The statement of the problem : Find the numbers $r \in (0,\infty)$ for which exists a $z \in \mathbb C$ such that $|z| = r$ and $|1+z^2|=2r$ .
My approach : I tried a geometric solution in which I looked for complex numbers such that circles of radius 2r have such numbers, but I got stuck.
Any and all proofs will be helpful. Thanks a lot!
On
In polar form, the two conditions mean that $$ (r^2\cos2\theta+1)^2+(r^2\sin2\theta)^2=4r^2. $$ That is, $$ r^4-4r^2+1=-2r^2\cos2\theta. $$ Therefore, a solution $z=r(\cos\theta+i\sin\theta)$ exists if and only if $-2r^2\le r^4-4r^2+1\le2r^2$.
The first inequality can be rewritten as $(r^2-1)^2\ge0$, which is always true. The second one is equivalent to $r^4-6r^2+1\le0$ which holds iff $3-\sqrt{8}\le r^2\le3+\sqrt{8}$, i.e., iff $\sqrt{3-\sqrt{8}}\le r\le\sqrt{3+\sqrt{8}}$.
A solution based on analytic geometry:
Render the equations as
$|z^2+1|=2|z|$
Square both sides and express the squared norms in terms of real and imaginary parts:
$(x^2-y^2+1)^2+4x^2y^2=4(x^2+y^2)$
$(x^4+2x^2y^2+y^4)-(2x^2\color{blue}{+6y^2})+1=0$
Compare this with
$(x^4+2x^2y^2+y^4)-(2x^2\color{blue}{+2y^2})+1=(x^2+y^2-1)^2$
and observe:
$(x^2+y^2-1)^2=4y^2$
We now take square roots of both sides and discover the relationship is the union of two circles!
$x^2+y^2-1=\pm2y$
The circle with the $\pm$ sign on the right side chosen as $+$ may be recast as
$x^2+(y^2-2y)=1,$
and upon completing the square in the $y$ terms we find the center of the circle is at $(0,1)$ and the radius is $\sqrt2$:
$x^2+(y-1)^2=2.$
If the $\pm$ sign is chosen as $-$ we similarly get a center at $(0,-1)$ and radius also $\sqrt2$
$x^2+(y+1)^2=2.$
Taking the difference between the radius and the center-to-origin distance, both of which terms are the same for both circles, we get $\sqrt2-1$ as the minimal distance from the origin and this represents the minimum allowed value of $r=|z|=\sqrt{x^2+y^2}$. The corresponding sum $\sqrt2+1$ is the maximum value of $r$.