Divergence in the Complex Plane

Question

Consider $z_n=(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})^n$.

Prove that $z_n$ does not converge.

My approach: it is easy to see that $z_n = e^{i\cdot n \frac{\pi}{4}}$. So $z_n$ is the eighth root of unity and is just rotating around the unit circle at angles of $\pi /4$. But I have to prove this using the $\epsilon$ definition of convergence of sequences.

I attack this problem the same way we approach proving that the harmonic sequence $(-1)^n$ does not converge.

I say that $z_n$ converges to some complex number $L$. Therefore, by the definition of convergence of sequences, $|z_n-L|<\epsilon$ for all $n>N$.

Let $\epsilon =1/2$.

Therefore, $|z_n-L|<1/2$, for $n>N$.

Let $n,m>N$ and $n=8k,m=8k+4$. This implies:

$|1-L|<1/2$ and $|1+L|<1/2$.

Now, is it okay for me to conclude that since the shortest distance between $1+0i$ and $-1+0i$ is $2$, no two discs with radius of at most $1/2$ that contain $1+0i$ and $-1+0i$ will ever intersect.

I think the final statement is kind of incomplete. How should I end this proof?

Answer 1

Because $z_{4n}=-1$ and $z_{8n}=1$

Let $\lim\limits_{n\rightarrow+\infty}z_n=L$.

Thus, by your work $$1>|1-L|+|1+L|\geq|1-L+1+L|=2,$$ which is contradiction.

Id est, our assuming was wrong.

Answer 2

Observe that $\left(\frac1{\sqrt 2}+i\frac1{\sqrt 2}\right)^4=\left(1cis \frac{\pi}4\right)^4=-1$ it follows \begin{align*} z_{n+4}-z_n&=\left(\frac1{\sqrt 2}+i\frac1{\sqrt 2}\right)^4z_n-z_n\\ &=-z_n-z_n\\ &=-2z_n \end{align*} Then, for any $0<\varepsilon<2$ we have \begin{align*} |z_{n+4}-z_n|&=|-2z_n|=2>\varepsilon \end{align*} So $\{z_n\}$ does not converge.

Answer 3

If a sequence converges then all the subsequences have to converges to the same limit (i.e. the convergence of the subsequences at the same limit is a necessary condition).

But in this case since

$$\lim_{n\rightarrow+\infty}z_{8n}=1 \neq\lim_{n\rightarrow+\infty}z_{4+8n}=-1$$

the sequence does not converges.