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?
 A: 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.
A: Let $a = re^{i \theta}$ 
Then, $a = b^2 = (a^2)^2 = a^4 = r^4e^{i (4 \theta)}$
What we need is, first $r=r^4$ 
That is, $r =0$ or $r = 1$ (since $r$ must be real)
$r=0$ is trivial, discard that.
Next, we need $e^{i \theta} = e^{i(4 \theta)}$
That is, $4\theta = \theta$ or $4 \theta = \theta + 2n\pi, n\in\mathbb Z\setminus \{0\}.$
You were missing this $\boldsymbol{2 n \pi}$ part in your attempt.
From here, we get $\theta =0 $ or $3\theta = 2n\pi$ 
$\theta=0$ means real numbers, discard that.
That leaves us with $3\theta = 2n\pi$ 
For $n=1, \theta = 2\pi/3$
For $n=2, \theta = 4\pi/3$
Other $n$ values can be discarded as for those $\theta \not \in [0, 2 \pi)$
Thus we're left with $e^{i (2\pi/3)}$ and $e^{i (4\pi/3)}$
They are both solutions, and these are the only non-real solutions for $a = b^2$ and $b = a^2$. They are the two complex cube roots of unity.
In real numbers, $a=b=0$ and $a=b=1$ form another solutions as we discovered above. The entire solution set in complex numbers is 
$(a, b) \in \{(0,0), (1,1), 
(\omega, \omega^2), 
(\omega^2, \omega)\}$
where $\omega = e^{i (2\pi/3)}$
A: 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.
A: As user Henry says in a comment, $a^2=b$ and $b^2=a$ imply $a^4=a$, i.e. $a$ is a zero of the polynomial:
$$p(x)=x^4-x$$
And conversely, every zero of that polynomial will give a solution for $a$, setting $b:=a^2$.
Now $p(x)$ factors in $\mathbb Z[x]$ as
$$p(x) = x \cdot (x-1) \cdot (x^2+x+1)$$
Which gives the two real solutions $(a,b)=(0,0)$ and $(a,b)=(1,1)$. The quadratic term does not factor in $\mathbb R$, but it does e.g. in $\mathbb C$, as
$$x^2+x+1 = (x+\frac12 -\frac{i}{2}\sqrt{3})(x+\frac12 +\frac{i}{2}\sqrt{3}) = (x-e^{i\frac{\pi}{3}})(x-e^{i\frac{2\pi}{3}})$$
The two zeros there are the two non-trivial cube roots of unity, often called $\zeta_3$ and $\bar \zeta_3$ (which is $ =\zeta_3^2$). They give the non-real solutions
$$(a,b) = (\zeta_3, \bar \zeta_3) \text{ and } (a,b) = (\bar \zeta_3, \zeta_3).$$
A: 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.
