Meaning of $x^2 \in \mathbb{R^2}$

Question

In this post someone suggested:

"$z\mapsto z^2$"

where both $z$ and $z^2$ are in $\mathbb{S^1}$ (unless I missed something).

Does this mean that they suggested: $z^2 = (z_1^2, z_2^2)$?

If not, what is it?

Did you see that $z\in\mathbb{C}$ in that question? So $z^2$ is simply the complex multiplication $z \cdot z$. — Martin R, Oct 06 '21 at 14:57

score 2 · Accepted Answer · answered Oct 06 '21 at 15:02

2

$$z\in S^1 \implies z\in\mathbb C\implies z^2\in\mathbb C;$$$$\\z\in\mathbb C\kern.6em\not\kern-.6em\implies z^2\in\mathbb C^2.$$

answered Oct 06 '21 at 15:02

ryang

Actually $z \in S^1$ implies $z^2 \in S^1$ as well. – principal-ideal-domain Oct 06 '21 at 20:47
@principal Well yes of course; that’s consistent with what I said. – ryang Oct 07 '21 at 04:21
Of course that's consistent ;-). Both is true and truth is consistent! – principal-ideal-domain Oct 07 '21 at 08:21

score 1 · Answer 2 · answered Oct 06 '21 at 15:00

1

As in the comments is already pointed out $z^2$ means $z\cdot z$.

You suggestest $(z_1^2,z_2^2)$. That's not how the multiplication in $\mathbb C$ works. It's actually $(x^2-y^2,2xy)$ if $z=(x,y)=x+iy$.

answered Oct 06 '21 at 15:00

2 Answers2