Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)? Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)?
My answer is not possible because for $a^2$ to be equal to $b$ means that the argument of $b$ is twice of arg(a) and for $b^2$ to be equal to $a$ means that arg(a) = 2.arg(b) but the answer is it is possible.
How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?

It is entirely possible. Consider the complex conjugate cube roots of one. $\omega = -\frac 12 + \frac{\sqrt 3}{2}i$ and $\overline \omega = -\frac 12 - \frac{\sqrt 3}{2}i$
Note that $\omega = (\overline \omega)^2$, but also $\overline \omega = \omega^2$.
There is no contradiction here, because from the two equations, you get $\omega = \omega^4 = \omega^3\omega = 1\cdot \omega$.
Going by arguments, $\mathrm{arg}(\omega) = \frac{2\pi}{3}$ while $\mathrm{arg}(\overline \omega) = \frac{4\pi}{3}$. Clearly, $2\cdot \frac{2\pi}{3} = \frac{4\pi}{3}$ but also $2\cdot \frac{4\pi}{3} = \frac{8\pi}{3} \pmod{2\pi} = \frac{2\pi}{3}$. Essentially, you're forgetting that arguments work modulo $2\pi$.
Note that that pair is the only possible solution. If you write $x = y^2$ and $y = x^2$, then $x = (x^2)^2 = x^4 \implies x^4 - x = 0 \implies x(x^3-1)=0 \implies x = 0$ or $x^3 - 1 = 0$. The latter implies $(x-1)(x^2 + x + 1)=0 \implies x=1$ or $x^2 + x + 1=0$. Solving the quadratic gives you the complex conjugate roots of one. Therefore, $x = 0, 1, \omega, \overline \omega$ are the only solutions. If you impose the additional constraint that $x \neq y$, you're left with the only solutions being $\omega$ and $\overline \omega$, as already stated.
