Check whether or not $\{z \in \mathbb{C} \mid z^5=1\}$ is a group with respect to addition.

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Let $S=\{z \in \mathbb{C} \mid z^5=1\}$

We know that, $$S=\left\{\cos(\frac{2\pi j}{5})+\iota \sin(\frac{2\pi j}{5})\mid j=0,1,2,3,4\right\}$$ and that $S$ is a group w.r.t. multiplication of complex numbers. So $1 \in S$.

Under Addition

Since $1 \in S$, $1+1=2$ must be in $S$. But we know that $2^5=32 \neq 1$. This implies that $2 \notin S$ and hence $S$ is not closed under addition and consequently not a group under addtion.

Please let me know if my conclusion and\or approach is right.

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That's fine. One counterexample, even using the same value twice, is enough. Indeed, since you've refuted closure, this set doesn't even form a magma under addition.