Is the set of complex solutions to $x^2+y^2 = 1$ isomorphic to $\mathbb{C}^*$?

Question

In his article about Grothendieck, Edward Frenkel states that the set of complex solutions to the equation $x^2+y^2 = 1$ is "a plane with one point removed." I'm curious how this can be made precise. Since the plane minus the origin is $\mathbb{C}^*$, it can be viewed as the algebraic variety cut out from $\mathbb{C}^2$ by the equation $zw=1$. Is Frenkel saying that these two varieties are isomorphic?

To be precise, my question is:

Are $\{(x,y) \in \mathbb{C}^2 : x^2+y^2 = 1\}$ and $\{(z,w) \in \mathbb{C}^2 : zw = 1\}$ isomorphic as complex algebraic varieties?

The accepted answer to this question shows that the above subsets of $\mathbb{C}^2$ are homeomorphic, but it's not at all clear to me how one would find polynomial functions realizing this homeomorphism.

Answer 1

You mean polynomial functions like $z = x + iy$, $w = x - iy$?

Answer 2

Hint. Substituting $iy$ for $y$ transforms $x^2+y^2=1$ into $x^2-y^2=1$, i.e. $(x+y)(x-y)=1$. Now let $z=x+y$ and $w=x-y$.