perplexus.info :: Just Math : A symmetric relation

A symmetric relation (Posted on 2014-11-19)

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

Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Solution

Comment 5 of 5 |

z_1 = z_2^2

z_2 = z_1^2

z_1 = (z_1^2)^2

z_1(z_1^3 - 1) = 0

z_1 = 0, 1, -1/2 + sqrt(3)/2 i, -1/2 - sqrt(3)/2 i

Plug into second equation to get corresponding z_2.

Edited on November 19, 2014, 10:03 pm

Posted by Bractals on 2014-11-19 21:54:44

Please log in:

Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info