Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Sept 24, 2017 23:08:14 GMT
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Post by Admin on Sept 25, 2017 11:01:05 GMT
Gary
That all seems fine to me. I didn't do this suggested exercise when I was studying chapter 9, but looking at it now I would have defined $P=1$ and $Q=(z-p)(z-q)$. Same final result.
Also, since $z^2+2az+1$ is a quadratic with real coefficients the roots are of the form $\alpha\pm\beta$ and so
$q-p=\alpha+\beta-(\alpha-\beta)=2\beta=2\frac{\sqrt{4a^2-4}}{2}=2\sqrt{a^2-1}$.
Same result just less algebra.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 25, 2017 16:29:57 GMT
Also, since $z^2+2az+1$ is a quadratic with real coefficients the roots are of the form $\alpha\pm\beta$ and so $q-p=\alpha+\beta-(\alpha-\beta)=2\beta=2\frac{\sqrt{4a^2-4}}{2}=2\sqrt{a^2-1}$. Same result just less algebra. Vasco I had both derivations, but for me the short one took a little more thought to realize it satisfied all the conditions for p and q, so I checked it against pq = 1. Gary |
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Post by mondo on Jun 19, 2023 7:20:37 GMT
I wonder how author concludes "Since the singularities $p$ and $q$ of the integrand satisfy $pq = 1$ only one of them lies inside $C$". In order to answer this question we first need to know what are the singularities. Using $\frac{dz}{z^2+2az+1}$ the singularity happens at roots of the denominator which are $\pm (a^2-1)^{1/2} - a$ but how does it lead to the $pq = 1$ conclusion, and the $Res[]$ formula?
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Post by Admin on Jun 19, 2023 13:41:36 GMT
Mondo
Given a quadratic equation
$Az^2+Bz+C=0$ where $z,A,B,C$ are complex numbers, we know from the theory of polynomials (or we can easily prove) that the sum of the two roots is $-B/A$ and the product of the two roots is $C/A$.
We also know that if the quadratic has real coefficients ($A,B,C$ real), then the roots are either both real or are complex conjugates.
Vasco
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Post by mondo on Jun 19, 2023 18:18:30 GMT
Right, but how do we know only one of them lies inside $C$? If they both lie on the $C$ itself then the product would also be $1$ right?
If both lies inside $C$ then the product would be less than $<1$, if outside then $> 1$
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Post by mondo on Jun 19, 2023 18:24:13 GMT
Additionally, in the $Res[]$ formula at the top of 438 author calculated the residue at a point $q$, how do we know this is the one inside and not $p$?
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Post by Admin on Jun 19, 2023 18:31:12 GMT
Mondo
It doesn't matter which you choose.
Vasco
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Post by mondo on Jun 19, 2023 18:45:01 GMT
Mondo It doesn't matter which you choose. Vasco How can it not matter if we evaluate an integral round $C$ so we should not care about any singularity outside this unit circle right? |
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Post by Admin on Jun 19, 2023 18:55:02 GMT
Mondo
What I mean is we can choose to call $p$ the one inside and $q$ the one outside or vice versa.
Vasco
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Post by mondo on Jun 19, 2023 19:12:28 GMT
Ok, I also have a question to this: Gary That all seems fine to me. I didn't do this suggested exercise when I was studying chapter 9, but looking at it now I would have defined $P=1$ and $Q=(z-p)(z-q)$. Same final result. Also, since $z^2+2az+1$ is a quadratic with real coefficients the roots are of the form $\alpha\pm\beta$ and so $q-p=\alpha+\beta-(\alpha-\beta)=2\beta=2\frac{\sqrt{4a^2-4}}{2}=2\sqrt{a^2-1}$. Same result just less algebra. Vasco Why can you write $q-p=\alpha+\beta-(\alpha-\beta)$? By $\alpha\pm\beta$ you mean there are two roots $\alpha$ and $\beta$ and they may be of opposite signs or equal signs right? |
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Post by Admin on Jun 19, 2023 19:18:31 GMT
Mondo
Because the roots are $q=\alpha+\beta$ and $p=\alpha-\beta$,.
Vasco
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Post by mondo on Jun 19, 2023 19:43:48 GMT
Right, I got it right after I posted. $p,q = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
So the things I am not sure about:
1. $p,q$ lie on a real axis, they are real numbers because coefficients of our quadratic are all real right?
2. The fact that $pq = 1$ means either they are both $1$ or inverse of each other. They can't be one because then we see our quadratic is not $0$ right? Hence we conclude we have two roots one inside and the other outside the unit circle.
3. For the same reason we are sure the order of the pole is $1$?
Thank you.
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Post by Admin on Jun 19, 2023 19:57:16 GMT
Mondo
Great!
Vasco
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Post by mondo on Jun 19, 2023 21:26:37 GMT
Thank you Vasco. One more question for this page 438 - at the very bottom $[1 - \frac{(\pi z)^2}{6} + ...]^{-1} $ got transformed to $[1 + \frac{(\pi z)^2}{6} + ...]$ is it a type or some transformation I don't see?
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