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Show that $C=\{z\in \mathbb{C} \mid |z|=1\}$ is a group under complex multiplication.

I'm a little confused because isn't the identity the only element with order $1$? What is this set?

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  • $\begingroup$ What have you done so far? Which properties have you already shown? (identity element, closure under multiplication and inverse?) $\endgroup$ – Clement C. May 15 '13 at 2:42
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    $\begingroup$ The notation $\vert z\vert$ means the absolute value of the complex number $z$, not its order in a group, so you're looking at the set of complex numbers with absolute value $1$. $\endgroup$ – Keenan Kidwell May 15 '13 at 2:44
  • $\begingroup$ Keenan- Sorry, my professor uses that notation for order of an element, but I know that is what he means in the question $\endgroup$ – Quizzical May 15 '13 at 2:44
  • $\begingroup$ Clement- Also, I just need to show closure under inverses. I have the others. $\endgroup$ – Quizzical May 15 '13 at 2:45
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    $\begingroup$ Hint: if $z=a+ib\in C$, $(a+ib)(a-ib)=a^2+b^2=1$. But again, the polar form is more convenient for this. $\endgroup$ – Clement C. May 15 '13 at 2:47
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Hint: prove that if you multiply two unitary complex numbers then the result is also an unitary number.

I suggest you to learn about polar representation of a complex number. This could make the solution easier.

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  • $\begingroup$ You're right, that does make it easier $\endgroup$ – Quizzical May 15 '13 at 2:50
  • $\begingroup$ No need for polar representation, just that $\bar{zw}=\bar z \bar w$. $\endgroup$ – lhf Jun 9 '14 at 19:02
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Regarding your second question: in $\mathbb{C}$, you have a lot of elements with modulus $1$ — all $e^{i\theta}$ for $\theta\in[0,2\pi)$.

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  • $\begingroup$ Module or modulus? $\endgroup$ – oldrinb May 15 '13 at 2:44
  • $\begingroup$ modulus, indeed. $\endgroup$ – Clement C. May 15 '13 at 2:45
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We first show that $G = \mathbb{C}^* = \mathbb{C} − \{0\}$ under complex multiplication forms a group.

  • Closure:

Let $z = a + bi$ and $w = c + di$ be both in $\mathbb{C}^∗$. Then we have $zw = (a + bi)(c + di) = ac − bd + (ad + bc)i$. To show that this element is in $G$, we compute $$(ac − bd)^2 + (ad + bc)^2 = a^2c^2 − 2abcd + b^2 d^2 + a^2 d^2 + 2abcd + b^2c^2 \\= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 \\= a^2(c^2 + d^2) + b^2(c^2 + d^2) \\= (a^2 + b^2)(c^2 + d^2).$$ Now $z, w ∈ G$ imply that a $a^2+b^2 > 0$ and $c^2+d^2 > 0$, whence $(ac−bd)^2+(ad+bc)^2 > 0$ and $zw ∈ G$.

  • Associativity:

To show associativity, let $z = a + bi, w = c + di$ and $u = e + fi$. Then

$$(zw)u = [(a + bi)(c + di)](e + f i) = [(ac − bd) + (ad + bc)i](e + f i) \\= [(ac − bd)e − (ad + bc)f] + [(ac + bd)f + (ad + bc)e]i \\= (ace − bde − adf − bcf) + (acf − bdf + ade + bce)i \\= [(a(ce − df) − b(cf + de)] + [a(cf + de) + b(ce − df)]i \\= (a + bi)[(ce − df) + (cf + de)i] \\= (a + bi)[(c + di)(e + f i)] \\= z(wu).$$

  • Identity Element and Inverse:

It is not difficult to check that $1 + 0i$ acts as a unit in multiplication and that

$$(a + bi)^{−1} = \frac{1}{a+bi} =\frac{a−bi}{(a+bi)(a−bi)} = \frac{a−bi}{a^2+b^2} = \frac{a}{a^2+b^2} − \frac{b}{a^2+b^2} i ∈ G$$ , since $$\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2 = \left(\frac{a^2+b^2}{(a^2+b^2)^2}\right) = \frac{1}{a^2+b^2} > 0$$

Since all the group axioms hold, $(G,\cdot)$, is a group under multiplication.

Ok, this is a group, now we let see that Circle Group is a Group.

Let $K$ be the set of all complex numbers of unit modulus: $K={z∈\mathbb{C}:|z|=1}$

  • Closure:

Then the circle group $(K,\cdot)$ is an uncountably infinite abelian group under the operation of complex multiplication. $$ z,w ∈ K ⟹ |z| = 1=|w| ⟹ |zw| = |z||w| ⟹ zw ∈ K $$
So $(S,\cdot)$ is closed.

  • Associativity, comes from complex multiplication is associative.

  • Identity : From Complex multiplication identity is one we have that the identity element of $K$ is $1+0i$.

  • Inverses

We have that $|z|=1⟹\frac{1}{|z|}=\left|\frac{1}{z}\right|=1$. But $z\cdot\frac{1}{z}=1+0i$. So the inverse of $z$ is $\frac{1}{z}$.

  • Commutative: We have that Complex Multiplication is Commutative

So $K$ is a subgroup of $G$ under complex multiplication.

We can assert that from complex multiplication is commutative it also follows from Subgroup of Abelian Group is Abelian that $K$ is an abelian group.

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Every complex numbers could be written as $re^{i\theta}$, where $r\ge0$ and $\theta\in[0,2\pi[$. In particular $r$ is the modulus of that complex number, i.e. $r=|re^{i\theta}|$. So it will be $r=1$. Hence $$ C=\{z\in\mathbb C\;:\;|z|=1\}=\{e^{i\theta}\;:\;\theta\in[0,2\pi[\} $$ which is a group under multiplication, since

  • Contains a unity which is $1=e^{i\cdot0}$
  • Every element admits inverse $(e^{i\theta})^{-1}=e^{i(-\theta)}$ which is already in $C$.
  • It's closed under mutiplication: $e^{i\theta}e^{i\psi}=e^{i(\theta+\psi)}$ and moreover the multiplication is associative.
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