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I want to answer part(b) here in this question:

Let $C^{*}$ denote the group of nonzero complex numbers under multiplication, and $S^{1} \subset C^{*}$ the subgroup of complex numbers of length one. Torsion elements of $C^{*}$ are called roots of unity.

(a) Show that $Tor(C^{*}) \subset S^1.$ Now give a simple reason that $Tor(C^{*}) \neq S^1.$

(b) Define $z \in S^{1} $ by $z = \frac{3}{5} + i \frac{4}{5}.$ Show that $z \notin Tor(\mathbb{C^*}).$

My question is:

I get this hint to answer it: "Show that the real and imaginary parts of $(3 + 4i)^n$ are congruent to 3 and 4 modulo 5, respectively."

But I do not understand how proving this hint will answer the question? could anyone help me in answering this question, please?

Note that: part $(a)$ answer can be found here A simple reason that $Tor(C^{*}) \neq S^1.$

Arctic Char
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  • Suppose $z^n=1$ for some some $n\in \Bbb N$. Then you have $5^n=(3+4i)^n$. Now compare the real and imaginary parts using the hint. – cqfd Sep 14 '20 at 07:29
  • Do I have to convert my $z$ to the polar form?@ShiveringSoldier –  Sep 14 '20 at 18:04
  • For me it is clear that $3 \equiv 3 (\mod 5)$ and $4 \equiv 4 (\mod 5)$ is that what you meant to say ? @ShiveringSoldier –  Sep 14 '20 at 18:10
  • @ShiveringSoldier I know that the imaginary part is 0. does that help in anything? Do we want to show that the argument of the given $z$ may not satisfy $\theta=\frac{2m\pi}{n}$ for some $n\in \Bbb N$ with $m\in \Bbb Z$ and so it does not belong to $Tor(\mathbb{C}^{*}).$ If so how the 3 numbers (3,4,5) will create this $\theta$? –  Sep 14 '20 at 18:19
  • In this case, stay away from polar form. It will merely confuse you. – Lubin Sep 14 '20 at 18:32

1 Answers1

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Any $z\in \Bbb S^1$ has modulus 1 and thus can be written as $$ z=e^{i\theta}. $$

Such a number is a root of unity if $\exists n$ such that $z^n=e^{i\theta n}=1$. This occurs if and only if $n\theta\in2\pi\Bbb Z$.

Can you conclude?

It is easy to find a real number $\theta$ such that $n\theta$ doesn't belong to $2\pi\Bbb Z$. That is, the equality $$ n\theta=2\pi m $$ is never true, whatever $n,m\in\Bbb Z$ you take.

For example $\theta=1$ does the job.

So for such $\theta$, one never gets $$ z^n=1 $$ so, such a $z=e^{i\theta}$ doesn't belong to $\operatorname{Tor}(\Bbb C^*)$, but stay obviously in $\Bbb S^1$.

Joe
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  • I will think and let you know if I can conclude or not. –  Sep 14 '20 at 07:58
  • No, I am unable to conclude. –  Sep 14 '20 at 18:01
  • I know that the imaginary part is 0. does that help in anything? Do we want to show that the argument of the given $z$ may not satisfy $\theta=\frac{2m\pi}{n}$ for some $n\in \Bbb N$ with $m\in \Bbb Z$ and so it does not belong to $Tor(\mathbb{C}^{*}).$ –  Sep 14 '20 at 18:17
  • I think the purely arithmetic method suggested in the hint is much more productive and explanatory. – Lubin Sep 14 '20 at 18:34