Question about the convexity of a set in $\mathbb{C}$

Question

I want to know if $N$ is convex set of $\mathbb{C}$ or not. I hope to show that it is not convex. $$N=\{\lambda\in \mathbb{C};\;\exists (x_n)=(a_n,b_n,c_n)\in \mathbb{C}^3\;\;\hbox{such that}\;\;|a_n|^2+|b_n|^2=1\;\;\hbox{and}\;\;\displaystyle\lim_{n\longrightarrow\infty}(a_n\overline{b_n}+b_n\overline{a_n})=\lambda\},$$ And thank you for your help..

Answer 1

Your set $N$ is just the interval $[-1,1]$, which is convex.

Each number $a_n\overline{b_n}+\overline{a_n}b_n$ is a real number. Therefore, $\lambda\in\mathbb R$. Besides,$$\bigl|a_n\overline{b_n}+\overline{a_n}b_n\bigr|=2\bigl|\operatorname{Re}\bigl(a_n\overline{b_n}\bigr)\bigr|\leqslant2|a_n|.|b_n|\leqslant|a_n|^2+|b_n|^2=1.$$So, $N\subset[-1,1]$.

On the other hand, if $\lambda\in[-1,1]$, then $\lambda=\sin\theta$, for some $\theta\in\mathbb R$. But then $\lambda=2\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)$. So, take $a_n=\sin\left(\frac\theta2\right)$ and $b_n=\cos\left(\frac\theta2\right)$. Then $|a_n|^2+|b_n|^2=1$ and$$a_n\overline{b_n}+\overline{a_n}b_n=2\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)=\sin\theta=\lambda.$$