Is any non-zero scalar of the unit disc also compact?

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There is the circle group which is specifically a set of points comprising the boundary of the unit disc in the complex plane which turns out to be a compact set.

However, is the set of points forming a circle of any arbitrary radius also compact in the complex plane?

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Yes, either because of Heine-Borel: a closed and bounded (in absolute value ) subset of $\Bbb C$ is compact, or by showing that all maps $f(z)= \alpha z$ are continuous (and have a continuous inverse when $\alpha \neq 0$ in $f^{-1}(z)=\frac{1}{\alpha}z$) and so preserve compactness (the continuous image of a compact set is compact as well).

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Sure. Any two circles in the complex plane are homeomorphic. So, if one of them is compact, then all of them are.