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Are these groups? If so show it, and if not provide a counterexample.

The set of all complex numbers $x$ that have absolute value $1$, with operation multiplication. Recall that the absolute value of a complex number $x$ written in the form $x = a +bi$, with $a$ and $b$ real, is given by $|x| = |a+bi| = (a^2 + b^2)^{1/2}$.

The set of all complex numbers $x$ that have absolute value $1$, with operation addition.

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    The properties of a group are 1) closure, 2) identity, and 3) inverse. Have you tested these cases against these properties? Where are you stuck?2012-04-09

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Hint: for $z,w\in\mathbb{C}$, $|zw|=|z||w|$ but $|z+w|\leq|z|+|w|$.

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    @user28615 What are the parts o$f$ the axiom you have proved? Have you seen that there is an identity in this set? Can you recognise this set with a $f$amiliar object from your pre-calculus algebra courses?2012-04-09
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Since you know that you're already working with complex numbers, I think it will be easier if you work with the polar representation, so that you can operate and manipulate better.