Question

the complex solution of (z+i)^2011=z^2011 lie on
a. a circle
b.an ellipse
c. a hyperbola
d. a straight line

Answer 1

z = x + i y
  1 / z = (x - i y)/(x²+y²)
   i / z = ( i x + y) /(x²+y²)
(z + i)²⁰¹¹  = z²⁰¹¹
=>  [ (z + i)/z ]²⁰¹¹ = 1
=>  [ 1 + i/z ]²⁰¹¹  = 1
=>  [ 1 + y/(x²+y²)  + i x/(x²+y²) ]²⁰¹¹ = 1 
=>  [ (x²+y²+y)/(x²+y²)  +  i x/(x²+y²) ]²⁰¹¹ = 1

The complex number on the LHS can be expressed in terms of magnitude A and exp(i theta).  Then A^2011 = 1.   So  A = 1.

Hence,     (x²+y²+y)²  +  x²  =  (x²+y²)²
                (x²+y²)² + y² + 2 (x²+y²) y  + x² = (x²+y²)²
                (x² + y²) (1 + 2 y) = 0
         => y = -1/2

Ans:   A straight line.

Answer 2

z = x + i y
1 / z = (x - i y)/(x²+y²)
i / z = ( i x + y) /(x²+y²)
(z + i)²⁰¹¹ = z²⁰¹¹
= [ (z + i)/z ]²⁰¹¹ = 1
= [ 1 + i/z ]²⁰¹¹ = 1
= [ 1 + y/(x²+y²) + i x/(x²+y²) ]²⁰¹¹ = 1
= [ (x²+y²+y)/(x²+y²) + i x/(x²+y²) ]²⁰¹¹ = 1
The number on the LHS can be expressed in terms of magnitude. Then a...2011 = 1. .:. a = 1.

(x²+y²+y)² + x² = (x²+y²)²
(x²+y²)² + y² + 2 (x²+y²) y + x² = (x²+y²)²
(x² + y²) (1 + 2 y) = 0
straight line.