### Instruction: Find all values of $z$ such that $z^4 - 4z^2 + 3 = 0$.  Enter all the solutions, separated by commas.### Response: This is a polynomial equation of degree 4, but I notice that it has only even powers of $z$.
That makes me think of a substitution: let $u = z^2$.
Then the equation becomes $u^2 - 4u + 3 = 0$.
This is a quadratic equation, and I can factor it as $(u - 3)(u - 1) = 0$.
This means that $u = 3$ or $u = 1$ are the solutions for $u$.
But I want to find the solutions for $z$, so I need to undo the substitution.
If $u = z^2$, then $z = \pm \sqrt{u}$.
So for each value of $u$, there are two possible values of $z$.
When $u = 3$, $z = \pm \sqrt{3}$.
When $u = 1$, $z = \pm 1$.
These are the four solutions for $z$.