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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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closed as off-topic by Rafflesia arnoldii, Claude Leibovici, PVAL, heropup, Shuchang Aug 14 at 7:05

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As I mentioned on your earlier question, it is not considered polite here to tell other users to do something. Your question does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. –  Zev Chonoles Apr 8 '12 at 23:47
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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? –  Tpofofn Apr 9 '12 at 0:03

2 Answers 2

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

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I am not really sure how to go about this one. That is why I posted it on here. I know the properties of a group and how to tell if a set is one or not, but I am not sure how to go about this problem using the definitions of a group. –  user28615 Apr 9 '12 at 0:03
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@user28615 What are the parts of the axiom you have proved? Have you seen that there is an identity in this set? Can you recognise this set with a familiar object from your pre-calculus algebra courses? –  user21436 Apr 9 '12 at 0:10

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.

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