I am trying to solve this equation in sage $$ \sqrt{-4 \, z^{2} + 2 \, \sqrt{-4 \, z^{2} + 1} - 1} = 0. $$ But when I try the code
var('z')
eq = sqrt(-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1) == 0
solve(eq,z)
I get
[z == -1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1), z == 1/2*sqrt(2*sqrt(-4*z^2 + 1) - 1)]
Is there any way to actually solve it in sage?
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_a = sqrt(-4*z^2 + 2*sqrt(-4*z^2 + 1) - 1) == 0
sage: 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: 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: eq_c = eq_b + 4*z^2 + 1
sage: 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: eq_d = eq_c^2
sage: eq_d
-16*z^2 + 4 == (4*z^2 + 1)^2
Solve the new equation.
sage: sols_d = solve(eq_d, z)
sage: 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_a.subs(s)) for s in 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)]
Asked: 2017-12-31 02:00:00 +0100
Seen: 1,305 times
Last updated: Dec 31 '17
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To better specify the question, do you want to solve for $z$ in $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, other?
One thing to notice is that $\sqrt{\operatorname{expression}} = 0$ is equivalent to $\operatorname{expression} = 0$. This gets you rid of one of the square roots.
This has two solutions in $\mathbb C$, $\pm \sqrt{2\sqrt{3}-3}/2$.
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