Why does $\left (\frac{3+4i}{5} \right)^n$ not converge? 
Why does $$\left (\frac{3+4i}{5} \right)^n$$ not converge?

I have calculated $\left |\left (\frac{3+4i}{5} \right)^n \right|$ and I got $1$. But in the Solutions they wrote without further comment that this sequence does not converge.
I understood that $|\cdot|\neq0$ is not a necessary condition for convergence because $|(-1)^n|=1$ but does not converge.
But I still don't know how I can prove that the named sequence is not convergent.
 A: Beacuse $$\left (\frac{3+4i}{5} \right)^n  = (\cos \phi +i\sin \phi)^n = \cos (n\phi) + i\sin (n\phi)$$
neither $\cos (n\phi)$ nor $\sin (n\phi)$ converge.
A: Let $z \in \mathbb{C}$ such that the sequence $(z^n)_{n \in \mathbb{N}}$ converges to a limit $l \in \mathbb{C}$.
Then, write $z^{n-1}  z = z^n$, and let $n$ tend to $+\infty$. You get $lz=l$.
The first possibility is that $l=0$ (in your case, it is impossible because $|z|^n=1$ so $|l|$ has to be $1$).
The second possibility is, dividing by $l$, that $z=1$.
In particular, the only number $z$, with $|z|=1$, such that $(z^n)$ converges, is $z=1$.
A: Your points all lie on the unit circle (since they're just repeated applications of the point $(3/5,4/5)$ on the complex plane).
This thus purely rotates the plane by some positive angle $\phi.$ It should be clear that rotating the plane as many times as you wish does not make the initial vector settle down anywhere, but just rotate every time by the angle $\phi$ forever.
A: One approach is as follows: if $|x| = 1$, then
$$
\left|\left(\frac{3 + 4i}{5} \right)x - x\right| = 
|x| \cdot \left|\left(\frac{3 + 4i}{5} \right) - 1\right| = \left|\frac{-2 + 4i}{5}\right| = \frac{2\sqrt{5}}{5}.
$$
Conclude that $|(\frac{3+4i}{5})^{n+1} - (\frac{3+4i}{5})^n| = \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}} > \frac{2}{\sqrt{16}} = \frac 12$ for all $n$, which (by whichever definition of convergence you're comfortable with) is enough to conclude that the sequence $(\frac{3+4i}{5})^n$ fails to converge.
One definition is the $\epsilon$-$N$ definition of convergence, which is

A sequence $a_n$ converges to limit $L$ if, for all $\epsilon > 0$, there exists an $n > N$ such that $|a_n - L| < \epsilon$ whenever $n > N$

I would say that the requirement fails since the sequence fails to be Cauchy.
