yes, sorry for more math speak. we take it for granted that e^(it) = cos(t) + i sin(t), which is a point of the unit circle in the complex plane. so the "circle group" means the group of unit length complex numbers, i.e. those of form cos(t) + i sin(t) = e^(it), for real t.
thus t-->e^(it) is a map from the additive group of reals to the multiplicative group U of unit complex numbers. hence by definition of the dual group, it is an element of the dual group of the reals. for any real s, t-->e^(its) is another one, and apparently all such occur this way, so this would identify the dual group of maps R-->U with the original group R. but i am really not at all proficient in this topic.
Thanks. Dont worry about the "maths speak". The rigour of your arguments reminds me to try to be more careful in how I phrase my arguments. Yes I enjoyed the theory of complex numbers when I was at University so I understand what you say.
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