Compactness of a set in the complex plane.

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Consider $$A=\Big\{(z_1,z_2) \in \Bbb{C}^2 : z_1^2+z_2^2=1\Big\}$$

Is $A$ compact?

My try: If $A$ is compact, then it is closed and bounded

But it is not bounded! Since $(n,\sqrt{1-n^2});n\in \Bbb{N}$ satisfies this and $\Bbb{N}$ is unbounded. So not compact.

Am I right? If not, any help?

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I personally dislike the use of the square root sign in this context (it stops being well-defined the moment you start using it on negative numbers), but that may be fixed by writing, say, $(n, i\sqrt{n^2-1})$ instead. However, not everyone are as opposed to writing $\sqrt{-1}$ as I am.

Apart from that it looks good. Well done!