Let $$M=\{\lambda\in \mathbb{C};\;\exists (x_n)=(a_n,b_n,c_n)\in \mathbb{C}^3\;\;\hbox{such that}\;\;|a_n|^2+|b_n|^2=1 \;\;\hbox{and}\;\;\displaystyle\lim_{n\longrightarrow+\infty}|a_n|^2+i|b_n|^2= \lambda\}.$$ My goal is to show that $M$ is not convex but perhaps it is convex. Could you please help me?
And thank you very much..
Yes, $M$ is convex. It is simply the line segment $L$ joining $1$ and $i$.
If $z\in L$, then $z=t+(1-t)i$, for some $t\in[0,1]$. So, take $a_n=\sqrt t$, $b_n=\sqrt{1-t}$, and $c_n=0$ (for any $n\in\mathbb N$). Then$$|a_n|^2+|b_n|^2={a_n}^2+{b_n}^2=t+(1-t)=1$$and $z=t+(1-t)i=\lim_{n\to\infty}|a_n|+|b_n|i$.
On the other hand, if $z\in M$, then\begin{align}z&=\lim_{n\to\infty}|a_n|^2+|b_n|^2i\\&=\lim_{n\to\infty}|a_n|^2+\bigl(1-|a_n|^2\bigr)i.\end{align}So, $z$ is the limit of a sequence of elements $L$, which is a closed subset of $\mathbb C$. Therefore, $z\in L$.