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?

There are the number of real solutions of the system
$$a^2-b^2=(a^2-b^2)^2-4a^2b^2\\2ab=4ab(a^2-b^2)\tag1$$ which come from
the equalities $$(a+bi)^2=(a^2-b^2)+2abi\\((a^2-b^2)+2abi)^2=(a^2-b^2)^2-4a^2b^2+4ab(a^2-b^2)i$$
The solution of system $(1)$ is easy and leaves to the only solutions, the two non real roots of unity.
