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Evaluate $int_=3fracdzz^3(z^10-2)$
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$$int_zfracdzz^3(z^10-2)$$
There are singularities at $z=0$ and $z^10=2iff z=sqrt[10]2e^fracipi k5 text where k=0,1,...9$ so we need to find $10$ residues? or have I got it wrong?
complex-analysis contour-integration residue-calculus
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
asked Jul 15 at 16:19
gbox
5,30851841
add a comment |Â
up vote
3
down vote
favorite
$$int_zfracdzz^3(z^10-2)$$
There are singularities at $z=0$ and $z^10=2iff z=sqrt[10]2e^fracipi k5 text where k=0,1,...9$ so we need to find $10$ residues? or have I got it wrong?
complex-analysis contour-integration residue-calculus
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
asked Jul 15 at 16:19
gbox
5,30851841
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$int_zfracdzz^3(z^10-2)$$
There are singularities at $z=0$ and $z^10=2iff z=sqrt[10]2e^fracipi k5 text where k=0,1,...9$ so we need to find $10$ residues? or have I got it wrong?
complex-analysis contour-integration residue-calculus
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
asked Jul 15 at 16:19
gbox
5,30851841
$$int_zfracdzz^3(z^10-2)$$
There are singularities at $z=0$ and $z^10=2iff z=sqrt[10]2e^fracipi k5 text where k=0,1,...9$ so we need to find $10$ residues? or have I got it wrong?
complex-analysis contour-integration residue-calculus
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
asked Jul 15 at 16:19
gbox
5,30851841
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
edited Jul 15 at 16:33
José Carlos Santos
114k1698177
114k1698177
asked Jul 15 at 16:19
gbox
5,30851841
asked Jul 15 at 16:19
gbox
5,30851841
asked Jul 15 at 16:19
gbox
5,30851841
5,30851841
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
Since there are no singularities outside $|z|=2^1/10$, the Cauchy Integral Theorem says that
$$
int_zfracmathrmdzz^3(z^10-2)
=int_zfracmathrmdzz^3(z^10-2)
$$
for all $rgt2^1/10$. As $rtoinfty$, the integral on the right obviously vanishes.
answered Jul 15 at 17:13
robjohn♦
258k26297612
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
1
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
add a comment |Â
up vote
3
down vote
This is easy by symmetry, without calculating any residues.
Say $f(z)=frac1z^3(z^10-2)$ and let $omega=e^2pi i/10$. Note that
$$f(omega z)=omega^-3f(z).$$It follows that the integral is $0$.
Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=int_zf(omega z),d(omega z)
=int_zomega^-3f(z)omega,dz=omega^-2I;$$since $omega^-2ne1$ this shows $I=0$.
(The "informal" part is the first equality above. To justify it, note that if $tmapstogamma(t)$ is a parametrization of $|z|=3$ then $tmapstoomegagamma(t)$ is also a parametrization.)
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
add a comment |Â
up vote
2
down vote
Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=infty$, where $displaystyle f(z)=frac1z^3(z^10-2)$ and then apply Cauchy's Residue Theorem.
First line says that, If $f$ has poles at $z=a_i(i=1,2,cdots ,n)$ then $displaystyle sum_i=1^n textRes(f;a_i) +textRes(f;infty)=0$.
answered Jul 15 at 16:24
Empty
7,81942154
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
1
Read the previous line again
– Empty
Jul 15 at 16:29
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
add a comment |Â
up vote
0
down vote
Actually, that would mean $11$ residues, but since $operatornameResleft(frac1z^3(z^10-2),0right)=0$…
On the other hand, if $z_k=sqrt[10]2e^frackpi i5$ ($kin0,1,ldots,9$), thenbeginalignoperatornameResleft(frac1z^3(z^10-2),z_kright)&=frac1z_k^310z_k^9\&=frac110z_k^12\&=fracsqrt[10]2^-1210expleft(-frac12kpi i5right)\&=fracsqrt[10]2^-1210expleft(-frac12pi i5right)^k.endalignSo, what you have to sum is a gemetric progression.
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Since there are no singularities outside $|z|=2^1/10$, the Cauchy Integral Theorem says that
$$
int_zfracmathrmdzz^3(z^10-2)
=int_zfracmathrmdzz^3(z^10-2)
$$
for all $rgt2^1/10$. As $rtoinfty$, the integral on the right obviously vanishes.
answered Jul 15 at 17:13
robjohn♦
258k26297612
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
1
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
add a comment |Â
up vote
4
down vote
Since there are no singularities outside $|z|=2^1/10$, the Cauchy Integral Theorem says that
$$
int_zfracmathrmdzz^3(z^10-2)
=int_zfracmathrmdzz^3(z^10-2)
$$
for all $rgt2^1/10$. As $rtoinfty$, the integral on the right obviously vanishes.
