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Metric spaces and the topology of complex plane

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Show that { cis k : k is a non-negative ineger} is dense in T = { z in C ( C here is complex plane) : |z| = 1 }. For which values of theta is { cis ( k*theta) : K is a non-negative integer} dense in T ?
P. S. cis k = cos k + i sin k, i here is square root of -1.
I want a full justification for each step or claim.

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Metric spaces and the topology of complex plane are investigated. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.

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a) Let's denote K = {z = exp(ik), k in N} - N = set of natural (whole) numbers
We need to prove that K is dense in T ={z in C, |z| = 1}

The set T can be defined as well T = {z in C, z = exp(i*theta), theta in [0, 2*pi)}
and exp(i*theta) = cos(theta) + i*sin(theta) according to Euler's formula.

In order to prove that K is dense in T, we have to show that every point of K is an accumulation point for T. In other words, we have to prove that, for every theta in [0, 2*pi) and eps >0, we can find 2 whole numbers (k) and (n) so that following condition is fulfilled:
|exp(i*k) - exp(i*(theta + 2*n*pi))| < eps (1)

Remark: we introduced the integer (n) because exp(i*theta) is congruent to exp(i*theta)*exp(i*2*n*pi) = exp(i*(theta + 2*n*pi)) on the unitary circle T.

The condition (1) used the topology of complex plane, induced by the standard metric:
...

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