Find an isomorphism between the group of orientation

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I need to find an isomorphism between the group of orientation preserving rigid motions of the plane (translations, rotations) and complex valued matrices of the form

a b
0 1

where |a|=1.

I defined an isomorphism where the rotation part goes to e^it with angle t and the translation by l=ax+by to b=a+bi. But the multiplication doesn't work out.
 
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I defined an isomorphism where the rotation part goes to e^it with angle t and the translation by l=ax+by to b=a+bi. But the multiplication doesn't work out.
The rotation part goes to eit? It's supposed to go to a matrix, eit is not a matrix. The translation goes to b=a+bi? I don't even know what this means. Is the b on the left side the same as the one on the right side? And again, a+bi is a number, not a matrix, so how can the translation go to a+bi? Moreover, what exactly do you mean by "the rotation part" and "the translation"? I assume you mean that any rigid motion can be decomposed some how into a translation part and rotation part. But have you proved that this is possible? And do you realize that if f is an arbitrary orientation preserving rigid motion, then it can be decomposed into a rotation and translation like so: f = rt for some rotation r and some translation t, and can also be decomposed f = r't', for some rotation r' and some translation t', but prima facie, r' need not equal r and t' need not equal t, so when you speak of "the rotation part" it's ambiguous until you say whether you're decomposing rotation-first or translation-first.

Once you write out something that's clear, unambiguous, and makes sense, we can suggest ways to get passed wherever you're getting stuck, but right now I don't know how to help you. Actually, you haven't even asked a question.
 
(a priori, not prima facie)
 
Every orientation preserving rigid motion can be written as \rho_{\theta}t_a where 0\leq\theta <2\pi and a=a_1x_1+a_2x_2. Define a map f(\rho_{\theta}t_a)=\left(\begin{array}{cc}{e^{i\theta}}&{a_1+a_2i}\\0&1\end{array}\right). Clear enough now?
 
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Okay, so you've tried one thing that doesn't work. What exactly do you want now? I ask because I'm having trouble thinking of a hint, so it would help me if I had a more specific question to answer. Do you know anything about fractional linear transformations, a.k.a. Mobius transformations?
 
I already solved it.
 
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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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