convexity of $M_1,M_2\subseteq \mathbb{C}^2$ Let us consider the following two subsets of $ \mathbb{C}^2$:
$$
M_1=\{(|a|^2+b\overline{c},|c|^2);\;(a,b,c) \in \mathbb{C}^3\;\;\hbox{such that}\;|a|^2+|c|^2=1\},
$$
and
$$
M_2=\{(|a|^2+b\overline{c},2|a|^2+|c|^2);\;(a,b,c) \in \mathbb{C}^3\;\;\hbox{such that}\;|a|^2+|c|^2=1\}
$$

What do you think about the convexity of $M_1$ and $M_2$? I want to show that  $M_1$ or $M_2$ is not a convex subset of $\mathbb{C}^2$.

Thank you!!
 A: Both $M_1$ and $M_2$ are convex.
For $M_1$ consider $c = 0$. Then $|a| = 1$ so we have $(|a|^2 + b\overline{c}, |c|^2) = (1,0)$.
Now consider $c \ne 0$ with $|c|^2 \in \langle 0, 1]$.
The first coordinate can be any complex number $z$. Indeed, for $b = \overline{c^{-1}}(z - |a|^2)$ we have $(|a|^2 + b\overline{c}, |c|^2) = (z,|c|^2)$. Therefore $\mathbb{C} \times \langle 0, 1] \cup \{(1, 0)\} \subseteq M_1$. Since $M_1 \subseteq \mathbb{C} \times [0,1]$, we see that in fact $M_1 = \mathbb{C} \times \langle 0, 1] \cup \{(1, 0)\}$. This is a convex set since $\mathbb{C} \times \langle 0, 1]$ is convex, and any line segment connecting $(1,0)$ to a point in $\mathbb{C} \times \langle 0, 1]$ is in $\mathbb{C} \times \langle 0, 1] \cup \{(1,0)\}$.
Similarly, for $M_2$ consider $c = 0$. Then $|a| = 1$ so we have $(|a|^2 + b\overline{c}, 2|a|^2 + |c|^2) = (1,2)$.
Now consider $c \ne 0$ with $|c|^2 \in \langle 0, 1]$ and $a \in \mathbb{C}$ such that $|a|^2 + |c|^2 = 1$.
The first coordinate can be any complex number $z$, regardless of $a$. Indeed, for $b = \overline{c^{-1}}(z - |a|^2)$ we have $(|a|^2+b\overline{c}, 2|a|^2 + |c|^2) = (z,1 + |a|^2)$. $|a|^2 \in [0, 1\rangle $ so $1 + |a|^2 \in [1, 2\rangle$. 
Therefore $\mathbb{C} \times [1, 2\rangle \cup \{(1, 2)\} \subseteq M_2$. Since $M_2 \subseteq \mathbb{C} \times [1,2]$, we see that in fact $M_2 = \mathbb{C} \times [1, 2\rangle \cup \{(1, 2)\} $. This is a convex set since $\mathbb{C} \times [1, 2\rangle \cup \{(1, 2)\}$ is convex, and any line segment connecting $(1,2)$ to a point in $\mathbb{C} \times [1, 2\rangle$ is in $\mathbb{C} \times [1, 2\rangle \cup \{(1, 2)\}$.
