Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Aug 23, 2016 23:21:58 GMT
Vasco, I ran across these exercises while working on spatial rotations as Mobius transformations in Chapter 6 and realized that I had not worked through them. I followed several tempting rabbit holes before finding the right path. I had always before worked out these relations using trig functions. It's interesting to see them worked out with similar triangles and algebra. nh.ch3.notes.exercises.p146.pdf (119.57 KB) Gary |
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Post by Admin on Aug 29, 2016 6:39:26 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Aug 30, 2016 4:58:41 GMT
Vasco,
Why did I write i/(i-Zi)? For some reason, I thought at the time that I had to write it as a complex number, which is silly with the absolute values on the LHS. I will look at the rest if your comments tomorrow.
Gary
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Post by mondo on Sept 29, 2022 3:25:06 GMT
I wonder what allows us to go from $\frac{|z|}{|z'|} = \frac{1}{1-Z}$ to $x+iy = \frac{X+iY}{1-Z}$. In other words why can we drop the lengths and start to treat this ratio as regular complex numbers?
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Post by Admin on Sept 29, 2022 8:02:29 GMT
I wonder what allows us to go from $\frac{|z|}{|z'|} = \frac{1}{1-Z}$ to $x+iy = \frac{X+iY}{1-Z}$. In other words why can we drop the lengths and start to treat this ratio as regular complex numbers? Mondo This is just a straightforward substitution: Given that $\frac{|z|}{|z'|} = \frac{1}{1-Z}$ substitute for $\frac{|z|}{|z'|}$ from this equation into the first equation on page 146 i.e. $z=\frac{|z|}{|z'|}z'$, and then write $x+iy$ for $z$ and $X+iY$ for $z'$ and we get: $x+iy = \frac{X+iY}{1-Z}$ Just simple algebra. Vasco |
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Post by mondo on Sept 29, 2022 16:28:46 GMT
Ok, so it is a substitution, yes it makes sense now. For a moment I thought there is some rule that allows us to drop the module and keep this formula valid.
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