Is this set $\{(x\overline{y},|x|^2)\,;\;(x,y)\in \mathbb{C}^2,\;|y|=1\}$ convex?

Question

Let the following subset of $\mathbb{C}^2$

$$F=\left\{(x\overline{y},|x|^2)\,;\;(x,y)\in \mathbb{C}^2,\;\text{and}\;|y|=1 \right\}.$$

Is $F$ a convex subset of $\mathbb{C}^2$.

Answer 1

No, $F$ is not convex. This is clear if we say $$F'=\{(z,|z|^2):z\in\Bbb C\}$$and note that

$F'=F$.

Proof: Say $p\in F$. Then $p=(x\overline y,|x|^2)$ where $|y|=1$; hence if we set $z=x\overline y$ then $p=(z,|z|^2)$, so $p\in F'$.

Conversely, say $p\in F'$, so $p=(z,|z|^2)$. If $x=z,y=1$ then $p=(x\overline y,|x|^2)$, so $p\in F$.