In $ \mathbb{C} $ we define the binary relations $ R_{j} $ , $ j=1,2,3,4. $
$ z_{1} R_{1} z_{2} \Leftrightarrow |z_{1}| = |z_{2}| $
$ z_{1} R_{2} z_{2} \Leftrightarrow arg(z_{1}) = arg(z_{2}) $ or $ z_{1} = z_{2} = 0 $
$ z_{1} R_{3} z_{2} \Leftrightarrow \bar{z_{1}} = z_{2} $
$ z_{1} R_{4} z_{2} \Leftrightarrow z_{1} = e^{i\phi}z_{2} $ , $ \phi \in \mathbb{R} $
Let X = { z | |z| = 1 }
Find $ R_{j}(\mathbb{R}), R_{j}(\mathbb{X}), R_{j}(i),R_{j}(\mathbb{R}),R_{j}\circ R_{k},R_{j}^{-1} $
Find which relation is an equivalence relation and in that case determine it's quotient set.
What is the elegant and right way to find those sets ? They might seem easy to guess but I am not that familiar with the demonstration.
Some answers : $ R_{1}(\mathbb{R}) = \mathbb{C} $ , $ R_{2}(\mathbb{R}) = \mathbb{R} $ , $ R_{3}(\mathbb{R}) = \mathbb{R} $ , $ R_{4}(\mathbb{R}) $ = { $ (a,b) \in \mathbb{R}^2 | b= a\tan\phi $ } , $ R_{1}(X) = X $, $ R_{2}(X) = \mathbb{C}^* $, $ R_{3}(X) = X $, $ R_{4}(X) = X $, $ R_{1}(i) = X $
And the only equivalent relations are the first 2 ( which is pretty obvious because they are Kernels )
You should think of $R_j(S)$ (where $S$ is some subset of $\Bbb{C}$) as the image of the set $S$ under the relation $R_j$. What it means is as follows: $$R_j(S)=\{z \in \Bbb{C}\, | \, s R_j z \,\,\text{ for some } s \in S\}$$
For example, if $S=\Bbb{R}$ and we are using relation $R_1$, then \begin{align*} R_1(\Bbb{R}) & =\{z \in \Bbb{C}\, | \, s R_j z \,\,\text{ for some } s \in \Bbb{R}\}\\ & =\{z \in \Bbb{C}\, | \, |s|=|z| \,\,\text{ for some } s \in \Bbb{R}\}\\ \end{align*}
Since every complex number $z=a+ib$ has magnitude $|z|=\sqrt{a^2+b^2}$, so we can say that the real number $\sqrt{a^2+b^2}$ is related to $z$ via the relation $R_1$. Thus $$R_1(\Bbb{R})=\Bbb{C}.$$
Hope you can take it from here.