If limlimits_(n→∞)z_n=a , then limlimits_(n→∞)|z_n|=|a| .…

If $\lim\limits_{n\to\infty}z_n=a$, then $\lim\limits_{n\to\infty}|z_n|=|a|$. Show by example that the converse is not true.

Step 1 · Proof of the direct statement

Given $z_n\to a$: for any $\varepsilon>0$, there is $N$ with $|z_n-a|<\varepsilon$ for $n>N$. By the reverse triangle inequality:

$$ \big||z_n|-|a|\big|\le|z_n-a|<\varepsilon \qquad(n>N) $$

so $|z_n|\to|a|$.

Step 2 · The converse fails

Take $z_n=(-1)^n$. Then $|z_n|=1$ for every $n$, so $|z_n|\to1$. But $z_n$ itself alternates forever between $-1$ and $1$ and never settles — $\lim_{n\to\infty}z_n$ does not exist.

$$ \boxed{|z_n|\to1, \text{ yet } z_n \text{ does not converge}} $$

A richer counterexample

Take $z_n=e^{in}$ (i.e. $\theta=1$ radian, and $1/(2\pi)$ is irrational). Then $|z_n|=1$ for all $n$, but by Weyl's equidistribution theorem the points $e^{in}$ are dense on the unit circle and never converge to a single value — a much more dramatic failure of the converse than the simple two-point oscillation of $(-1)^n$.

Why the converse must fail

Knowing $|z_n|\to|a|$ only pins down that $z_n$ eventually lies near the circle of radius $|a|$ — it says nothing about where on that circle $z_n$ sits. Convergence of $z_n$ itself additionally requires the argument to settle down, which $|z_n|\to|a|$ alone cannot guarantee.