What are the solutions to $x=\pm \sqrt y$ and $y=\pm \sqrt x$?

Question

What are the solutions to $x=\pm \sqrt y$ and $y=\pm \sqrt x$? Are there complex solutions? $x$ and $y$ are both variables. Working in the field of complexes.

Answer 1

Yes, there are solutions. If $x$ is a square root of $y$ and $y$ is a square root of $x$, then $x^4=y^2=x$. If $x^4=x$, then$$x\in\left\{0,1,-\frac12+\frac{\sqrt3}2i,-\frac12-\frac{\sqrt3}2i\right\},$$and each of these numbers is a solution:

• if $x=0$, take $y=0$;
• if $x=1$, take $y=1$,
• if $x=-\frac12+\frac{\sqrt3}2i$, take $y=-\frac12-\frac{\sqrt3}2i$;
• if $x=-\frac12-\frac{\sqrt3}2i$, take $y=-\frac12+\frac{\sqrt3}2i$;.

Answer 2

Your formulae solve to $x^2=y$ and $y^2=x$, thus inserting into onanother $x^4=(x^2)^2=x$, so we get 4 solutions, i.e. $x_1=0$ plus the 3 complex solutions of $x^3=1$.

--- rk

Answer 3

Obviously,

$$x=\pm\sqrt{\pm\sqrt x}.$$

In the polar representation, the modulus equation is

$$r=\sqrt[4]r,$$ solved by $r=0$ and $r=1$.

The argument equation is

$$\theta=\frac{\dfrac{\theta}2+m\pi}2+n\pi=\frac\theta4+k\frac\pi2$$ and

$$\theta=k\frac{2\pi}3$$ for $k=0,1,2$.

By the first equation, for a given $x$, the corresponding $y$ is $x^2$.

Answer 4

To find intersection points we square the given equations

$$ y=x^2,\,x=y^2\,$$

Eliminate $x$ and to find $y$ solutions,

$$ y^4-y=0,\, y= ( 0,1, \frac{-1}{2}\pm\frac{ \sqrt3i}{2})$$

Similarly the $x$ solutions.

Two Real points are:

$$ (0,0) , ( 1,1) $$

Two complex points are:

$$ [(\frac{-1}{2}, \frac{ \sqrt3i}{2}), (\frac{-1}{2}, \frac{- \sqrt3i}{2})]\quad$$ $$ [(\frac{-1}{2}+ \frac{- \sqrt3i}{2}), (\frac{-1}{2}, \frac{+ \sqrt3i}{2})].$$