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If we have two complex numbers $a, b \in \mathbb{C}$ such that $|a|=1$ and $|b|=1$ is $|a\cdot b|=1$ as well?

I am trying to determine if the set $\left(\{z\in\mathbb{C}:|z|=1\},\cdot\right)$ is a group.

I am not sure if it is closed under the binary operation $\cdot$. My intuition is that it is not closed under this operation. I may be misunderstanding something here, but any clarification and help would be helpful.

Git Gud
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Yousuf Soliman
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    Yes! And it is a very nice group :) – Avitus Jul 02 '13 at 16:09
  • These are just points on the unit circle and multiplication would be the same as adding the angles together with reductions if the angles are large enough, or am I missing what the geometric interpretation here would be? – JB King Jul 02 '13 at 19:15

4 Answers4

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\begin{eqnarray} |(x+iy)(u+iv)|^2 &=& |xu-yv+i(xv+yu)|^2 \\ &=& (xu-yv)^2+(xv+yu)^2 \\ &=& v^2 y^2+u^2 y^2+v^2 x^2+u^2 x^2 \\ &=& (x^2+y^2)(u^2+v^2) \\ &=& |x+iy|^2|u+iv|^2 \end{eqnarray}

copper.hat
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Yes! If $z=re^{2\pi i\theta}$, $z=se^{2\pi i\alpha}$ and $|z|=|w|=1$, then $r=s=1$ implies $zw=rse^{2\pi i(\alpha+\theta)}$, hence $|zw|=rs=1$.

BlackAdder
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In general, $|zw|^2 = z\,w\,\overline{z\,w} = z\,w\,\overline z \, \overline w = z\,\overline z \, \,w\overline w = |z|^2\,|w|^2$.

In particular, $|z|=1=|w|$ implies $|zw|=1$.

lhf
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It is the image of the subgroup $i \mathbb{R} \le (\mathbb{C}, +)$ under the homomorphism $$\mathrm{exp} : (\mathbb{C}, +) \rightarrow (\mathbb{C}^{\times}, *)$$ so it is a subgroup as well. This is essentially the same as the polar form arguments above.

Cocopuffs
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