z1 , z2 , z3 , , z4  ,  . . . . . . , zn-1 are the non real nth  roots of unity, then find the value of                                 (3-z1)-1 + (3-z2)-1 + (3-z3)-1 + . . . . . . + (3-zn-1)-1

                               

 

 

 

 

 

 

 

 

 


 

            

Asked by Sourav | 8th May, 2015, 12:12: PM

Expert Answer:

I f space Z subscript 1 comma space Z subscript 2 comma space Z subscript 3 comma space..... Z subscript n minus 1 end subscript space a r e space t h e space n o n space r e a l space n t h space r o o t s space o f space u n i t y comma space w e space c a n space w r i t e x to the power of n minus 1 equals open parentheses x minus 1 close parentheses open parentheses x minus Z subscript 1 close parentheses open parentheses x minus Z subscript 2 close parentheses open parentheses x minus Z subscript 3 close parentheses..... open parentheses x minus Z subscript n minus 1 end subscript close parentheses T a k i n g space log space o n space b o t h space s i d e s ln open parentheses x to the power of n minus 1 close parentheses equals ln open square brackets open parentheses x minus 1 close parentheses open parentheses x minus Z subscript 1 close parentheses open parentheses x minus Z subscript 2 close parentheses open parentheses x minus Z subscript 3 close parentheses..... open parentheses x minus Z subscript n minus 1 end subscript close parentheses close square brackets D i f f e r e n t i a t i n g space b o t h space s i d e s space w e space g e t fraction numerator n x to the power of n minus 1 end exponent over denominator x to the power of n minus 1 end fraction equals fraction numerator 1 over denominator x minus 1 end fraction plus fraction numerator 1 over denominator x minus Z subscript 1 end fraction plus fraction numerator 1 over denominator x minus Z subscript 2 end fraction plus..... plus fraction numerator 1 over denominator x minus Z subscript n minus 1 end subscript end fraction P u t t i n g space x equals 3 comma space w e space g e t fraction numerator n open parentheses 3 close parentheses to the power of n minus 1 end exponent over denominator 3 to the power of n minus 1 end fraction equals 1 half plus fraction numerator 1 over denominator 3 minus Z subscript 1 end fraction plus fraction numerator 1 over denominator 3 minus Z subscript 2 end fraction plus...... plus fraction numerator 1 over denominator 3 minus Z subscript n minus 1 end subscript end fraction  rightwards double arrow fraction numerator 1 over denominator 3 minus Z subscript 1 end fraction plus fraction numerator 1 over denominator 3 minus Z subscript 2 end fraction plus...... plus fraction numerator 1 over denominator 3 minus Z subscript n minus 1 end subscript end fraction equals fraction numerator n open parentheses 3 close parentheses to the power of n minus 1 end exponent over denominator 3 to the power of n minus 1 end fraction minus 1 half

Answered by satyajit samal | 11th May, 2015, 06:25: AM