### 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 the equation has only even powers of $z$, so I can try to make a substitution $u = z^2$ and see if that simplifies the problem.
If I do that, I get $u^2 - 4u + 3 = 0$, which is a quadratic equation that I can factor as $(u - 3)(u - 1) = 0$.
Then, $u = 3$ or $u = 1$ are the solutions for $u$, but I need to find the solutions for $z$.
To do that, I need to undo the substitution and take the square root of both sides, remembering to include both the positive and negative roots.
So, $z^2 = 3$ implies that $z = \pm \sqrt{3}$, and $z^2 = 1$ implies that $z = \pm 1$.
Therefore, the solutions for $z$ are $\sqrt{3}, -\sqrt{3}, 1, -1$.