So I remember scrolling through Youtube a while back when I saw this video by Blackpenredpen that was about all solutions to $$x^3=1$$Now, I will admit, I didn't think that much about it when I saw it at first, but it seems to have been bugging me because I have been thinking, "Is that even possible?" Here are my attempts at finding all solutions to$$x^3=1$$
Attempt $1$
Time: circa. $3-4$ weeks ago
We write this as$$x^3=1+0i\implies x=\sqrt[3]{1+0i}$$Now, I was thinking of using the $(a+bi)^{c+di}$ formula for this, but now that I think about it, I have no idea why I would have been doing that.
Attempt $2$
Time: circa. $4-5$ hours ago
We write $1$ as $$e^{2i\pi}=\cos(2\pi k)+i\sin(2\pi k)$$$$\implies1^{1/3}=\cos\left(\dfrac{2\pi}3\right)+i\sin\left(\dfrac{2\pi}3\right),\cos\left(\dfrac{4\pi}3\right)+i\sin\left(\dfrac{4\pi}3\right),1$$And so we have$$\cos\left(\dfrac{2\pi}3\right)\pm i\sin\left(\dfrac{2\pi}3\right)=-\dfrac12\pm\dfrac{i\sqrt2}2$$$$\therefore\sqrt[3]1=1,-\dfrac12\pm\dfrac{i\sqrt2}2$$
My question
Did I find the $3$ cube roots of $1$ correctly, or what are the $3$ cube roots of $1$?