All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
A symmetric relation (Posted on 2014-11-19) Difficulty: 3 of 5
Find a couple of complex numbers, such that one number is the square of the other and vice versa.

Rem: There will be more than one solution.

See The Solution Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 1 of 5
x=y^2 (1)
y=x^2 (2)

(1)-(2) gives
x-y=y^2-x^2
-(y-x)=(y+x)(y-x)

first we have y-x=0 as a solution
giving y=x thus
x=x^2 thus
x=0 or x=1 giving the solutions (0,0) and (1,1)

if y-x is not zero then we have
y+x=-1
y=-x-1 giving
-x-1=x^2
x^2+x+1=0
x=(-1+-sqrt(-3))/2
x=-1/2+i*sqrt(3)/2 or x=-1/2-i*sqrt(3)/2
giving
y=-1/2-i*sqrt(3)/2 and y=-1/2+i*sqrt(3)/2 respectively
thus we have three distinct solutions
(0,0)
(1,1)
(-1/2+i*sqrt(3)/2,-1/2-i*sqrt(3)/2)

  Posted by Daniel on 2014-11-19 08:28:56
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information