Complex analysis limit points question

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Homework Statement



Find the limit points of the set of all points z such that:
a.) z=1+(-1)^{n}\frac{n}{n+1} (n=1, 2, ...)
b.) z=\frac{1}{m}+\frac{i}{n} (m, n=+/-1, +/-2, ...)
c.) z=\frac{p}{m}+i\frac{q}{n} (m, n, p, q=+/1, +/-2 ...)
d.) |z|<1

Homework Equations



None.

The Attempt at a Solution



I'm unsure on a.
I'm also unsure on b. I think it's just a bunch of points starting at the line 1 above the real axis and going down. But not totally filled in, so I'd think there'd be no limit points. But then again maybe 0 is a limit point?
c.) I think it's the set of all pts Imz=0
d.) I think it's the set of all pts |z|=1

This is problem 1 on page 29 in Introductory Complex Analysis by Silverman...it's on google books.
http://books.google.com/books?id=Oy...ver&dq=introductory+complex+analysis#PPA29,M1
 
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What's your definition of limit points? All points in the set plus all accumulation points (where accumulation points are defined to be points where all neighborhoods containing them intersect with the original set), or just the latter? Based on your answers, I'm guessing the latter.

If it's just the latter, then your d) is correct. Did you take real analysis? If you did, limit points in C are conceptually identical to limit points in R.

For b), it's a bit tricky. First, let both m and n go to infinity. This will be a limit point, right? After you've done that, fix m (for example, take m = 3), and let n go to infinity. This will also be a limit point. Same thing if you fix n and let m vary. Once you're familiar with this method, you can do c), too.
 
So b would be 1/m, i/n, and 0?
And c would be 1+i (all go to infinity), 1, 0, p/m, and iq/n?

What about a? Are there no limit points? It doesn't seem to converge anywhere. Except maybe at 1 and 2, when n goes to infinity?
 
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