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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 ᴡᴏʀᴅs, Claude Leibovici, PVAL, heropup, Shuchang Aug 14 '14 at 7:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – ᴡᴏʀᴅs, Claude Leibovici, PVAL, heropup, Shuchang
If this question can be reworded to fit the rules in the help center, please edit the question.

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
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

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
@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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