answered Jul 15 at 17:13
robjohn♦
258k26297612
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
1
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Since there are no singularities outside $|z|=2^1/10$, the Cauchy Integral Theorem says that
$$
int_zfracmathrmdzz^3(z^10-2)
=int_zfracmathrmdzz^3(z^10-2)
$$
for all $rgt2^1/10$. As $rtoinfty$, the integral on the right obviously vanishes.
answered Jul 15 at 17:13
robjohn♦
258k26297612
Since there are no singularities outside $|z|=2^1/10$, the Cauchy Integral Theorem says that
$$
int_zfracmathrmdzz^3(z^10-2)
=int_zfracmathrmdzz^3(z^10-2)
$$
for all $rgt2^1/10$. As $rtoinfty$, the integral on the right obviously vanishes.
answered Jul 15 at 17:13
robjohn♦
258k26297612
answered Jul 15 at 17:13
robjohn♦
258k26297612
answered Jul 15 at 17:13
robjohn♦
258k26297612
answered Jul 15 at 17:13
robjohn♦
258k26297612
258k26297612
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
1
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
add a comment |Â
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
1
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
Why can’t we do the same with $f(z)=fracdzz^2+1$ at $|z|=2$
– gbox
Jul 15 at 17:27
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
@gbox: We can. Why do you think we can't?
– robjohn♦
Jul 15 at 18:18
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
So $int_=2fracdxz^2+1=0$ and not $2pi i (res(f,i)+res(f,-i))$?
– gbox
Jul 15 at 18:34
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
the residue of $frac1z^2+1$ at $i$ is $-frac i2$. The residue at $-i$ is $frac i2$, so the sum of the residues is $0$.
– robjohn♦
Jul 15 at 19:17
1
1
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
@gbox If you calculate those residues and that doesn't make you wish you'd calculated them before making that comment then you calculated them wrong.
– David C. Ullrich
Jul 15 at 19:17
add a comment |Â
up vote
3
down vote
This is easy by symmetry, without calculating any residues.
Say $f(z)=frac1z^3(z^10-2)$ and let $omega=e^2pi i/10$. Note that
$$f(omega z)=omega^-3f(z).$$It follows that the integral is $0$.
Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=int_zf(omega z),d(omega z)
=int_zomega^-3f(z)omega,dz=omega^-2I;$$since $omega^-2ne1$ this shows $I=0$.
(The "informal" part is the first equality above. To justify it, note that if $tmapstogamma(t)$ is a parametrization of $|z|=3$ then $tmapstoomegagamma(t)$ is also a parametrization.)
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
add a comment |Â
up vote
3
down vote
This is easy by symmetry, without calculating any residues.
Say $f(z)=frac1z^3(z^10-2)$ and let $omega=e^2pi i/10$. Note that
$$f(omega z)=omega^-3f(z).$$It follows that the integral is $0$.
Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=int_zf(omega z),d(omega z)
=int_zomega^-3f(z)omega,dz=omega^-2I;$$since $omega^-2ne1$ this shows $I=0$.
(The "informal" part is the first equality above. To justify it, note that if $tmapstogamma(t)$ is a parametrization of $|z|=3$ then $tmapstoomegagamma(t)$ is also a parametrization.)
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is easy by symmetry, without calculating any residues.
Say $f(z)=frac1z^3(z^10-2)$ and let $omega=e^2pi i/10$. Note that
$$f(omega z)=omega^-3f(z).$$It follows that the integral is $0$.
Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=int_zf(omega z),d(omega z)
=int_zomega^-3f(z)omega,dz=omega^-2I;$$since $omega^-2ne1$ this shows $I=0$.
(The "informal" part is the first equality above. To justify it, note that if $tmapstogamma(t)$ is a parametrization of $|z|=3$ then $tmapstoomegagamma(t)$ is also a parametrization.)
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
This is easy by symmetry, without calculating any residues.
Say $f(z)=frac1z^3(z^10-2)$ and let $omega=e^2pi i/10$. Note that
$$f(omega z)=omega^-3f(z).$$It follows that the integral is $0$.
Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=int_zf(omega z),d(omega z)
=int_zomega^-3f(z)omega,dz=omega^-2I;$$since $omega^-2ne1$ this shows $I=0$.
(The "informal" part is the first equality above. To justify it, note that if $tmapstogamma(t)$ is a parametrization of $|z|=3$ then $tmapstoomegagamma(t)$ is also a parametrization.)
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
edited Jul 15 at 17:30
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
answered Jul 15 at 17:16
David C. Ullrich
54.4k33583
54.4k33583
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=infty$, where $displaystyle f(z)=frac1z^3(z^10-2)$ and then apply Cauchy's Residue Theorem.
