Show that the equation |z^2+1|-2|z|=0 represents two circles…

Show that the equation $|z^2+1|-2|z|=0$ represents two circles and find their equations.

Step 1 · Set up in Cartesian form

Let $z=x+iy$. Then $z^2+1=(x^2-y^2+1)+2ixy$, so:

$$ |z^2+1| = \sqrt{(x^2-y^2+1)^2+4x^2y^2}, \qquad |z|=\sqrt{x^2+y^2} $$

The equation $|z^2+1|=2|z|$ becomes $\sqrt{(x^2-y^2+1)^2+4x^2y^2} = 2\sqrt{x^2+y^2}$.

Step 2 · Square both sides

$$ (x^2-y^2+1)^2+4x^2y^2 = 4(x^2+y^2) $$

Expanding $(x^2-y^2+1)^2 = (x^2-y^2)^2+2(x^2-y^2)+1$ and adding $4x^2y^2$:

$$ x^4+2x^2y^2+y^4 + 2x^2-2y^2+1 = 4x^2+4y^2 $$

Step 3 · Recognize $(x^2+y^2)^2$

Since $x^4+2x^2y^2+y^4=(x^2+y^2)^2$:

$$ (x^2+y^2)^2 + 2x^2-2y^2+1 = 4x^2+4y^2 \quad\Longrightarrow\quad (x^2+y^2)^2 - 2x^2 - 6y^2 + 1 = 0 $$

Step 4 · Rewrite in terms of $r^2=x^2+y^2$

Using $x^2=r^2-y^2$:

$$ r^4 - 2(r^2-y^2) - 6y^2 + 1 = 0 \quad\Longrightarrow\quad r^4-2r^2+1 = 4y^2 $$

$$ (r^2-1)^2 = 4y^2 \quad\Longrightarrow\quad r^2-1 = \pm2y $$

Step 5 · Two circle equations

Case $r^2-1=2y$: $$ x^2+y^2-2y-1=0 \quad\Longrightarrow\quad x^2+(y-1)^2=2 $$

Case $r^2-1=-2y$: $$ x^2+y^2+2y-1=0 \quad\Longrightarrow\quad x^2+(y+1)^2=2 $$

Both are circles of radius $\sqrt2$, centered at $(0,1)$ and $(0,-1)$ respectively — symmetric about the real axis, with centers $2$ apart on the imaginary axis.

Step 6 · Verify with sample points

On circle 1, take $z=i(1+\sqrt2)$ (i.e. $(0,1+\sqrt2)$):

$$ z^2 = -(1+\sqrt2)^2 = -(3+2\sqrt2), \qquad z^2+1 = -2-2\sqrt2, \qquad |z^2+1| = 2+2\sqrt2 $$

$$ 2|z| = 2(1+\sqrt2) = 2+2\sqrt2 \quad\checkmark $$

On circle 2, take $z=i(-1+\sqrt2)$:

$$ z^2 = -(3-2\sqrt2), \qquad z^2+1=-2+2\sqrt2, \qquad |z^2+1|=2\sqrt2-2 $$

$$ 2|z| = 2(\sqrt2-1) = 2\sqrt2-2 \quad\checkmark $$

(Both checks were also confirmed numerically by sampling many points around each full circle — the residual of $|z^2+1|-2|z|$ is zero to floating-point precision everywhere on both circles.)