Why does the principle value of $z^{\frac{1}{2}} := \sqrt{r}e^{i\frac{\theta}{2}}$ need to be restricted to one rotation in order to be continuous? I understand that the function $f(z) = z^{\frac{1}{2}}$ is multi-valued, since it has exactly two solutions for every $z$. However if I take the principle value square root function, expressed as
$$f(r,\theta) = \sqrt{r}e^{i\frac{\theta}{2}}$$
Then why does a branch cut need to be taken here? From what I've read, if I want this function to be continuous then I should restrict $\theta$. But as far as I understand, $e^z$ is single-valued for any complex number $z$. 

Yes, $f(z) = \exp(z)$ is single-valued. However the function we're considering here in terms of $z$ is $f(z) = \sqrt{|z|}\exp({i\frac{\arg(z)}{2}})$ which is not single-valued.  To see why, write $f(z)$ as
$$f(z) = \sqrt{|z|}\exp(i{\frac{Arg(z) + 2k\pi}{2}})$$
Where $k$ is an integer and $Arg(z)$ is the principle argument of $z$. Now for example, take $z=1$:
$$f(1) = \exp(i{k\pi})$$
And for multiple values of k,
$$\text{k = 0: }\;f(1) = \exp(0) = 1$$
$$\text{k = 1: }\;f(1) = \exp(i{\pi}) = -1$$
In fact if we take any $z$, we can find multiple values of $f(z)$ for different values of $k$. Therefore in order to make this function single-valued, we should only choose one value of $k$. That is, we should only consider one rotation.
