### Instruction: Find all values of $z$ such that $z^4 - 4z^2 + 3 = 0$.  Enter all the solutions, separated by commas.### Response: I notice that this equation is a quartic, but it only has even powers of $z$.
That makes me think of substituting $w = z^2$, so that the equation becomes a quadratic in terms of $w$.
If I do that, I get $w^2 - 4w + 3 = 0$, which I can factor as $(w - 1)(w - 3) = 0$.
This means that $w$ can be either $1$ or $3$, but I need to find the values of $z$ that satisfy this.
To do that, I can take the square root of both sides of the equations $w = 1$ and $w = 3$, keeping in mind that there are two possible values for each square root: one positive and one negative.
For $w = 1$, I get $z^2 = 1$, which means that $z = \pm 1$.
For $w = 3$, I get $z^2 = 3$, which means that $z = \pm \sqrt{3}$.
So the four values of $z$ that make the original equation true are $-1, 1, -\sqrt{3}, \sqrt{3}$.