1 | initial version |

To better specify the question, one would need to say if we want to solve for $z$ in $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, or other.

Here we examine how to solve over $\mathbb{C}$.

Define $z$ as a symbolic variable, and define the equation (as you did).

```
sage: z = SR.var('z')
sage: eq = sqrt(-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1) == 0
sage: solve(eq, z)
[z == -1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1), z == 1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1)]
```

First notice that $\sqrt{\operatorname{expression}} = 0$ is equivalent to $\operatorname{expression} = 0$. This gets you rid of one of the square roots.

```
sage: eqq = eq^2
sage: eqq
-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1 == 0
```

Then isolate the other square root.

```
sage: eqqq = eqq + 4*z^2 + 1
sage: eqqq
2*sqrt(-4*z^2 + 1) == 4*z^2 + 1
```

Remove it by squaring. Note that the new equation is implied by the initial one, but no longer equivalent to it.

```
sage: eqqqq = eqqq^2
sage: eqqqq
-16*z^2 + 4 == (4*z^2 + 1)^2
```

Solve the new equation.

```
sage: sols = solve(eqqqq, z)
sage: sols
[z == -1/2*sqrt(2*sqrt(3) - 3), z == 1/2*sqrt(2*sqrt(3) - 3), z == -1/2*sqrt(-2*sqrt(3) - 3), z == 1/2*sqrt(-2*sqrt(3) - 3)]
```

Check which of the solutions are solutions of the initial equation.

```
sage: [bool(eq.subs(s).simplify_full()) for s in sols]
[True, True, False, False]
```

2 | No.2 Revision |

To better specify the question, one would need to say if we want to solve for $z$ in $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, or other.

Here we examine how to solve over $\mathbb{C}$.

Define $z$ as a symbolic variable, and define the equation (as you did).

```
sage: z = SR.var('z')
sage:
```~~eq ~~eq_a = sqrt(-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1) == 0
sage: ~~solve(eq, ~~solve(eq_a, z)
[z == -1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1), z == 1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1)]

First notice that $\sqrt{\operatorname{expression}} = 0$ is equivalent to $\operatorname{expression} = 0$. This gets you rid of one of the square roots.

`sage: `~~eqq = eq^2
sage: eqq
~~eq_b = eq_a^2
sage: eq_b
-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1 == 0

Then isolate the other square root.

`sage: `~~eqqq = eqq ~~eq_c = eq_b + 4*z^2 + 1
sage: ~~eqqq
~~eq_c
2*sqrt(-4*z^2 + 1) == 4*z^2 + 1

Remove it by squaring. Note that the new equation is implied by the initial one, but no longer equivalent to it.

`sage: `~~eqqqq = eqqq^2
sage: eqqqq
~~eq_d = eq_c^2
sage: eq_d
-16*z^2 + 4 == (4*z^2 + 1)^2

Solve the new equation.

`sage: `~~sols = solve(eqqqq, ~~sols_d = solve(eq_d, z)
sage: ~~sols
~~sols_d
[z == -1/2*sqrt(2*sqrt(3) - 3), z == 1/2*sqrt(2*sqrt(3) - 3), z == -1/2*sqrt(-2*sqrt(3) - 3), z == 1/2*sqrt(-2*sqrt(3) - 3)]

Check which of the solutions are solutions of the initial equation.

`sage: `~~[bool(eq.subs(s).simplify_full()) ~~[bool(eq_a.subs(s)) for s in ~~sols]
~~sols_d]
[True, True, False, False]

Define the solutions of the original equation:

```
sage: sols_a = [s for s in sols_d if bool(eq_a.subs(s))]
sage: sols_a
[z == -1/2*sqrt(2*sqrt(3) - 3), z == 1/2*sqrt(2*sqrt(3) - 3)]
```

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