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What happens if I have an essential singularity and a pole for the same $z$?
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for instance $$dfracsin(dfrac1z)z$$
$z=0$ is a pole for the denominator
but $z=0$ is an essential singularity for the numerator too.
So how does it work ? i have two residues ? or it's different some way ?
complex-analysis trigonometry residue-calculus singularity
edited Jul 29 at 11:29
Lolita
52318
asked Jul 29 at 10:20
xmaionx
52
add a comment |Â
up vote
0
down vote
favorite
for instance $$dfracsin(dfrac1z)z$$
$z=0$ is a pole for the denominator
but $z=0$ is an essential singularity for the numerator too.
So how does it work ? i have two residues ? or it's different some way ?
complex-analysis trigonometry residue-calculus singularity
edited Jul 29 at 11:29
Lolita
52318
asked Jul 29 at 10:20
xmaionx
52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
for instance $$dfracsin(dfrac1z)z$$
$z=0$ is a pole for the denominator
but $z=0$ is an essential singularity for the numerator too.
So how does it work ? i have two residues ? or it's different some way ?
complex-analysis trigonometry residue-calculus singularity
edited Jul 29 at 11:29
Lolita
52318
asked Jul 29 at 10:20
xmaionx
52
for instance $$dfracsin(dfrac1z)z$$
$z=0$ is a pole for the denominator
but $z=0$ is an essential singularity for the numerator too.
So how does it work ? i have two residues ? or it's different some way ?
complex-analysis trigonometry residue-calculus singularity
edited Jul 29 at 11:29
Lolita
52318
asked Jul 29 at 10:20
xmaionx
52
edited Jul 29 at 11:29
Lolita
52318
edited Jul 29 at 11:29
Lolita
52318
edited Jul 29 at 11:29
Lolita
52318
52318
asked Jul 29 at 10:20
xmaionx
52
asked Jul 29 at 10:20
xmaionx
52
asked Jul 29 at 10:20
xmaionx
52
52
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$z=0$ is an essential singularity for your function too. with the expansion
$sin z=sum_n=0frac(-1)^nz^2n+1(2n+1)!$ we see
$$frac1zsin frac1z=sum_n=0frac(-1)^n(2n+1)!z^2n+2=frac1z^2-frac13!z^4+frac15!z^6-...$$
Can you find the residue?
answered Jul 29 at 10:25
Lolita
52318
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
thank you accepting this
– Lolita
Jul 29 at 11:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$z=0$ is an essential singularity for your function too. with the expansion
$sin z=sum_n=0frac(-1)^nz^2n+1(2n+1)!$ we see
$$frac1zsin frac1z=sum_n=0frac(-1)^n(2n+1)!z^2n+2=frac1z^2-frac13!z^4+frac15!z^6-...$$
Can you find the residue?
answered Jul 29 at 10:25
Lolita
52318
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
thank you accepting this
– Lolita
Jul 29 at 11:00
add a comment |Â
up vote
1
down vote
accepted
$z=0$ is an essential singularity for your function too. with the expansion
$sin z=sum_n=0frac(-1)^nz^2n+1(2n+1)!$ we see
$$frac1zsin frac1z=sum_n=0frac(-1)^n(2n+1)!z^2n+2=frac1z^2-frac13!z^4+frac15!z^6-...$$
Can you find the residue?
answered Jul 29 at 10:25
Lolita
52318
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
thank you accepting this
– Lolita
Jul 29 at 11:00
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$z=0$ is an essential singularity for your function too. with the expansion
$sin z=sum_n=0frac(-1)^nz^2n+1(2n+1)!$ we see
$$frac1zsin frac1z=sum_n=0frac(-1)^n(2n+1)!z^2n+2=frac1z^2-frac13!z^4+frac15!z^6-...$$
Can you find the residue?
answered Jul 29 at 10:25
Lolita
52318
$z=0$ is an essential singularity for your function too. with the expansion
$sin z=sum_n=0frac(-1)^nz^2n+1(2n+1)!$ we see
$$frac1zsin frac1z=sum_n=0frac(-1)^n(2n+1)!z^2n+2=frac1z^2-frac13!z^4+frac15!z^6-...$$
Can you find the residue?
answered Jul 29 at 10:25
Lolita
52318
edited Jul 29 at 10:30
answered Jul 29 at 10:25
Lolita
52318
answered Jul 29 at 10:25
Lolita
52318
answered Jul 29 at 10:25
Lolita
52318
52318
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
thank you accepting this
– Lolita
Jul 29 at 11:00
add a comment |Â
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
thank you accepting this
– Lolita
Jul 29 at 11:00
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
ah maybe because the series of sin(1/z) has infiniely z at the denominator , so if i multiply by 1/z we will continue to have infinitlky z at the denominator ( also with a major grade) is this the reason ?
– xmaionx
Jul 29 at 10:30
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
the residue is the coefficient of the one with lower grade ? so is 1 ?
– xmaionx
Jul 29 at 10:41
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
ahh so it would be the 1/z coefficient ?
– xmaionx
Jul 29 at 10:45
thank you accepting this
– Lolita
Jul 29 at 11:00
thank you accepting this
– Lolita
Jul 29 at 11:00
add a comment |Â
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