Here is my question. Let $z_n=r_ne^{i\theta_n}\ne 0$, where $\theta_n=Arg(z_n)$. Is it possible to have $\lim_{n\to\infty}=z_0=r_0e^{i\theta_0}$, but $\lim_{n\to\infty}r_n$ does not exist?
Thanks.
Update: Thanks to the nice hint from Kavi Rama Murthy, I've done this:
Using the triangle inequality, I can write
$$|z|=|z-w+w|\le|z|+|w|$$
which is equivalent to:
$$|z|-|w|\le |z-w|$$
Secondly, I can write
$$|w|=|w-z+z|\le|w-z|+|z|$$
which is equivalent to
$$|w|-|z|\le|w-z|=|z-w|$$
or
$$-|z-w|\le |z|-|w|$$
Thus, I've shown that
$$-|z-w|\le |z|-|w|\le |z-w|$$
which means that
$$||z|-|w||\le |z-w|$$
Now, suppose that $\lim_{n\to\infty}z_n=z_0$. Let $\epsilon>0$. There exists an $N$ so that
$$|z_n-z_0|<\epsilon$$
for all $n\ge N$. Using the inequality above,
$$||z_n|-|z_0||\le |z_n-z_0|<\epsilon$$
Therefore, $\lim_{n\to\infty}|z_n|=|z_0|$. Thus, we've shown that
$$\lim_{n\to\infty}z_n=z_0\qquad\text{implies}\qquad \lim_{n\to\infty}|z_n|=|z_0|$$
Now, since
$$\lim_{n\to\infty}r_ne^{i\theta_n}=r_0e^{i\theta_0}$$
this implies that
\begin{align*}
\lim_{n\to\infty}|r_ne^{i\theta_n}|&=|r_0e^{i\theta_0}|\\
\lim_{n\to\infty}|r_n||e^{i\theta_n}|&=|r_0||e^{i\theta_0}|\\
\lim_{n\to\infty}|r_n|(1)&=|r_0|(1)\\
\lim_{n\to\infty}|r_n|&=|r_0|\\
\end{align*}
But since $r_n$ and $r_0$ are positive numbers, this gives us
$$\lim_{n\to\infty}r_n=r_0$$
Best Answer
If $r_ne^{i\theta_n} \to r_0 e^{i\theta_0}$ then $r_n \to r_0$ becasuse $z \to |z|$ is a continuous function.