When is $\sqrt{z^2} = -z$ for a complex number $z$? One of my problems requires determining for which $z \in \mathbb{C}$ is $\sqrt{z^2} = z$ true, and for which $\sqrt{z^2} = -z$. The former can be found by noting that if $z = \left|z\right|e^{i\varphi_z}$ with $\varphi_z \in (-\pi, \pi]$, then $\sqrt{z} = \sqrt{\left|z\right|}e^{i\varphi_z/2}$. Thus only if $\varphi_{z^2} = 2\varphi_z \in (-\pi, \pi] \Longleftrightarrow \varphi _z \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right]$ does $\sqrt{z^2} = z$ hold. But how can we find with a similar reasoning when $\sqrt{z^2} = -z$?
Answer in my reading material is that as a complex number is either nonnegative or negative, it follows that if $\sqrt{z^2} = z$ for $\varphi _z \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right]$, then necessarily $\sqrt{z^2} = -z$ for $\varphi _z \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right]$. However I don't quite find this as a satisfactory answer as firstly, it is not a direct reasoning with the argument of a complex number and secondly, the requirement $\varphi \in \left(-\pi, \pi\right]$ is not satisfied. On the other hand I haven't come up with anything useful for determining the required bounds for $\varphi_z$. So how should the question of determining where $\varphi_z$ needs to live, in order for $\sqrt{z^2} = -z$ be satisfied, be formed?
 A: Every non-zero complex number $\DeclareMathOperator{\Arg}{Arg}w$ has a unique representation as $w=r\exp(i\theta)$ if we require that $r>0$ and $\theta\in(-\pi,\pi]$. Then we can define $\sqrt{w}$ as $\sqrt{r}\exp(i\theta/2)$, where $\sqrt{r}$ denotes the positive square root of $r$.
Suppose that $z$ is a complex number such that $z=r\exp(i\theta)$, with $r>0$ and $\theta\in(-\pi,\pi]$. Then, $z^2=r^2\exp(i(2\theta))$. There are three possible cases we must consider:

*

*If $2\theta$ is the principal argument of $z^2$ (that is, if
$2\theta\in(-\pi,\pi]$), then $\sqrt{z^2}=r\exp(i\theta)=z$. So
$\sqrt{z^2}=z$ if $\theta\in(-\pi/2,\pi/2]$. Actually, $\sqrt{z^2}=z$
if and only if $\theta\in(-\pi/2,\pi/2]$, but that remains to be proven.

*If $2\theta\in(-2\pi,-\pi]$, then the principal argument of $z^2$ is
$2\theta+2\pi$. Hence,
$\sqrt{z^2}=\sqrt{r^2\exp(i(2\theta+2\pi))}=r\exp(i(\theta+\pi))=-r\exp(i\theta)=-z$. So $\sqrt{z^2}=-z$ if $\theta\in(-\pi,-\pi/2]$.

*If $2\theta\in(\pi,2\pi]$, then the principal argument of $z^2$ is
$2\theta-2\pi$, and so
$\sqrt{z^2}=\sqrt{r^2\exp(i(2\theta-2\pi))}=r\exp(i(\theta-\pi))=-r\exp(i\theta)=-z$. So $\sqrt{z^2}=-z$ if $\theta\in(\pi/2,\pi]$.

In summary,

*

*$\sqrt{z^2}=z$ if and only if $z=0$ or has a principal argument $\theta\in(-\pi/2,\pi/2]$.

*$\sqrt{z^2}=-z$ if and only if $z=0$ or has a principal argument $\theta\in(-\pi,-\pi/2]\cup(\pi/2,\pi]$.

Warning: while this procedure does define a single-valued square root function in the complex plane, this comes at a cost: $\sqrt{z}$ is discontinuous along the negative real axis, and in order to define $\sqrt{z}$, we had to make an arbitrary choice about the "principal" argument of $z$. Moreover, the radical rule $\sqrt{z}\sqrt{w}=\sqrt{zw}$ is true if and only if $\Arg(z)+\Arg(w)=\Arg(zw)$. On the plus side, this function does define $\sqrt{-1}=i$ rather than $\sqrt{-1}=-i$, and our choice of principal square root is consistent with that for nonnegative reals.
