Why is $oint frac{1}{z^{2}}dz = 0$, where $z$ is complex?
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Consider the complex function $f(z) = frac{1}{z^{2}}$. How come the closed integral about the centre vanishes, i.e. $int_{partial B_{r}(0)} f(z) dz = 0$, where $r>0$. The function clearly has a pole of second order at $z=0$ and therefore isn't holomorphic on a disc centred at zero.
complex-analysis
asked Jan 16 at 11:37
playdisplaydis
153
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add a comment |
$begingroup$
Consider the complex function $f(z) = frac{1}{z^{2}}$. How come the closed integral about the centre vanishes, i.e. $int_{partial B_{r}(0)} f(z) dz = 0$, where $r>0$. The function clearly has a pole of second order at $z=0$ and therefore isn't holomorphic on a disc centred at zero.
complex-analysis
asked Jan 16 at 11:37
playdisplaydis
153
$endgroup$
$begingroup$
This thread may also help you, math.stackexchange.com/questions/1165813/…
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– Someone86
Jan 16 at 11:49
1
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Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic.
$endgroup$
– Vasily Mitch
Jan 16 at 11:50
add a comment |
$begingroup$
Consider the complex function $f(z) = frac{1}{z^{2}}$. How come the closed integral about the centre vanishes, i.e. $int_{partial B_{r}(0)} f(z) dz = 0$, where $r>0$. The function clearly has a pole of second order at $z=0$ and therefore isn't holomorphic on a disc centred at zero.
complex-analysis
asked Jan 16 at 11:37
playdisplaydis
153
$endgroup$
Consider the complex function $f(z) = frac{1}{z^{2}}$. How come the closed integral about the centre vanishes, i.e. $int_{partial B_{r}(0)} f(z) dz = 0$, where $r>0$. The function clearly has a pole of second order at $z=0$ and therefore isn't holomorphic on a disc centred at zero.
complex-analysis
complex-analysis
asked Jan 16 at 11:37
playdisplaydis
153
asked Jan 16 at 11:37
playdisplaydis
153
asked Jan 16 at 11:37
playdisplaydis
153
asked Jan 16 at 11:37
playdisplaydis
153
asked Jan 16 at 11:37
playdisplaydis
153
153
$begingroup$
This thread may also help you, math.stackexchange.com/questions/1165813/…
$endgroup$
– Someone86
Jan 16 at 11:49
1
$begingroup$
Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic.
$endgroup$
– Vasily Mitch
Jan 16 at 11:50
add a comment |
$begingroup$
This thread may also help you, math.stackexchange.com/questions/1165813/…
$endgroup$
– Someone86
Jan 16 at 11:49
1
$begingroup$
Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic.
$endgroup$
– Vasily Mitch
Jan 16 at 11:50
$begingroup$
This thread may also help you, math.stackexchange.com/questions/1165813/…
$endgroup$
– Someone86
Jan 16 at 11:49
$begingroup$
This thread may also help you, math.stackexchange.com/questions/1165813/…
$endgroup$
– Someone86
Jan 16 at 11:49
1
1
$begingroup$
Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic.
$endgroup$
– Vasily Mitch
Jan 16 at 11:50
$begingroup$
Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic.
$endgroup$
– Vasily Mitch
Jan 16 at 11:50
add a comment |
2 Answers
2
active
oldest
votes
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Three answers:
The integral evaluates to the residue, which is the coefficient of $z^{-1}$ in the Laurent expansion at $0$ of the function. But this is clearly zero, as the Laurent expansion of $z^{-2}$ around $z_0=0$ is again $z^{-2}$, so the coefficient of $z^{-1}$ is $0$.
The function has an antiderivative: $-z^{-1}$ which is continuous on the path you are integrating over.
Evaluating the integral directly via polar coordinates $z=re^{ivarphi}$ gives zero.
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
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add a comment |
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The fact that integrals of holomorphic functions along ["nice"] closed curves are $0$ does not mean those non-holomorphic functions cannot be integrated to be $0$.
To calculate this, just use the parametrization $z = r mathrm e^{mathrm i t}, t in [0, 2pi)$.
Alternatively, we consider the entire function $f(z) equiv 1$. Then by the Cauchy integral formula and the derivative formula,
$$
f'(z) = frac {1!}{mathrm i2pi}int_{partial B_r (z)} frac {f(zeta)}{(zeta - z)^2} ,mathrm dzeta,
$$
then your integral is just $f'(0) = 0$.
answered Jan 16 at 11:56
xbhxbh
6,3201522
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Three answers:
The integral evaluates to the residue, which is the coefficient of $z^{-1}$ in the Laurent expansion at $0$ of the function. But this is clearly zero, as the Laurent expansion of $z^{-2}$ around $z_0=0$ is again $z^{-2}$, so the coefficient of $z^{-1}$ is $0$.
The function has an antiderivative: $-z^{-1}$ which is continuous on the path you are integrating over.
Evaluating the integral directly via polar coordinates $z=re^{ivarphi}$ gives zero.
