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 $$\{a-bi,a+bi,z\},$$ consisting of different complex numbers forms a multiplicative group. Justify the existence or non-existence of such numbers. This 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. $$a\pm bi\notin\mathbb C\setminus \mathbb R$$ because there would be two equal elements, therefore, $z=\pm1$ and $$a\pm bi=\frac{1}{a\mp bi}$$ $$\implies a^2+b^2=1,$$ so the set looks like this: $$\left\{a-\sqrt{1-a^2}i,a+\sqrt{1-a^2}i,1\right\}$$ Every critic is helpful.

  • $\begingroup$ It's a very nice effort. By the way, could you prove $\;(a-\sqrt{1-a^2})^2\;$ is in that set? Not to mention that you didn't specify what $\;a\;$ is...It'd seem like someone being asked this exercise should know a little better $\;\Bbb C\;$ , and then know that the solutions of $\;z^2=1\;$ are such a multiplicative group with three elements. $\endgroup$ – DonAntonio Nov 1 '19 at 8:55

It makes no sense to write that $a\pm bi\notin\mathbb C$. On the other hand, it is correct that $a+bi$ and $a-bi$ must be the inverse of each other and that it follows from it that $a^2+b^2=1$. But that does not mean that that any pair $(a,b)$ of real number such that $a^2+b^2$ will do. For instance, $\{1,i,-i\}$ is not a group.

Actually, the only solution is$$\left\{-\frac12+\frac{\sqrt3}2i,-\frac12-\frac{\sqrt3}2i,1\right\}.$$This follows from the fact that the numbers $-\frac12+\frac{\sqrt3}2i$, $-\frac12-\frac{\sqrt3}2i$, and $1$ are the only complex numbers whose cube is equal to $1$.


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