A Riemann surface for the function $f(z) = z^{1/p} + z^{1/q}$ When considering the Riemann surface for the function $f(z) = z^{1/p} + z^{1/q}$, where $p,q$ are positive integers, I always thought that it consists of $pq$ branches. However, then I analyzed the function $$f(z) = z^{1/2} + z^{1/4}.$$ It has a single branch cut on the negative real axis. I would expect it to have 8 branches. But, when I start to move on the unit circle around the origin from the point $z=1$, I found that I have to encircle the origin only four times to get back to the original value $f(1) = 2$. Specifically, on the branch, the function changes according to $$\sqrt{z} + \sqrt[4]{z} \to -\sqrt{z} + i \sqrt[4]{z} \to \sqrt{z} - \sqrt[4]{z} \to -\sqrt{z} -i \sqrt[4]{z} \to \sqrt{z} + \sqrt[4]{z}$$
Am I missing some additional branch cuts? What is going on here?
 A: It's nearly the same story as when you construct Riemann surface for a function $\sqrt{z^2}$: it's two $\mathbb C$s glued along $0$, and going around the ramification point will never bring you from one sheet to another — you have to pass through it to change sheets. The difference between, say, yours $z^{1/2}+z^{1/4}$ and $z^{1/2}+z^{1/5}$ is that 2 and 5 are relatively prime, while 2 and 4 are not. What's essential in understanding monodromy of multivalued functions is that when you focus on a particular ramification point, gluing is controlled by homomorphism $\mathbb Z \to Sym(sheets)$, where $\mathbb Z$ is fundamental group of little disc minus point.  But the image is restricted by the form of a function — sheets cannot be shuffled arbitrarily by going around, and possible permutations are restricted by the fact that you cannot, say, pass from $(z^{1/2}, z^{1/4})$ sheet to $(z^{1/2}, iz^{1/4})$ one by traversing along any path missing zero — the parity of square and quartic roots of unity should remain same! If root powers were 2 and 5, then corresponding shuffling group contains a copy of $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 5 \mathbb Z$, but this one equals $\mathbb Z / 10 \mathbb Z$ because 2 and 5 are relatively prime, and loop going around origin corresponds to element $(1, 1)$ which is exactly its generator. And in your case to traverse along all 8 sheets in one sweep you'd need the group $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 4 \mathbb Z$ to be generated by one element, and it's impossible. 
There's nice book about cuts, branches (and using them to prove Abel's theorem) in PIY (prove it yourself) style: V. B. Alekseev, Abel’s Theorem in Problems & Solutions
