Can it be concluded that $z_ n \to z$ as $n \to \infty\ $?

Question

Let $\{z_n\}_{n \geq 1}$ be a sequence of complex numbers and $z \in \mathbb C$ be such that for a given $\varepsilon \gt 0$ there exists $k \in \mathbb N$ such that for all $n \geq k$ we have $$|z_n - z| \lt \varepsilon \sqrt {1 + |z_n|^2}.$$

From here can it be concluded that $z_n \to z$ as $n \to \infty\ $?

Actually I have found this argument in the lecture note I am following but I am unable to figure that out. Any help in this regard would be warmly appreciated.

Thanks for investing your valuable time.

Answer 1

$$|z_n - z| \lt \varepsilon \sqrt {1+|z_n|^2} \iff\frac{|z_n-z|}{\sqrt {1+|z_n|^2}}<\varepsilon,$$ this means $$\lim_{n\to\infty}\frac{|z_n-z|}{\sqrt {1+|z_n|^2}}=0.$$ We claim that:

$\{z_n\}$ is bounded.

Otherwise, there exists subsequence $\{z_{n_k}\}$ such that $$\lim_{k\to\infty}z_{n_k}=\infty,$$ but this implies $$\lim_{k\to\infty}\frac{|z_{n_k}-z|}{\sqrt {1+|z_{n_k}|^2}}=1.$$ This is a contradiction.

Suppose $0\leq|z_n|\leq M$, then $$0\leq\frac{|z_n-z|}{\sqrt {1+M^2}}\leq\frac{|z_n-z|}{\sqrt {1+|z_n|^2}}\to0,$$ which implies $$\lim_{n\to\infty}z_n=z.$$

Answer 2

Kavi Rama Murthy's idea for proving the boundedness is a little bit easier.

First observe $$\sqrt{1+|z_n|^2}\leq\sqrt{1+2|z_n|+|z_n|^2}=1+|z_n|$$ By triangle inequality, we have for all $n\geq k(\varepsilon)$ $$|z_n|\leq|z|+|z_n-z|\leq |z|+\varepsilon\sqrt{1+|z_n|^2}\leq |z|+\varepsilon(1+|z_n|) $$ By substracting $\varepsilon |z_n|$, we get $$(1-\varepsilon)|z_n|\leq |z|+\varepsilon $$ Maybe, let $\varepsilon:=\frac 12$, then for all $n\geq k(\frac 12)$ holds $$|z_n|\leq 2|z|+1. $$ The rest can be done, like Riemann already did.