First line says that, If $f$ has poles at $z=a_i(i=1,2,cdots ,n)$ then $displaystyle sum_i=1^n textRes(f;a_i) +textRes(f;infty)=0$.
answered Jul 15 at 16:24
Empty
7,81942154
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
1
Read the previous line again
– Empty
Jul 15 at 16:29
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
add a comment |Â
up vote
2
down vote
Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=infty$, where $displaystyle f(z)=frac1z^3(z^10-2)$ and then apply Cauchy's Residue Theorem.
First line says that, If $f$ has poles at $z=a_i(i=1,2,cdots ,n)$ then $displaystyle sum_i=1^n textRes(f;a_i) +textRes(f;infty)=0$.
answered Jul 15 at 16:24
Empty
7,81942154
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
1
Read the previous line again
– Empty
Jul 15 at 16:29
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=infty$, where $displaystyle f(z)=frac1z^3(z^10-2)$ and then apply Cauchy's Residue Theorem.
First line says that, If $f$ has poles at $z=a_i(i=1,2,cdots ,n)$ then $displaystyle sum_i=1^n textRes(f;a_i) +textRes(f;infty)=0$.
answered Jul 15 at 16:24
Empty
7,81942154
Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=infty$, where $displaystyle f(z)=frac1z^3(z^10-2)$ and then apply Cauchy's Residue Theorem.
First line says that, If $f$ has poles at $z=a_i(i=1,2,cdots ,n)$ then $displaystyle sum_i=1^n textRes(f;a_i) +textRes(f;infty)=0$.
answered Jul 15 at 16:24
Empty
7,81942154
edited Jul 25 at 15:35
answered Jul 15 at 16:24
Empty
7,81942154
answered Jul 15 at 16:24
Empty
7,81942154
answered Jul 15 at 16:24
Empty
7,81942154
7,81942154
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
1
Read the previous line again
– Empty
Jul 15 at 16:29
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
add a comment |Â
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
1
Read the previous line again
– Empty
Jul 15 at 16:29
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
Why "it is sufficient to evaluate the singularity of $f$ at $z=infty$" is there a theorem about it?
– gbox
Jul 15 at 16:28
1
1
Read the previous line again
– Empty
Jul 15 at 16:29
Read the previous line again
– Empty
Jul 15 at 16:29
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
@gbox See the edited answer.
– Empty
Jul 15 at 16:35
add a comment |Â
up vote
0
down vote
Actually, that would mean $11$ residues, but since $operatornameResleft(frac1z^3(z^10-2),0right)=0$…
On the other hand, if $z_k=sqrt[10]2e^frackpi i5$ ($kin0,1,ldots,9$), thenbeginalignoperatornameResleft(frac1z^3(z^10-2),z_kright)&=frac1z_k^310z_k^9\&=frac110z_k^12\&=fracsqrt[10]2^-1210expleft(-frac12kpi i5right)\&=fracsqrt[10]2^-1210expleft(-frac12pi i5right)^k.endalignSo, what you have to sum is a gemetric progression.
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
Actually, that would mean $11$ residues, but since $operatornameResleft(frac1z^3(z^10-2),0right)=0$…
On the other hand, if $z_k=sqrt[10]2e^frackpi i5$ ($kin0,1,ldots,9$), thenbeginalignoperatornameResleft(frac1z^3(z^10-2),z_kright)&=frac1z_k^310z_k^9\&=frac110z_k^12\&=fracsqrt[10]2^-1210expleft(-frac12kpi i5right)\&=fracsqrt[10]2^-1210expleft(-frac12pi i5right)^k.endalignSo, what you have to sum is a gemetric progression.
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Actually, that would mean $11$ residues, but since $operatornameResleft(frac1z^3(z^10-2),0right)=0$…
On the other hand, if $z_k=sqrt[10]2e^frackpi i5$ ($kin0,1,ldots,9$), thenbeginalignoperatornameResleft(frac1z^3(z^10-2),z_kright)&=frac1z_k^310z_k^9\&=frac110z_k^12\&=fracsqrt[10]2^-1210expleft(-frac12kpi i5right)\&=fracsqrt[10]2^-1210expleft(-frac12pi i5right)^k.endalignSo, what you have to sum is a gemetric progression.
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
Actually, that would mean $11$ residues, but since $operatornameResleft(frac1z^3(z^10-2),0right)=0$…
On the other hand, if $z_k=sqrt[10]2e^frackpi i5$ ($kin0,1,ldots,9$), thenbeginalignoperatornameResleft(frac1z^3(z^10-2),z_kright)&=frac1z_k^310z_k^9\&=frac110z_k^12\&=fracsqrt[10]2^-1210expleft(-frac12kpi i5right)\&=fracsqrt[10]2^-1210expleft(-frac12pi i5right)^k.endalignSo, what you have to sum is a gemetric progression.
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
answered Jul 15 at 16:33
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
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