Principal square root of complex numbers

Let $\sqrt.:= r^{1/2}[cos(\theta/2)+isin(\theta/2)], 0 \leq \theta < 2\pi$
define the the particular square root of a complex number.

For what values of z does the equation $\sqrt{z^2} = z$ hold?

I am really sorry, but this question has me stumped and I have no idea how to proceed, hence I couldn't show any working. If someone could please give me a hint.

Answer

The problem arises when $\theta>\pi$. Let $\theta =\pi + \delta$ where $0<\delta <\pi$. Then, we have

$$z^2=r^2e^{i2\delta}$$

on the branch for which arguments are restricted between $0$ and $2\pi$. Then, the square root of $z^2$ is

$$\sqrt{z^2}=re^{i\delta}=re^{i(\theta -\pi)}=-re^{i\theta}\ne z=re^{i\theta}$$

Therefore, the relationship $\sqrt{z^2}=z$ is valid only for $0\le \arg (z) <\pi$.