Solving this problem of convexity of $N\subseteq \mathbb{C}^2$ Let us consider the following set of $ \mathbb{C}^2$:
$$
N=\{(|a|^2-|b|^2,2\Re e(a\overline{b}));\;x =(a,b) \in \mathbb{C}^2\;\;\hbox{and}\;|a|^2+|b|^2=1\}
$$

I want to prove that $N$ is convex or to construct a counter-example which show that $N$ is not a convex set of $ \mathbb{C}^2$.

We see that $N\subseteq \mathbb{R}^2$. I try to show that $N$ is not convex:
$(1; 0),\,(-1; 0)\in N$ but their midpoint $(0; 0)\in N$.
Thank you
 A: Your set $N$ is simply the closed unit disk, which is convex.
If $a,b\in\mathbb C$ are such that $|a|^2+|b|^2=1$, then$$a=r(\cos\alpha+i\sin\alpha)\text{ and }b=\sqrt{1-r^2}(\cos\beta+i\sin\beta),$$for some $r\in[0,1]$ and some $\alpha,\beta\in\mathbb R$. Therefore $|a|^2-|b|^2=r^2-(1-r^2)=2r^2-1\in[-1,1]$. On the other hand, $2\operatorname{Re}\bigl(a\overline b\bigr)=2r\sqrt{1-r^2}\cos(\alpha-\beta)$. So, if $(x,y)\in N$, then, for some $r\in[0,1]$, $x=2r^2-1$ and $y\in\left[-2r\sqrt{1-r^2},2r\sqrt{1-r^2}\right]$. Note that$$x=2r^2-1\iff r=\sqrt{\frac{x+1}2}$$and that therefore$$y\in\left[-2r\sqrt{1-r^2},2r\sqrt{1-r^2}\right]\iff y\in\left[-\sqrt{1-x^2},\sqrt{1-x^2}\right].$$So, this proves that $N$ is a subset of the closed unit disk.
On the other hand, if $(x,y)$ belongs to that disk, set $r=\sqrt{\frac{x+1}2}$. Then$$|y|\leqslant\sqrt{1-x^2}=2r\sqrt{1-r^2}.$$So, $y=2r\sqrt{1-r^2}\cos\gamma$, for some $\gamma\in\mathbb R$. So, take $a=r(\cos\gamma+i\sin\gamma)$ and $b=\sqrt{1-r^2}$.
