Problem of convexity in $ \mathbb{C}$ Consider the following set
\begin{eqnarray*}
S
&=&\Big\{\lambda\in \mathbb{C};\;\exists (x_n,y_n)\in \mathbb{C}^2\,;\;\;\;|x_n+y_n|=1, \\
&&\phantom{++++++++++}\;\displaystyle\lim_{n\longrightarrow+\infty}|x_n|^2+x_n\overline{y_n}= \lambda\;\;\hbox{and}\;\;\displaystyle\lim_{n\longrightarrow+\infty}|x_n|=\frac{1}{2}\Big\}.
\end{eqnarray*}

I want to show that $S$ is not convex.



*

*If $(x_n,y_n)=(1/2,1/2)$ then $1/2\in S$.

*If $(x_n,y_n)=(-1/2,3/2)$ then $-1/2\in S$. 
I hope to show that $0\notin S$.
Thank you for your help.
 A: As you have shown $\lambda_1=1/2$ and $\lambda_2=-1/2$ both are in $S$. If $S$ were convex then for every $\alpha\in(0,1)$ we must have $\alpha\lambda_1+(1-\alpha)\lambda_2\in S$. In particular for $\alpha=1/2$ it follows that $0\in S$. By definition of $S$ this implies that there exists $(x_n,y_n)\in \mathbb{C}^2$ such that $|x_n+y_n|=1, \lim_n(|x_n|^2+x_n\overline{y}_n)=0$ and $\lim_n|x_n|=1/2$. From these conditions we get the following implications
$$\lim_nx_n\overline{y}_n=-\lim_n|x_n|^2=-1/4\Rightarrow\lim_n|x_n||\overline{y}_n|=|\lim_nx_n\overline{y}_n|=1/4\Rightarrow \lim_n|\overline{y}_n|=1/2$$
since the absolute value function $|\cdot|$ is continuous. On the other hand
$$|x_n+y_n|=1\Rightarrow (x_n+y_n)(\overline{x}_n+\overline{y}_n)=1\Rightarrow|x_n|^2+x_n\overline{y}_n+\overline{x}_ny_n+|y_n|^2=1$$
This implies
$$\lim_n(\overline{x}_ny_n+|y_n|^2)=1\Rightarrow |\lim_n\overline{x}_ny_n+|y_n|^2|=1\Rightarrow\lim_n|\overline{x}_ny_n|+\lim_n|y_n|^2\geqslant 1$$
by triangle inequality. Using $\lim_n|\overline{x}_n|=\lim_n|x_n|=1/2$ and $\lim_n|y_n|=\lim_n|\overline{y}_n|=1/2$ one gets
$$\lim_n|\overline{x}_ny_n|+\lim_n|y_n|^2=\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2^2}=\frac{1}{2}<1$$
which is a contradiction.
