Q: Explain why the spiralling sheets of Z^a's Riemann surface join back together after n sheets where m/n=a where there is no common factor between m and n and a.

A:
Z^a = (r^a) exp{i theta a}
= exp{i theta a}

Where we use r = 1 for simplicity.

(Z^a)start = exp{i 0 a} = 1 (1)

Where we are starting at theta = 0.

(Z^a)n turns = exp{i n 2pi a} = 1
(Z^a)n turns = exp{i n 2pi m/n}
(Z^a)n turns = exp{i 2pi m} = 1 (2)

So, it is clear that they rejoin as (1) = (2). But, why would they not rejoin earlier? Because there are no common factors between m and n. Consider the case after q turns where q < m.

(Z^a)q turns = exp{i q 2pi m/n}

For (Z^a)q turns to equal 1, qm/n must be an integer, which is impossible if there are no common factors between m and n.

Your small text is too small to read. Can you please edit your post to make it readable?
Thanks

I've made it a bit bigger, it should be a bit easier to read now.

It seems that it is the fundamental property of any Riemann surface that in it
for a point z.(excluding branch points)
If we assume this property of then
where 'a' is real.
From the solution to Exercise [7.1] using polar coordinates,this condition amounts to
where k is an integer for all non-zero z.
If is rational and is in its lowest terms i.e. then the above condition implies
where t is an integer.
Hence, a closed loop around the the origin must go through 'tn' turns. We thus have a spiral ramp of 'n' turns with the sheets joined at the point '0'.
The reason for the surface to have 'n' sheets is quite simple.The map
where is one-many.(It takes every non-zero point in to n-points.)
We define a Riemann surface for this map so that the map

is one-one.We now have a parameter that distinguishes one root from another namely the 'sheet'.The surface for is also a spiral ramp of 'n' sheets because
and w is just another complex number.So the structure of the Riemann surface is unaltered.
I am not sure why an irrational 'a' means infinite sheets.Probably it is because that irrationals can be arbitrarily approximated by rationals but I couldn't prove it.

Real numbers are the limits of integral numbers when the division is shrinking to zero, i.e. and . Hence

Sameed, where did you get the integrals? is only true for integer, not real. The simplest and most elegant solution has already been given by lukewm.

The point z in my discussion is not a point in the complex plane. It is in fact a point on the Riemann surface 'S'. The result you quoted is true for points in the complex plane.

A bit confusion here. Previously you've got . Hence is on complex plane; on Riemann surface. Or is it the other way round?