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[問題] HELP !! complex number 同 prob 各1條

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HELP !! complex number 同 prob 各1條

Q1) Find the four zeros of the polynomial  z^4+4

Q2) In order to enter into a competition , 12 people have to be divided into 3 groups of 4 each . How many different divisions are possible ?

計埋 , plx
   

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Q1) z^4 = -4 = 4(-1) = 4[ cos(pi) + isin(pi) ] = 4e^(ipi)
之後應該識做吧...
【DSE Maths】中大教育文憑 × 數學系畢業◆助你奪*升Lv~
首次以課題分班,10月開班招生,快d黎睇下~

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引用:
原帖由 Simon 於 2012-11-18 08:25 PM 發表
Q1) z^4 = -4 = 4(-1) = 4[ cos(pi) + isin(pi) ] = 4e^(ipi)
之後應該識做吧...
睇完 , 都係唔識
help

成年冇掂MATH , 突然計番

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z^4 = 4e^(i pi)
z = sqrt(2) e^[ i (pi + 2kpi) / 4 ], k = 0, 1, 2, 3

k = 0:
z = sqrt(2) e^(i pi/4) = sqrt(2) [ cos(pi/4) + isin(pi/4) ] = sqrt(2) [1/sqrt(2) + i/sqrt(2)] = 1 + i

k = 1:
z = sqrt(2) e^[ i (3pi/4) ] = sqrt(2) [ cos(3pi/4) + isin(3pi/4) ] = sqrt(2) [-1/sqrt(2) + i/sqrt(2) = -1 + i

k = 2:
z = sqrt(2) e^[ i (5pi/4) ] = sqrt(2) [ cos(5pi/4) + isin(5pi/4) ] = sqrt(2) [-1/sqrt(2) - i/sqrt(2)] = -1 - i

k = 3:
z = sqrt(2) e^[ i (7pi/4) ] = sqrt(2) [ cos(7pi/4) + isin(7pi/4) ] = sqrt(2) [1/sqrt(2) - i/sqrt(2)] = 1 - i
【DSE Maths】中大教育文憑 × 數學系畢業◆助你奪*升Lv~
首次以課題分班,10月開班招生,快d黎睇下~

導師簡介:按此進入

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Q2 :Name 3 groups as A,B & C.
We assign people in these groups accordingly
12C4=495  ways to group A , then 8C4 = 70 ways to group B , then 4C4 = 1 way to group C
Thus , 495x70x1 = 34650 ways to assign people in these groups .
Since the label of each group is arbitrary ,
The total # of ways to separate them into 3 groups = 34650/3! = 5775

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