Let $z \in \mathbb C$, $|z| = 1$. Assume the sequence $a_n = z^n$ is convergent. Prove $z = 1$.

Question

Let $z \in \mathbb C$, $|z| = 1$. Assume the sequence $a_n = z^n$ is convergent. Prove $z = 1$.

The case $z = 1$ implies convergence of $a_n$ is easy to prove. It is also easy to prove that $z = -1$ implies divergence of $a_n$.

So clearly $z$ must be a complex number on the complex unit circle of radius 1, and I must prove divergence for any $z \ne \pm 1$.

My idea is: $z = a + ib$ and $|z| = 1$ can be written in polar form $e^{i \theta} = cos(\theta) + i sin(\theta)$. Then $a^n = e^{i \theta n} = cos(\theta n) + i sin(\theta n)$ for $n \in \mathbb N$. Now we may assume $\theta \neq \pi, 2\pi$, but how do I verify that the sequence does in fact diverge in this case ?

Answer 1

If $z^n$ converges, then $\left|z^{n+1}-z^n\right|\to 0$

But you also know $\left|z^{n+1}-z^n\right|=\left|z^n\right|\left|z-1\right|=|z|^n\left|z-1\right|=|z-1|$

So $|z-1|\to 0$, that is, $z=1$

Answer 2

Hint Let $z= e^{2 \pi it}$ with $0 \leq t <1$

Case 1: t rational. Prove that $z$ is a root of unity, thus $z^n$ is periodic. Prove that in this case the period is $1$ thus $z^{n+1}=z^n$.

Case 2: $t$ irrational. Use Dirichclet principle to prove that $z^n$ is dense on the unit circle.

The idea is that if you split the circle in $n$ equal parts, along $z,z^2,..,z^{n+1}$ you can find $z^i$ and $z^j$ with $i<j$ in the same part. Then multiplication by $z^{j-i}$ moves you around the circle, in the same direction, by the same quantity, which is smaller than one part. Repeated multiplication lands you in every part of the circle.