complex analysis

Question

If z= x+iy , then the number of solutions of the equation   

 

z² = z¯ is

 

a) 1      b) 2        c) 3            d) infinite

 

where z¯ = x- iy

please solve it, i m confused

Answer 1

z = x + iy, where x and y are real numbers

_

z = x - iy

      _                  

z² - z  = 0     So, (x + iy)² - (x - iy) = 0

 

                         x² + (2xy)i + i²y² - x + iy = 0

 

                        (x² - x - y²) + (2xy + y)i = 0 = 0 + 0i

 

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

 

So, we have x² - x - y² = 0 and 2xy + y = 0

 

From the second equation, y(2x+1) = 0   So, y = 0 or x = -1/2

 

If y = 0, then, from the first equation, x² - x = 0

 

                                                    So, x(x-1) = 0   x = 0 or x = 1

 

 0 + 0i (also known as 0) is a solution and 1 + 0i (also known as 1) is a solution as well.

 

If x = -1/2, then 1/4 + 1/2 - y² = 0.   So, y² = 3/4.

 

y = √3/2 or -√3/2

 

So, (-1/2) + (√3/2)i and (-1/2) - (√3/2)i are solutions

 

ANSWER:  There are a total of 4 solutions: 0 (ie, 0+0i), 1 (ie, 1+0i),                              (-1/2) + (√3/2)i, and (-1/2) - (√3/2)i

 

NOTE:  There are 4 solutions (not an answer choice!!).  Was it assumed in the problem that z ≠ 0 + 0i??  

Answer 2

If you expand each side you have x²+2ixy-x-y²+iy=0 now you can solve this equation as a quadratic which has two solutions.

 

Jim