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麥旋風 發表於 2005-3-16 10:33:35

Pure 03/I/Q12

究竟佢諗緊乜……
(a)(ii)一片迷惘……

久違 發表於 2005-3-16 13:42:35

呢隻數我死背:o
次次都係咁出:icon101:

久違 發表於 2005-3-16 13:44:40

part a死背
part b i 都幾明顯:icon101:
唔知咁樣答唔答到你:icon101:

[ Last edited by 久違 on 16-3-2005 at 02:20 PM ]

久違 發表於 2005-3-16 14:04:04

講真
呢到咁多步
你唔明邊到?:icon101:

久違 發表於 2005-3-16 14:12:01

仲有...a(ii)係咪少左d野??:icon101:
所以你先做唔到:icon101:

[ Last edited by 久違 on 16-3-2005 at 02:21 PM ]

小妖 發表於 2005-3-16 14:45:18

少o左個 =

麥旋風 發表於 2005-3-16 16:18:22

Originally posted by 久違 at 2005-3-16 02:12 PM:
仲有...a(ii)係咪少左d野??:icon101:
所以你先做唔到:icon101:

[ Last edited by 久違 on 16-3-2005 at 02:21 PM ]


Originally posted by 小妖 at 2005-3-16 02:45 PM:
少o左個 =

依度我明既:icon101:

麥旋風 發表於 2005-3-16 16:18:40

Originally posted by 久違 at 2005-3-16 01:44 PM:
part a死背
part b i 都幾明顯:icon101:
唔知咁樣答唔答到你:icon101:

[ Last edited by 久違 on 16-3-2005 at 02:20 PM ]
咁我明白啦:icon101:

北倫敦之王 發表於 2005-3-16 16:57:33

MI+Complex number:o

Raul 發表於 2005-3-16 21:15:23

見到咁都唔想做:icon101:
抄marking算:icon101:

凍奶茶 發表於 2005-3-17 07:08:52

a.i) z^2n=-1,
   cos(pi) + isin(pi)=-1,
   hence cos(pi) + isin(pi) is a root of z^2n, so as cos(pi +2kpi) + isin(pi   +2kpi), for k=0,1,..........
   to find the roots of z, apply the De Moivre's Them.
   z^2n = cos(pi + 2kpi) + isin(pi +2kpi)
   z=cos((pi +2kpi)/2n) + isin((pi+2kpi)/2n),
   the power 2n indicates there are 2n roots in total, so k takes value of   0,1,...2n-1

a.ii) first note that the roots of any z occurs in conjugate pair,
      first of the solution just proof that in details,
      then since z1, z2,.. zk, are roots of z, muitiple of (z-z1)=(z-z2)=(z-zk)=0
      mutiple of all (z-zk) still=0=LHS,
      and then couple more of maths skills will solve the prob..

記憶糢糊﹐有錯請指正:o

麥旋風 發表於 2005-3-17 11:41:46

Originally posted by Raul at 2005-3-16 09:15 PM:
見到咁都唔想做:icon101:
抄marking算:icon101:
我都唔想做:icon101:
不過我就係唔想抄
抄完自己都唔明
晒時間晒精神:o

麥旋風 發表於 2005-3-17 11:44:41

Originally posted by 凍奶茶 at 2005-3-17 07:08 AM:
a.i) z^2n=-1,
   cos(pi) + isin(pi)=-1,
   hence cos(pi) + isin(pi) is a root of z^2n, so as cos(pi +2kpi) + isin(pi   +2kpi), for k=0,1,..........
   to find the roots of z, apply the D ...
雖然都係唔係好明
但好感動
thx:icon099:

kawai 發表於 2005-3-17 12:21:42

PURE果然唔係屬於我的世界:icon023:

久違 發表於 2005-3-17 18:50:09

Originally posted by 麥旋風 at 2005-3-16 04:18 PM:

咁我明白啦:icon101:
你知唔知04年既grade???
上年好難...做到我死:icon099::icon101:

久違 發表於 2005-3-17 19:10:59

你試下自己做一次z^2n-1=0先
呢個易掌握d
先做03年
03個題比較煩:icon101:

麥旋風 發表於 2005-3-17 20:08:01

Originally posted by 久違 at 2005-3-17 06:50 PM:

你知唔知04年既grade???
上年好難...做到我死:icon099::icon101:
04年勁反常
雨雲話oscar話今年仲難
因為最後一年
溫完=冇溫

但我都唔知上年個grade:icon101:

麥旋風 發表於 2005-3-17 20:09:43

Originally posted by 久違 at 2005-3-17 07:10 PM:
你試下自己做一次z^2n-1=0先
呢個易掌握d
先做03年
03個題比較煩:icon101:
(a)(i)就做左架啦
不過(a)(ii)唔知點解佢(marking)會咁樣諗野姐
但做左唔知2000定係2001果條開始明白
原來pure 真係要背野:icon101:

Thx anyway:icon099:

久違 發表於 2005-3-17 20:10:05

Originally posted by 麥旋風 at 2005-3-17 08:08 PM:

04年勁反常
雨雲話oscar話今年仲難
因為最後一年
溫完=冇溫

但我都唔知上年個grade:icon101:
我覺得唔會連難兩年掛:icon099:

Sol_Campbell 發表於 2005-3-17 22:03:13

Originally posted by 久違 at 2005-3-17 08:10 PM:

我覺得唔會連難兩年掛:icon099:
上年邊份paper難?邊個topic??

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