How to reconcile the identity $\left( e^{i \theta} \right)^{1/2} = e^{i \theta/2}$ with the fact that $a^2 = b$ is solved by $a = \pm\sqrt{b}$ So I worked along the lines of the following:
$$
\left( \cos \left( \theta \right) + i \sin \left( \theta \right) \right)^{\alpha} = \left( e^{i \theta} \right)^{\alpha} = e^{i (\theta \alpha)} = \cos \left( \theta \alpha \right) + i \sin \left( \theta \alpha \right)
$$
with $i$ the imaginary number, $\theta$ real and $\alpha$ real and $\sin , \cos$ the normal trig functions
However if we take $\alpha = \frac{1}{2}$ a fellow member saw that
$$
\left( \cos \theta + i \sin \theta \right)^{1/2} = \left\{ \begin{array}{l} \cos \left( \frac{\theta}{2} \right) + i \sin \left( \frac{\theta}{2} \right) \\ \cos \left( \frac{\theta}{2} + \pi \right) + i \sin \left( \frac{\theta}{2} + \pi \right) \end{array}\right.
$$
which, from my perspective comes down to the fact that $a^2 = b$ can be solved as $a = \sqrt{b} \lor a = -\sqrt{b}$. I feel as though either I'm being very silly (as per usual) or there's something deep going on here that I'm missing

Tpofofn's hint is huge:
$$\bullet\;\;\;\left(\cos\frac\theta2+i\sin\frac\theta2\right)^2=\cos^2\frac\theta2-\sin^2\frac\theta2+2i\cos\frac\theta2\sin\frac\theta2=\cos\theta+i\sin\theta$$
$$\bullet\;\;\left(\cos\left(\frac\theta2+\pi\right)+i\sin\left(\frac\theta2+\pi\right)\right)^2=\left(-\cos\frac\theta2-i\sin\frac\theta2\right)^2=\ldots$$
You can complete the exercise above and see that both leftmost expressions above are square roots of the same complex number $\,\cos\theta+i\sin\theta\;\ldots$
