Proof for proving infinite order of complex numbers

Question

I have question about my proof. I could not tell whether it is sufficient enough since my professor approached it differently.

The problem:

Let $z \in \mathbb{C}^{*}$. If $|z| \neq 1$, prove that the order of $z$ is infinite.

My proof: (by contradiction)

Let $z = r\cos(\theta)+r\sin(\theta)=r\operatorname{cis}(\theta)$, where $r> 0, \theta \in [0, 2\pi]$. Since $|z| \neq 1$, then $r \neq 1$.

Suppose that the order of $z$ is finite i.e. $\exists m \in \mathbb{Z}_{+} s.t. z^{m}=1$. Then, observe that:

$z^{m}=r^{m}\operatorname{cis}(m\theta)$, so

$|z^{m}|=|1| \implies \sqrt{r^{2m}\operatorname{cis}^{2}(m\theta)}=1 \implies r^{m}=1$.

However, since $m$ is the least positive integer that $z^{m}=1$ i.e. $m> 0$. Then, $r$ has to be $1$. Yet, this contradicts the assumption that $r\neq 1$.

Hence, we proved that the order of $z$ can’t be finite.

My question:

Is this proof complete? I did not do the way my professor talked about, which using induction and state that $z \neq 1$ is equivalent to $z \in \mathbb{Q}^{*}$.

Any suggestion or different approach is highly appreciated.

Answer 1

The proof is good, but has several redundant steps.

Suppose $z$ has finite order. Then there exists an integer $m>0$ such that $z^m=1$. Hence $|z|^m=1$ which implies $|z|=1$.

No contradiction, but “contrapositive”: if $z$ has finite order, then $|z|=1$.

Answer 2

The proof looks fine, but I always prefer to do things without contradiction where possible.

Let $z \in \mathbb{C}^*$, and consider the sequence of reals whose $n$th term is $|z^n| = |z|^n$.

If $|z| > 1$, then this sequence is strictly increasing, and hence in particular the sequence $z^n$ cannot repeat.

If $|z| < 1$, then this sequence is strictly decreasing, so $z^n$ does not repeat.