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.
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 $1$.