# Construction of a group of complex numbers [duplicate]

Groups are rarely mentioned during the first semester, but they appear in linear algebra exams. This is an example and my idea which undoubtedly lacks formal language:

If possible, find numbers $$a,b \in \mathbb R, z\in \mathbb C$$ so that the set $$S=\{a-bi,a+bi,z\},$$ consisting of different complex numbers forms a multiplicative group. Justify the existence or non-existence of such numbers. This is my attempt, please, correct me. As a student, I have to be $$mathematically$$ $$literate$$.

Closure under multiplication, the existence of identity and inverse element imply one of the three elements is 1 an the other two are inverse. $$a\pm bi\notin \mathbb R$$ because there would be two equal elements, therefore, $$z=1$$ and $$a\pm bi=\frac{1}{a\mp bi}$$ Since the set is closed under multiplication, any power of each element must be in that set. Let me take an arbitrary number x and its inverse: $$x^z\in S, z\in \mathbb Z \implies x^z=\frac{1}{x} \implies x^{z+1}=1.$$ If I write x in the trigonometric form: $$x=\lvert x\rvert(\cos\alpha + i\sin\alpha)$$ I have to find an odd exponent so that
$$x^z=1, x\ne\pm1.$$ $$\implies$$ for example: $$x^z=\lvert x\rvert(\cos3\alpha+i\sin3\alpha)=1$$ For $$\alpha=-\frac{\pi}{3}$$: Is the solution: $$S=\{\frac{1}{2}+\frac{\sqrt3}{2}i,\frac{1}{2}-\frac{\sqrt3}{2}i,1\}$$ Every critic is helpful.

• I don't understand you: you erased your prior question, which was identical with this one...just that this time you added to your question the answer you got there and that you erased. Why?! This seems to be very whimsical and very, very not nice. – DonAntonio Nov 1 '19 at 11:33
• @LauraŽupčić, being defensive and antagonising experienced users of a community, especially when you're a new user asking for help, is never a good idea. Deleting posts and reposting, for the sake of "attracting attention", isn't constructive and definitively not useful for the community. The "attracting attention" argument is false anyway, since you did get a correct answer on the deleted post (which has meanwhile been reinstated) before any of the answers provided here. – Randy Savage Nov 2 '19 at 3:42
• Laura, it is much better solution to edit your question in response, not rewrite everything. If you delete question that already has an answer, you are invalidating someones work and spent time. That's why others and I voted to undelete it. In the meantime you wrote this new question, so now we have two of the same, which is not ideal. The idea here is that math.stackexchange should also serve as sort of an archive, so we would very much like to keep things as tidy as possible. The thing going for you is that your question is well-received, otherwise these two questions would be shut down. – Ennar Nov 2 '19 at 10:35
• You are welcome, glad to help. If you feel that edit doesn't suffice, in the new question you can always link the old one with an explanation why you needed a new question. That way everything is more transparent. Just bear in mind that this is a community with some well established rules and following them to the best of your ability is showing respect to people who share their knowledge freely, many of which are professional mathematicians that work in academia. Just keep up with well-structured questions and you will fit in nicely. – Ennar Nov 2 '19 at 12:53

Groups are mentioned in linear algebra courses, as any vector space is an abelian group, and also invertible matrices form so called global linear group. However, your question has little to do with linear algebra, and I sincerely doubt it would appear on any linear algebra exam. It is more appropriate for introductory course in group theory/algebra.

That said, let's take a look at what you wrote.

You correctly conclude that if such a group exists, then $$z = 1$$ and that $$\omega = a + bi$$ and $$\overline\omega = a - bi$$ must be inverses to each other.

Again, you are correct that any integer power of a group element must be contained in the group, but I can't really follow your line of thought here. You are looking at solutions of $$x^n = 1$$ (it is customary to use $$n$$ for integers and $$z$$ for complex numbers) which is certainly correct idea, but then out of nowhere, you set $$n = 3$$. Again, this is correct, but you don't explain anywhere why this is significant.

