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)

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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

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