Convergence of Sequence in $\mathbb{C}$

Question

Suppose that $x_n\to x\in\mathbb{C}$. Is it true that $|x_n\bar{x}-\bar{x}_n x|\to 0$? I was wondering if my proof was correct for this statement or whether this statement is true at all. \begin{align} |x_n\bar{x}-\bar{x}_nx|&= |x_n\bar{x}+x\bar{x}-x\bar{x}+{x}_n\bar{x}_n-{x}_n\bar{x}_n-\bar{x}_nx|\\ &=|\bar{x}(x_n-x)+\bar{x}_n(x_n-x)+|x|^2-|x_n|^2|\\ &\leq |x||x_n-x|+|x_n||x_n-x|+||x_n|^2-|x|^2| \end{align} which goes to zero since $x_n\to x$.

Answer 1

That's a very elegant solution, I wouldn't change a thing!

Just for informativeness, a more brute force way would be to write $x_n=a_n+ib_n$ and $x=a+ib$. You can then know that $a_n\to a$ and $b_n\to b$, and rewrite

$$\begin{align} |x_n\bar x - \bar x_n x| &= |(a_n + ib_n)(a-ib)-(a_n-ib_n)(a+ib)|\\ &=|a_n a - ia_nb+ib_na+b_nb - a_na-ia_nb+ib_na-b_nb|\\ &=|2b_na - 2a_nb| = 2|b_n a - a_n b| \end{align}$$ and the limit of that is then $$2|b_na-a_nb|\to 2|ba-ab|=0$$

Answer 2

Your solution is correct (although I would not use the letter “x” for complex numbers).

If $z_n \to z$ then both $\bar z_n z$ and $z_n \bar z$ converge to $\bar z z = |z|^2$, so that the difference converges to zero.

Your calculation can be slightly simplified and improved: $$ \begin{align} |\bar z_n z - z_n \bar z| &= |\bar z_n z - \bar z z + z \bar z - z_n \bar z| \\ &\le |\bar z_n z - \bar z z |+ |z\bar z - z_n \bar z| \\ & = 2 |z_n - z| \cdot |z| \, . \end{align} $$