<s>[INST] 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?
 [/INST] The arguments $\arg(a)$ and $\arg(b)$ are only defined up to a multiple of $2\pi$. Or, if we require that $\arg(z) \in (-\pi,\pi]$ for all $z$, then it's not necessarily the case that $\arg$ is multiplicative: we might end up having to add or subtract $2\pi$.
If we go with the first option, we know that $\arg(a) \equiv 2\arg(b) \pmod{2\pi}$ and $\arg(b) \equiv 2\arg(a) \pmod{2\pi}$, but this is not a problem. Substituting, we get $\arg(a) \equiv 4 \arg(a) \pmod{2\pi}$, or $3\arg(a) \equiv 0 \pmod{2\pi}$; this is possible and requires $\arg(a)$ to be a multiple of $\frac{2\pi}{3}$.
Or, if we go with the second option, we should consider alternatives to $\arg(a) = 2\arg(b)$ and $\arg(b) = 2\arg(a)$. For example, if we have $\arg(a) = 2\arg(b) + 2\pi$ and $\arg(b) = 2\arg(a) - 2\pi$, then we get $\arg(a) = 4\arg(a) - 2\pi$, which leads to $\arg(a) = \frac{2\pi}{3}$ as the answer.
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