -3i + 2 + (-7 + 10i ) = -5 + 7i <--- 呢度相信你應該無問題~

如果以後有咩類似既問題 又或者係初中數學,334,香港中學文憑考試既問題, 歡迎e-mail 我 :[email protected]

(9+6i)/(3i)
分子和分母各自乘 i
[(9+6i)i]/[(3i)i]
(9i +6i^2)/(3i^2)
(9i +6(-1))/(3(-1))
(9i -6)/(-3)
-3i +2

Another example,
(9+6i)/(1+ 3i)
分子和分母各自乘 (1 – 3i), the conjugate of (1+ 3i)
[(9+6i)(1- 3i)]/[(1+ 3i)(1- 3i)]
Another example
(9+6i)/(7- 3i)
[(9+6i) /(7+ 3i)]/[(7- 3i) )/(7+ 3i)]

2010-09-12 05:23:06 補充：
Correction: (The / in front of (7 +3i) should not be there.)
Another example:
(9+6i)/(7- 3i)
[(9+6i) (7+ 3i)]/[(7- 3i) (7+ 3i)]
(63 + 69i -18)/(49 - 9i^2)
(45 + 69i)/58
45/58 +69/58i

(-7+10i)+(9+6i)/(3i)
= -7 + 10i + 9/3i + 6i/3i
= -7 + 10i + 3/i + 2
= -7 + 10i + 3i/(-1) + 2
= -7 + 10i - 3i + 2
= -5 + 7i


2010-09-11 21:02:13 補充：
i = √(-1)
3/i = 3/√(-1) = 3*√(-1) / √(-1) * √(-1) = 3√(-1) / (-1) = -3√(-1)