Equivalence relation, complex numbers: x equiv y iff x^3 = y^3 for x,y complex

[Faker97]

Let the relation ~ on $\displaystyle \mathbb{C}$ be defined by:

. . . . .$\displaystyle x\, \sim\, y\, \Longleftrightarrow\, x^3\, =\, y^3$

...for $\displaystyle x,\, y\, \in\, \mathbb{C}.$ You may assume that ~ is an equivalent relation.

Find all the elements of the equivalence classes $\displaystyle \left[e^{i\pi / 4}\right]\, =\, \left\{z\, \in\, \mathbb{C}\, :\, z\, \sim\, e^{i\pi /4}\right\}$ and $\displaystyle \left[-2\right]\, =\, \left\{z\, \in\, \mathbb{C}\, :\, z\, \sim \, -2\right\}$

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Any ideas on how to progress with this? Do you cube and add 2pi to the arguments? For the second part is it acceptable to rewrite -2 in exponential form and then continue from there? Thanks

https://gyazo.com/c0663bbc6abe1224ba5ea32de4568cec

 

[HallsofIvy]

Yes, do exactly what the problem tells you to do. Since two complex numbers, x nd y, are equivalent if and only if $\displaystyle x^3= y^3$, to find numbers equivalent to y you have to find all solutions to $\displaystyle x^3= y^3$.

If $\displaystyle y= e^{i\pi/4}$ then you need to find all solutions to $\displaystyle x^3= e^{3i\pi/4}$. If y= 2 then $\displaystyle y^3= 8$ so you need to find all solutions to $\displaystyle x^3= 8$.