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True or False: the isometries $f(z)=\dfrac{5z+19}{z+4}$ and $g(z)=\dfrac{z}{28z+1} $ are conjugate. Explain your answer.

I'm lost as to how to proceed with this question. I don't know what 'conjugate' isometries are. Any help would be appreciated.

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closed as too localized by Andres Caicedo, Noah Snyder, Gerry Myerson, Amzoti, vonbrand May 5 '13 at 3:12

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Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. –  Antonio Vargas Apr 26 '13 at 18:45
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@Antonio’s view of MSE quality standards is not universally shared; I consider ‘I’m lost as to how to proceed’ adequate context. –  Brian M. Scott Apr 26 '13 at 23:30
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I don't consider "I'm lost as to how to proceed" to be adequate context. To begin with, it doesn't indicate whether or not OP knows what "conjugate" means. How can one give an answer useful to OP, if one doesn't even know whether the problem is just one of not knowing the definitions? –  Gerry Myerson Apr 27 '13 at 23:56
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@GerryMyerson - Apologies, I wasn't readily near a terminal this weekend. But yes, I'm not sure what exactly is meant by "conjugate" isometries, hence the lost comment. I will try to improve the wording on future questions. –  devcoder Apr 28 '13 at 19:09
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Why not try to improve the wording on this question? –  Gerry Myerson Apr 29 '13 at 9:08
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1 Answer 1

For $f$ and $g$ to be conjugate, there must be some transformation $$h(z)=\frac{az+b}{cz+d}$$ with $ad-bc\ne 0$ such that $g=h^{-1}\circ f\circ h$. Without loss of generality, we may assume that any such $h$ has determinant $ad-bc=1$. (Why?)

If two isometries (in normalized form with determinant $1$, as yours are) are conjugate, then the squares of their traces are equal. Let's prove this.

Let $$g_1(z)=\frac{\alpha z+\beta}{\gamma z+\delta}$$ and $h$ as above with $\alpha\delta-\beta\gamma=1$ and $ad-bc=1.$ Then $h^{-1}(z)=\frac{dz-b}{-cz+a}$, so $$\begin{align}\left(h^{-1}\circ g_1\circ h\right)(z) &= \left(h^{-1}\circ g_1\right)\left(\frac{az+b}{cz+d}\right)\\ &= h^{-1}\left(\frac{(\alpha a+\beta c)z+(\alpha b+\beta d)}{(\gamma a+\delta c)z+(\gamma b+\delta d)}\right)\\ &= \frac{\bigl(d(\alpha a+\beta c)-b(\gamma a+\delta c)\bigr)z+\bigl(d(\alpha b+\beta d)-b(\gamma b+\delta d)\bigr)}{\bigl(-c(\alpha a+\beta c)+a(\gamma a+\delta c)\bigr)z+\bigl(-c(\alpha b+\beta d)+a(\gamma b+\delta d)\bigr)},\end{align}$$ and so the trace of $h^{-1}\circ g_1\circ h$ is $$\begin{align}\text{tr}(h^{-1}\circ g_1\circ h) &= d(\alpha a+\beta c)-b(\gamma a+\delta c)-c(\alpha b+\beta d)+a(\gamma b+\delta d)\\ &= \alpha ad+\beta cd-\gamma ab-\delta bc-\alpha bc-\beta cd+\gamma ab+\delta ad\\ &= \alpha ad-\alpha bc+\beta cd-\beta cd-\gamma ab+\gamma ab+\delta ad-\delta bc\\ &= \alpha(ad-bc)+\delta(ad-bc)\\ &= \alpha+\delta.\end{align}$$ On the other hand, multiplying the numerator and denominator by $-1$ gives us the only other normalized form (why are there no more of these?) for $h^{-1}\circ g_1\circ h,$ which has opposite trace. Thus, if $g_1,g_2$ are normalized conjugate isometries, then $$\text{tr} g_1=\pm\text{tr}g_2,$$ and so $$(\text{tr}g_1)^2=(\text{tr}g_2)^2.$$ The converse also holds (though this is trickier to prove, and not relevant to the problem at hand). The upshot for you, here, is that the isometries are both normalized, but their traces' squares aren't equal, and thus, they are not conjugate.

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Must $h$ be a fractional linear transformation? Are we assuming "conjugate" means "conjugate in the group of fractional linear transformations"? Could the maps be conjugate via some $h$ that is not an FLT? –  Gerry Myerson Apr 27 '13 at 5:33
    
The OP has been asking several questions lately about the group of FLT's, so I am assuming that is the context here, too. If that isn't what is meant, then the OP will hopefully clarify that. –  Cameron Buie Apr 27 '13 at 12:25
    
@CameronBuie - Could you elaborate on what FLT is ? All I could find on google about it is Fermat's Last Theorem, but this is not related to that. –  devcoder Apr 27 '13 at 16:50
    
@devcoder: Fractional Linear Transformation. Functions of the form of $h$ in my answer. –  Cameron Buie Apr 27 '13 at 17:03
    
@devcoder: Were you able to get anywhere with my hint? If you didn't get all the way to the solution, let me know what you did, and what you were able to conclude, and I'll expand my answer. –  Cameron Buie Apr 29 '13 at 14:22
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