An easy way to justify this is with little bit of group theory. Since your group has $$3$$ elements, for any group element $$x$$ we must have $$x^3 = 1$$. This equation has precisely $$3$$ distinct solutions in $$\mathbb C$$, so if your group exists, its elements are third roots of unity. And these do form a subgroup of $$\mathbb C$$ (which you didn't prove), since they are closed under multiplication and inverses: $$(\alpha\beta)^3 = \alpha^3\beta^3 = 1\cdot 1 = 1,\ (\alpha^{-1})^3 = (\alpha^3)^{-1} = 1^{-1} = 1.$$

We can also arrive at the same conclusion without this general fact from group theory.

Note that $$\omega^2\in\{1,\omega,\overline\omega\}$$. Consider cases:

1. $$\omega^2 = 1 \implies \omega = \pm 1.$$ Contradiction.
2. $$\omega^2 = \omega \implies \omega = 1.$$ Contradiction.
3. $$\omega^2 = \overline\omega \implies \omega^3 = \omega\overline\omega = 1$$. Also, $$\overline\omega^3 = \overline{\omega^3} = \bar 1 = 1$$ and $$1^3 = 1$$, so $$\{1,\omega,\overline\omega\}$$ are the three solutions of the equation $$x^3 = 1$$, as before.

And finally, what you wrote aren't solutions to $$x^3 = 1$$. Note that $$(\frac{1}{2}+\frac{\sqrt3}{2}i)^3 = -1$$. What you want is not the angle $$\alpha = -\pi/3$$, since if you triple it you get $$3\alpha = -\pi$$, and the argument of $$1$$ is $$0$$, not $$-\pi$$. The trick is to write $$1 = \cos 2\pi + i\sin 2\pi$$, so you want $$\alpha = 2\pi/3.$$ With that you get $$\omega = \cos \frac{2\pi}3 + i \sin \frac{2\pi}3,\ \overline\omega = \omega^2 = \cos \frac{4\pi}3 + i \sin \frac{4\pi}3,\ z = \omega^3 = \cos \frac{6\pi}3 + i \sin \frac{6\pi}3 = 1.$$

• I see, also, thank you for the feedback. We have just done some tasks related to subspaces yesterday, but in general, I haven't heard of groups at the lectures so far, only in workbooks. – Praskovya2.718281828 Nov 1 '19 at 11:22
• Why would $\omega=1$ be a contradiction? The set $\{1,\omega,\overline\omega\}=\{1\}$ consists of different complex numbers and is a multiplicative group. – Randy Savage Nov 1 '19 at 23:38
• @Goran Malic, "consisting of different complex numbers" very likely means $\omega$, $\overline\omega$ and $z$ are distinct. Your interpretation would render the word "different" completely useless in that sentence. – Ennar Nov 2 '19 at 1:17
• @Ennar, yes, all the complex numbers in $\{1\}$ are different to each other, since 1 is the only complex number there. – Randy Savage Nov 2 '19 at 1:19
• @Ennar, exactly, the word "different" is useless, but that is an issue of semantics and possibly mis-translation to English. There are no restrictions on $a$, $b$ and $z$ that would force us to exclude $z=\omega=\overline\omega$. – Randy Savage Nov 2 '19 at 1:53

Yes, your derivation shows you have the right idea.

You could make it clearer that you have found all the solutions as follows.

If a complex number $$w$$ is in a finite group then so are $$w,w^2,w^3,...$$ and so $$w$$ must represent a rotation of some angle $$\theta$$. For a group of order 3,the only possibilities are $$\theta \in$$ {$$0,\pi /3, 2\pi /3$$}.

This leads immediately to the group that you have found.

• Note that they haven't found all the solutions. They have found all the solutions where $a+bi, a-bi, z$ are distinct. Assuming $b$ and $-b$ are considered the same solution. – Arthur Nov 1 '19 at 10:32
• Thank you very much, I appreciate your comment. – Praskovya2.718281828 Nov 1 '19 at 10:36
• @Arthur I think the OP's proof was OK - they did say "different complex numbers". – S. Dolan Nov 1 '19 at 10:39
• The only possibilities are $\{0, 2\pi/3, 4\pi/3\}$. $\theta = \pi/3$ generates group of order $6$. – Ennar Nov 1 '19 at 11:13