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標題:
complex2
發問:
Let a,b and c be complex numbers with modulus equal to 1. Show that if a+b+c=0, then ab+bc+ca=0
最佳解答:
|a| = |b| = |c| = 1 aa' = bb' = cc' = 1 (a' = conjugate of a) So, a + b + c = 0 (a + b + c)' = 0 a' + b' + c' = 0 1/a + 1/b + 1/c = 0 (bc + ca + ab)/abc = 0 ab+bc+ca=0
Sorry, I am not sure if my answer is correct. You can treat it as a reference. In order to let you see clearly, I change them to capital alphabets. AB + BC + CA = 0 'A' in AB is a double figure. That means if A=5, B=4, AB=54 So The value of A in AB is actually 10 times. Same as 'B' in BC and 'C' in CA, they are also double figures. AB + BC + CA = (10A + B) + (10B + C) + (10C + A) = 10A + B + 10B + C + 10C + A = 10A + A + 10B + B + 10C + C = 11A + 11B + 11C = 11(A + B + C)......(1) Since A + B + C = 0. Therefore Substitute A + B + C = 0 into (1), AB + BC + CA = 11(0) AB + BC + CA = 0 (proved)
complex2
發問:
Let a,b and c be complex numbers with modulus equal to 1. Show that if a+b+c=0, then ab+bc+ca=0
最佳解答:
|a| = |b| = |c| = 1 aa' = bb' = cc' = 1 (a' = conjugate of a) So, a + b + c = 0 (a + b + c)' = 0 a' + b' + c' = 0 1/a + 1/b + 1/c = 0 (bc + ca + ab)/abc = 0 ab+bc+ca=0
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其他解答:Sorry, I am not sure if my answer is correct. You can treat it as a reference. In order to let you see clearly, I change them to capital alphabets. AB + BC + CA = 0 'A' in AB is a double figure. That means if A=5, B=4, AB=54 So The value of A in AB is actually 10 times. Same as 'B' in BC and 'C' in CA, they are also double figures. AB + BC + CA = (10A + B) + (10B + C) + (10C + A) = 10A + B + 10B + C + 10C + A = 10A + A + 10B + B + 10C + C = 11A + 11B + 11C = 11(A + B + C)......(1) Since A + B + C = 0. Therefore Substitute A + B + C = 0 into (1), AB + BC + CA = 11(0) AB + BC + CA = 0 (proved)
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