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    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Polar solution | Comment 4 of 5 |
We want x=y^2 and y=x^2 thus x = x^4
One solution is x=0
others arise if
x³ = 1

In polar form
x = r(cos(θ)+isin(θ))
x³ = r³(cos(3θ)isin(3θ)) = 1

which requires r³=1 and 3θ=360º*n (n an integer)
so r=1
and θ can be one of 3 values
if n=0, θ=0º
if n=1, θ=120º
if n=2, θ=240º
(if n is over 2, these repeat)

Solutions:
(0,0) (1,1)  ((cos(120)+isin(120)), (cos(240)+isin(240))

  Posted by Jer on 2014-11-19 10:40:05
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 (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information