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
Three answers:
The integral evaluates to the residue, which is the coefficient of $z^{-1}$ in the Laurent expansion at $0$ of the function. But this is clearly zero, as the Laurent expansion of $z^{-2}$ around $z_0=0$ is again $z^{-2}$, so the coefficient of $z^{-1}$ is $0$.
The function has an antiderivative: $-z^{-1}$ which is continuous on the path you are integrating over.
Evaluating the integral directly via polar coordinates $z=re^{ivarphi}$ gives zero.
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
$endgroup$
add a comment |
$begingroup$
Three answers:
The integral evaluates to the residue, which is the coefficient of $z^{-1}$ in the Laurent expansion at $0$ of the function. But this is clearly zero, as the Laurent expansion of $z^{-2}$ around $z_0=0$ is again $z^{-2}$, so the coefficient of $z^{-1}$ is $0$.
The function has an antiderivative: $-z^{-1}$ which is continuous on the path you are integrating over.
Evaluating the integral directly via polar coordinates $z=re^{ivarphi}$ gives zero.
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
$endgroup$
Three answers:
The integral evaluates to the residue, which is the coefficient of $z^{-1}$ in the Laurent expansion at $0$ of the function. But this is clearly zero, as the Laurent expansion of $z^{-2}$ around $z_0=0$ is again $z^{-2}$, so the coefficient of $z^{-1}$ is $0$.
The function has an antiderivative: $-z^{-1}$ which is continuous on the path you are integrating over.
Evaluating the integral directly via polar coordinates $z=re^{ivarphi}$ gives zero.
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
answered Jan 16 at 11:46
b00n heTb00n heT
10.5k12335
10.5k12335
add a comment |
add a comment |
$begingroup$
The fact that integrals of holomorphic functions along ["nice"] closed curves are $0$ does not mean those non-holomorphic functions cannot be integrated to be $0$.
To calculate this, just use the parametrization $z = r mathrm e^{mathrm i t}, t in [0, 2pi)$.
Alternatively, we consider the entire function $f(z) equiv 1$. Then by the Cauchy integral formula and the derivative formula,
$$
f'(z) = frac {1!}{mathrm i2pi}int_{partial B_r (z)} frac {f(zeta)}{(zeta - z)^2} ,mathrm dzeta,
$$
then your integral is just $f'(0) = 0$.
answered Jan 16 at 11:56
xbhxbh
6,3201522
$endgroup$
add a comment |
$begingroup$
The fact that integrals of holomorphic functions along ["nice"] closed curves are $0$ does not mean those non-holomorphic functions cannot be integrated to be $0$.
To calculate this, just use the parametrization $z = r mathrm e^{mathrm i t}, t in [0, 2pi)$.
Alternatively, we consider the entire function $f(z) equiv 1$. Then by the Cauchy integral formula and the derivative formula,
$$
f'(z) = frac {1!}{mathrm i2pi}int_{partial B_r (z)} frac {f(zeta)}{(zeta - z)^2} ,mathrm dzeta,
$$
then your integral is just $f'(0) = 0$.
answered Jan 16 at 11:56
xbhxbh
6,3201522
$endgroup$
add a comment |
$begingroup$
The fact that integrals of holomorphic functions along ["nice"] closed curves are $0$ does not mean those non-holomorphic functions cannot be integrated to be $0$.
To calculate this, just use the parametrization $z = r mathrm e^{mathrm i t}, t in [0, 2pi)$.
Alternatively, we consider the entire function $f(z) equiv 1$. Then by the Cauchy integral formula and the derivative formula,
$$
f'(z) = frac {1!}{mathrm i2pi}int_{partial B_r (z)} frac {f(zeta)}{(zeta - z)^2} ,mathrm dzeta,
$$
then your integral is just $f'(0) = 0$.
answered Jan 16 at 11:56
xbhxbh
6,3201522
$endgroup$
The fact that integrals of holomorphic functions along ["nice"] closed curves are $0$ does not mean those non-holomorphic functions cannot be integrated to be $0$.
To calculate this, just use the parametrization $z = r mathrm e^{mathrm i t}, t in [0, 2pi)$.
Alternatively, we consider the entire function $f(z) equiv 1$. Then by the Cauchy integral formula and the derivative formula,
$$
f'(z) = frac {1!}{mathrm i2pi}int_{partial B_r (z)} frac {f(zeta)}{(zeta - z)^2} ,mathrm dzeta,
$$
then your integral is just $f'(0) = 0$.
answered Jan 16 at 11:56
xbhxbh
6,3201522
answered Jan 16 at 11:56
xbhxbh
6,3201522
answered Jan 16 at 11:56
xbhxbh
6,3201522
answered Jan 16 at 11:56
xbhxbh
6,3201522
6,3201522
add a comment |
add a comment |
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$begingroup$
This thread may also help you, math.stackexchange.com/questions/1165813/…
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– Someone86
Jan 16 at 11:49
1
$begingroup$
Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic.
$endgroup$
– Vasily Mitch
Jan 16 at 11:50