 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
```
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

```
11 years ago

```							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

```
11 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.
```
11 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions