If $$|z^2 - 1| = |z|^2 + 1$$, then $$z$$ lies on

Let the complex number be expressed in its Cartesian form:

$$ z = x + iy $$

Substitute this form into the definition of the square of the absolute value of z:

$$ |z|^2 = x^2 + y^2 $$

Now, find an expression for the square of z:

$$ z^2 = (x + iy)^2 $$

$$ z^2 = x^2 + 2ixy + (iy)^2 $$

$$ z^2 = x^2 - y^2 + i2xy $$

Substitute this value into the left side of the given condition equation:

$$ |z^2 - 1| = |(x^2 - y^2 - 1) + i2xy| $$

Calculate the modulus of this expression by summing the squares of the real and imaginary parts under a square root:

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

Now write out the full original equation using these expressions:

$$ \sqrt{(x^2 - y^2 - 1)^2 + (2xy)^2} = x^2 + y^2 + 1 $$

Square both sides of the equation to eliminate the radical sign:

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

Expand the trinomial on the left side of the equality:

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

Combine this with the remaining term on the left side:

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

Expand the trinomial on the right side of the equality:

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

Equate the simplified left side expression and the expanded right side expression:

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

Cancel out all the identical terms present on both sides of the equation:

$$ -2x^2 = 2x^2 $$

Move all terms to one side to solve for x:

$$ 4x^2 = 0 $$

$$ x^2 = 0 $$

$$ x = 0 $$

Since the real part of the complex number is 0, the number is purely imaginary. This condition means that the point always lies on the vertical line of the complex plane.

Final Answer:

The complex number z lies on the imaginary axis.