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Post by Admin on Feb 15, 2017 7:02:05 GMT
Following discussions with Gary we have concluded that there is an error in the statement of this exercise in the book on page 337 (paperback edition year 2000 with corrections).
In part (ii) the expression for $\widetilde{\mathcal{M}}$ should read:
$\widetilde{\mathcal{M}}=\dfrac{-a\bar{z}-b}{c\bar{z}+d}$ where $a,b,c,d$ are real, and $(ad-bc)>0$.
This is after having defined a Möbius transformation $M$ as
$M=\dfrac{az+b}{cz+d}$, using the usual notation and where again $a,b,c,d$ are real, and $(ad-bc)>0$.
Vasco
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Deleted
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Post by Deleted on Aug 1, 2017 4:54:29 GMT
Wouldn't it be simpler (and more natural) to just say the inequality sign in (ii) needs to be reversed? That makes more sense to me, especially as the mistake is then reduced to nothing more than a typo for a single symbol. At any rate that was my conclusion after a lengthy (and obviously unsuccessful) struggle to get the wrong answer before knowing this forum existed!
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Post by Admin on Aug 1, 2017 11:39:54 GMT
Hi Version 1Version 2We (Gary and I) discussed the best way of doing this earlier this year when we were working on this exercise. I agree with you that it's possible to correct the exercise in the way you suggest. However in my solution to part (ii) I use a Mobius transformation ($M$ not $\mathcal{M}$) using the standard notation from the book and this leads to a solution to part (ii) for $\mathcal{M}$ with a negative sign in front of $a$ and $b$, and so as well as changing the inequality, it is necessary to redefine $a$ as $-a$ and $b$ as $-b$. (see my solution for Version 2 via the link above). Notice that part (iii) uses the same definitions for $a,b,c,d$ as those used in part (ii). So we decided it was better to use the standard definition of a Mobius transformation and alter the result accordingly without redefining the coefficients. The inequality is then the same as it is for the Mobius transformation. An alternative approach we looked at was to define the Mobius transformation using non-standard notation (see my solution for Version 1 via the link above). This leads to the solution in the book for $\mathcal{M}$, but with the inequality reversed. In the end it's a matter of personal choice which of the two ways above you prefer or any other way for that matter. Vasco PS Don't forget that in the exercise $M$ represents a Mobius transformation but $\mathcal{M}$ does not. |
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