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srichaitanya srichaitanya Grade:
        

1/(w+a)+1/(w+b)+1/(w+c)=2w2


and


1/(w2+a)+1/(w2+b)+1/(w2+c)=2w


what is


1/(a+1)+1/(b+1)+1/(c+1)=


A)2


B)-2


C)w2


D)w


where w is a complex cube root of unity.


 

8 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear srichaitanya


let f(z) =1/(z+a)+1/(z+b)+1/(z+c)  a function in z


    given f(w) =2w2


and        f(w2)=2w


clearly one of the possible function is


    f(z)=2z2


 


so f(1)=2





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Badiuddin

8 years ago
mycroft holmes
266 Points
										

I wonder what is the basis for assuming that f(z) = z2 here.


 


Multiplying the first relation by w, we get


 


w/w+a + w/w+b + w/w+c = 2


 


or a/w+a + b/w+b + c/w+c = 1


 


Similarly from the second relation, we get a/a+w2 + b/b+w2 + c/c+w2 = 1


 


Thus, w and w2 are the roots of a/(x+a) + b/(x+b) + c/(x+c) = 1


 


which simplifies to x3 - (ab+bc+ca)x -2abc = 0. This means that the sum of roots of this cubic is zero. Let the third root be r.


 


Then r + w+w2 = 0 which means r = 1.


 


Thus a/a+1 + b/b+1 + c/c+1 = 1 or 1/a+1 + 1/b+1 + 1/c+1 = 2

7 years ago
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