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# z^n=(z+1)^n. roots of the equation lie on the linea) 2x+1=0b)2x-1=0c)x+1=0d)x-1=0

12 years ago

let w=(z+1)/z , then wn=1 and w not equal to 1

w = [ cos(2rπ)+i sin(2rπ) ]1/n

= cos(2rπ/n)+i sin(2rπ/n)    ( de moivres theorem )

w  = cos(2rπ/n)+i sin(2rπ/n)    where r=1,2,3,.......n-1

(z+1)/z  = cos(2rπ/n)+i sin(2rπ/n)

1/z  = -1+ cos(2rπ/n)+i sin(2rπ/n)

= -2 sin2(rπ/n) + 2i sin(rπ/n).cos(rπ/n)

=  2i sin(rπ/n)[ cos(rπ/n)+i sin(rπ/n)]

=>   x +iy = z = 1 / {2i sin(rπ/n).[ cos(rπ/n)+i sin(rπ/n)]}

=  [ cos(rπ/n)+i sin(rπ/n)]-1 /  [2i sin(rπ/n)]

=  [ cos(rπ/n) -i sin(rπ/n)] / [2i sin(rπ/n)]               ( De moivres theorem )

= -1/2  - i.cot(rπ/n)/2  where   r = 1,2,3,.......n-1

this shows that all roots lie on 2x+1=0

12 years ago

Any solution to the equation zn = (z+1)n satisfies |z| = |z+1| i.e. the point is equidistant from (0,0) and (-1,0).

This means it lies on the perpendicular bisector of the line joining (0,0) and (-1,0) i.e. x = -1/2 or 2x+1 = 0.