MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
palash awasthi Grade:
        

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

8 years ago

Answers : (2)

Ramesh V
70 Points
										

The answer is A


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


 


 


 


 
8 years ago
mycroft holmes
266 Points
										

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.

8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details