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I'm trying to find the order of the cyclic subgroup of the group $\mathbb{C}^\times$ (non zero complex numbers) generated by $(1+i)$. Well, I know that in polar form, the angle of $1 + i$ is $\pi/4$. So angle of $(1+i)^2$ is $\pi/2$, etc. So this makes me think this cyclic subgroup has order 8.

My book says something along the lines of:

"$\lvert 1+i\rvert$ is $\sqrt2$, so the order of this cyclic subgroup is infinite."

Totally confused on this explanation...

Any help would be much appreciated!

Thanks, Mariogs

Thomas
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anon_swe
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    But the distance from the origin is $\sqrt2$, and this multiplies, so as you take higher powers, the distance goes to infinity too. – Lubin Feb 19 '14 at 18:13
  • HINT: $(1+i)^4=(1-1+2i)^2=-4$ and so $-4\in\langle i+1\rangle$. – Bobby Feb 19 '14 at 18:15

2 Answers2

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You are right that when you take higher powers of $1+i$ then the argument just keeps cycling through $\pi / 4, \pi / 2 , 3\pi/4, \dots$. But the modulus of the higher powers will grow without bound. So you can't find an $n>1$ such that $(1+i)^n = 1$. (Remember that in a finite cyclic group for each element $x$ you can find an $n$ such that $x^n = 1$.)

Thomas
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Recall that $|z^n| = |z|^n$. We have $|1+\mathrm{i}|=\sqrt{2}$ and so $|(1+\mathrm{i})^n|=|1+\mathrm{i}|^n = (\sqrt{2})^n$.

If $g_n = (1+\mathrm{i})^n$ then $g_i \neq g_j$ for $i \neq j$ because $|g_i| \neq |g_j|$.

In words: each power will give a different complex number, because they will have different moduli.

Fly by Night
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