Solve Binary Operations Question: Prove Isomorphism

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So I have really been struggling with this question. The original question said: The map \varphi:Z->Z defined by \varphi(n)=n+1 for n in Z is one to one and onto Z. For (Z, . ) onto (Z,*) (i am using . for usual multiplication) define * and show that * makes phi into an isomorphism.
I know that the operation must be m*n=mn-m-n+2. But I get stuck in proving that the operations are preserved. When I do \varphi(m.n) i get mn+1. and i can't get \varphi(m). \varphi(n) to work. I think I am doing something wrong. Can anyone help?
 
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Never mind! i just got it to work!
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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