Unexpected result from Euler's formula

Think about it this way: if we say that $a=-1$, it follows that $a^2=1$. From there, if you take the square root, you get that $a=1$, meaning that $-1=1$. Clearly this is ridiculous. Roots are multivalued. The square root of a positive real number has two results: one positive, one negative. Similarly, there are three complex numbers whose cube is equal to $1$. Your mistake lies in assuming that $1^{1/3}$ is well-defined. It is true that $1^3=1$ and that $(-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i)^3=1$.

You are making the classical mistake of generalizing the exponentiation rule $(a^b)^c=a^{bc}$ to complex numbers: it doesn't hold !

No, that looks good! The three cube roots of unity are indeed

$$1, -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i.$$

Picture unit vectors in the complex plane at angles $0, 120, 240$ degrees.

You may have looked at your answer and scratched your head as to how weird it looked. The polar representation is compact and has polar symmetry: $z=re^{i\theta}.$ If $r=1$ then the only thing that changes when you square $z$ is the angle.

But slap the Cartesian coordinate system on top of that number, and you're forced to evaluate the $x$ and $y$ components. You're forcing square graph paper on a circle, and stuff doesn't line up. The $x$ and $y$ components are going to be, in general, more complicated (uglier?) when you do the conversion.