Gary
GaryVasco
Posts: 3,352 
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Post by Gary on May 26, 2017 14:58:44 GMT
Vasco,
Would you agree that in line 3, the equation, $\nu[L, p]$ should be $\nu[f(L), p]$ ?
Gary
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Post by Admin on May 26, 2017 15:22:32 GMT
Vasco, Would you agree that in line 3, the equation, $\nu[L, p]$ should be $\nu[f(L), p]$ ? Gary Gary Yes, without actually solving the exercise I would say that it makes more sense the way you suggest, because $\nu[L, p]$ is zero. Vasco PS Should it also say: "..., if $f(L)$ does not contain $p$..." I have also taken the liberty of editing your heading to refer to exercise 21 not 20 |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 26, 2017 19:02:33 GMT
Vasco, Would you agree that in line 3, the equation, $\nu[L, p]$ should be $\nu[f(L), p]$ ? Gary Gary Yes, without actually solving the exercise I would say that it makes more sense the way you suggest, because $\nu[L, p]$ is zero. Vasco PS Should it also say: "..., if $f(L)$ does not contain $p$..." I have also taken the liberty of editing your heading to refer to exercise 21 not 20 Vasco, Thank you. Your PS was to be my next question or suggestion. I think it might be part of a deleted thought that there could be a root outside L. I couldn't make sense of it with $f(L)$. Gary |
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Post by Admin on Sept 11, 2017 16:31:08 GMT
Gary
I now think that exercise 21 should be rewritten as follows:
Let $\mathcal{R}(L)$ be the net rotation of $f(z)$ round $p$ as $z$ traverses a loop $L$. For example if $p$ does not lie on $f(L)$ then
$\nu[f(L),p]=\frac{1}{2\pi}\mathcal{R}(L)$
By taking this formula...their multiplicities. (This last paragraph remains unchanged)
Vasco
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Post by Admin on Sept 18, 2017 13:01:00 GMT
Gary
In my post above I forgot to say that in the exercise as stated in the book, after correcting "... if $L$ does not contain $p$ then ..." to read
"...if $f(L)$ does not contain $p$ then ...", the reader might still be confused by the use of 'does not contain' to mean that $p$ is not a member of the set $f(L)$.
In other words that it does not lie on the contour $f(L)$.
The phrase 'does not contain' could be interpreted to mean that $p$ lies outside $f(\Gamma)$, but this interpretation does not make sense because
$\nu[L,p]=\frac{1}{2\pi}\mathcal{R}(L)$ whether $p$ is outside or inside $L$.
Also in the last paragraph of the exercise the loop is referred to as $\Gamma$ rather than $L$, causing further confusion, and so I think that the whole exercise would be better written as follows
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 18, 2017 14:44:23 GMT
Gary In my post above I forgot to say that in the exercise as stated in the book, after correcting "... if $L$ does not contain $p$ then ..." to read "...if $f(L)$ does not contain $p$ then ...", the reader might still be confused by the use of 'does not contain' to mean that $p$ is not a member of the set $f(L)$. In other words that it does not lie on the contour $f(L)$. The phrase'does not contain' could be interpreted to mean that $p$ lies outside $f(\Gamma)$, but this interpretation does not make sense because $\nu[L,p]=\frac{1}{2\pi}\mathcal{R}(L)$ whether $p$ is outside or inside $L$. Also in the last paragraph of the exercise the loop is referred to as $\Gamma$ rather than $L$, causing further confusion, and so I think that the whole exercise would be better written as follows Vasco Vasco, I agree with both suggestions. $\Gamma$ appears as the loop symbol in both the preceding and following problems. I had not noticed the inconsistency. Gary |
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