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標題:

complex no.

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• F.5 M2 about Determinants~~
• 靚女通常係米寸d-
• Complex analysis

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solve (z-1)^5 + z^5 = 0

最佳解答:

Note this is a question concerning complex number. 圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn01Feb011700.jpg?t=1233478911 圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn02Feb011701.jpg?t=1233478916 2009-02-01 17:04:15 補充： For the answer given by dt50000, z = 1/2, just simply put k = 2 into the expression and you can get z = 1/2. But this is a complex number question. As the question is a polynomial with degree 5, so 5 complex roots should be found. 2009-02-01 17:04:42 補充： If you cannot see the picture clearly, please visit: http://s182.photobucket.com/albums/x4/A_Hepburn_1990/?action=view&current=A_Hepburn01Feb011700.jpg http://s182.photobucket.com/albums/x4/A_Hepburn_1990/?action=view&current=A_Hepburn02Feb011701.jpg 2009-02-01 17:06:27 補充： Of course you may not express the answer in standard form. Just apply some trigometric relation to simply the result.

其他解答:

#002個答案真係超型……完全理解唔到……=.="|||||(z-1)^5 + z^5 = 0 z^5+z^5=1 2z^5=1 z^5=0.5 z=0.1 驗證 : (z-1)^5 + z^5 = (0.1-1)^5+0.1^5 = (0.5-1)+0.5 = -0.5+0.5 = 0|||||(z-1)^5 + z^5 = 0 (z-1)^5 = -z^5 z-1 = -z 2z=1 z=1/2