Proving $Res(f'/f,z)$ is an integer for f analytic on a domain $Omega$ and $z in Omega$
$begingroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
$endgroup$
add a comment |
$begingroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
$endgroup$
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
add a comment |
$begingroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
$endgroup$
Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) not = 0$ and so we know that $Res(f'/f,z) = frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
complex-analysis residue-calculus
complex-analysis residue-calculus
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 0:29
Richard VillalobosRichard Villalobos
1757
1757
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
add a comment |
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
2
2
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45
add a comment |
1 Answer
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oldest
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$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
$endgroup$
The statement $Resleft(frac{f'}{f}, zright) = = frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then begin{align*}
Resleft(frac{g}{h},z_0right) &= (k+1)frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)}
end{align*}
If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore begin{align*}
Resleft(frac{f'}{f},z_0right) = (k+1)frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1
end{align*}
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
answered Dec 13 '18 at 2:47
J. PistachioJ. Pistachio
488212
488212
add a comment |
add a comment |
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$begingroup$
No, try with $f(z) = z^2$ and the residue at $0$. The correct argument is that $f(z) = (z-a)^n h(z)$ with $h(a) ne 0$ thus $f'(z)/f(z) = ...$
$endgroup$
– reuns
Dec 12 '18 at 0:45