z^n=(z+1)^n. roots of the equation lie on the line

a) 2x+1=0

b)2x-1=0

c)x+1=0

d)x-1=0

Question

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

Ramesh V · Answer

The answer is A let w=(z+1)/z , then w n =1 and w not equal to 1 w = [ cos(2r&pi;)+i sin(2r&pi;) ] 1/n = cos(2r&pi;/n)+i sin(2r&pi;/n) ( de moivres theorem ) w = cos(2r&pi;/n)+i sin(2r&pi;/n) where r=1,2,3,.......n-1 (z+1)/z = cos(2r&pi;/n)+i sin(2r&pi;/n) 1/z = -1+ cos(2r&pi;/n)+i sin(2r&pi;/n) = -2 sin 2 (r&pi;/n) + 2i sin(r&pi;/n).cos(r&pi;/n) = 2i sin(r&pi;/n)[ cos(r&pi;/n)+i sin(r&pi;/n)] => x +iy = z = 1 / {2i sin(r&pi;/n).[ cos(r&pi;/n)+i sin(r&pi;/n)]} = [ cos(r&pi;/n)+i sin(r&pi;/n)] -1 / [2i sin(r&pi;/n)] = [ cos(r&pi;/n) -i sin(r&pi;/n)] / [2i sin(r&pi;/n)] ( De moivres theorem ) = -1/2 - i.cot(r&pi;/n)/2 where r = 1,2,3,.......n-1 this shows that all roots lie on 2x+1=0

mycroft holmes · Answer

Any solution to the equation z n = (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.

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

z^n=(z+1)^n. roots of the equation lie on the line

a) 2x+1=0

b)2x-1=0

c)x+1=0

d)x-1=0

2 Answers

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