# Construction of multiplicative group of complex numbers

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'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. – 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 $$1$$.