There is no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$ [on hold]Is the function really analytic at $infty$?Find a single-valued analytic branch of $sqrtz^2-1$ in $mathbbC backslash [-1,1]$.Confusion with definition of poleA question about how to take a -1 out of mutiple-valued analytic function $z^alpha $ $0<alpha<1 $Definition of a branch point (Ahlfors)Branch cuts of $log z$Describe all real-valued functions which are analytic on $mathbbC$Exercise problem on existence of analytic function on a given domain which satisfies certain conditionsWhat is the integral division if $lim_z rightarrow a f(z)=0$?For which subsets of $mathbbC$ can a non-constant analytic function have an accumulation point of zeros?
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There is no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$ [on hold]
Is the function really analytic at $infty$?Find a single-valued analytic branch of $sqrtz^2-1$ in $mathbbC backslash [-1,1]$.Confusion with definition of poleA question about how to take a -1 out of mutiple-valued analytic function $z^alpha $ $0<alpha<1 $Definition of a branch point (Ahlfors)Branch cuts of $log z$Describe all real-valued functions which are analytic on $mathbbC$Exercise problem on existence of analytic function on a given domain which satisfies certain conditionsWhat is the integral division if $lim_z rightarrow a f(z)=0$?For which subsets of $mathbbC$ can a non-constant analytic function have an accumulation point of zeros?
$begingroup$
Ahlfors' Complex Analysis page 128:
For a more detailed discussion of isolated singularities, we consider the conditions (1) $lim_zto a|z-a|^alpha|f(z)|=0$, (2) $lim_zto a|z-a|^alpha|f(z)|=infty$, for real values of $alpha$. If (1) hold for a certain $alpha$, then it holds for all larger $alpha$, and hence for some integer $m$. Then $(z-a)^mf(z)$ has a removable singularity and vanishes for $z=a$. Either $f(z)$ is identically zero, in which case (1) holds for all $alpha$, or $(z-a)^mf(z)$ has a zero of finite order $k$. In the latter case it follows at once that (1) holds for all $alpha>h=m-k$, while (2) holds for all $alpha<h$. Assume now that (2) holds for some $alpha$; then it holds for all smaller $alpha$, and hence for some integer $n$. The function $(z-a)^nf(z)$ has a pole of finite order $l$, and setting $h=n+l$ we find again that (1) holds for $alpha>h$ and (2) for $alpha<h$. The discussion shows that there are three possibilities: (i) condition (1) holds for all $alpha$, and $f(z)$ vanishes identically; (ii) there exists an integer $h$ such that (1) holds for $alpha>h$ and (2) for $alpha<h$; (iii) neither (1) nor (2) holds for any $alpha$.
Case (i) is uninteresting. In case (ii) $h$ may be called the algebraic order of $f(z)$ at $alpha$. It is positive in case of a pole, negative in case of a zero, and zero if $f(z)$ is analytic but $ne0$ at $a$. The remarkable thing is that the order is always an integer; there is no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$.
Why is there no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$?
complex-analysis
edited yesterday
Parcly Taxel
43.7k1375104
asked 2 days ago
user398843user398843
689216
$endgroup$
put on hold as off-topic by Theo Bendit, José Carlos Santos, mrtaurho, Peter Foreman, Eevee Trainer 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Theo Bendit, José Carlos Santos, Peter Foreman, Eevee Trainer
add a comment |
$begingroup$
Ahlfors' Complex Analysis page 128:
For a more detailed discussion of isolated singularities, we consider the conditions (1) $lim_zto a|z-a|^alpha|f(z)|=0$, (2) $lim_zto a|z-a|^alpha|f(z)|=infty$, for real values of $alpha$. If (1) hold for a certain $alpha$, then it holds for all larger $alpha$, and hence for some integer $m$. Then $(z-a)^mf(z)$ has a removable singularity and vanishes for $z=a$. Either $f(z)$ is identically zero, in which case (1) holds for all $alpha$, or $(z-a)^mf(z)$ has a zero of finite order $k$. In the latter case it follows at once that (1) holds for all $alpha>h=m-k$, while (2) holds for all $alpha<h$. Assume now that (2) holds for some $alpha$; then it holds for all smaller $alpha$, and hence for some integer $n$. The function $(z-a)^nf(z)$ has a pole of finite order $l$, and setting $h=n+l$ we find again that (1) holds for $alpha>h$ and (2) for $alpha<h$. The discussion shows that there are three possibilities: (i) condition (1) holds for all $alpha$, and $f(z)$ vanishes identically; (ii) there exists an integer $h$ such that (1) holds for $alpha>h$ and (2) for $alpha<h$; (iii) neither (1) nor (2) holds for any $alpha$.
Case (i) is uninteresting. In case (ii) $h$ may be called the algebraic order of $f(z)$ at $alpha$. It is positive in case of a pole, negative in case of a zero, and zero if $f(z)$ is analytic but $ne0$ at $a$. The remarkable thing is that the order is always an integer; there is no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$.
Why is there no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$?
complex-analysis
edited yesterday
Parcly Taxel
43.7k1375104
asked 2 days ago
user398843user398843
689216
$endgroup$
put on hold as off-topic by Theo Bendit, José Carlos Santos, mrtaurho, Peter Foreman, Eevee Trainer 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Theo Bendit, José Carlos Santos, Peter Foreman, Eevee Trainer
$begingroup$
Does Ahlfors specify exactly what tending to $0$ or $infty$ "like" another function means explicitly? I could take a guess, but a bit more context would be good.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@TheoBendit I upload a new picture that includes a previous paragraph. I am not sure why he points out that.
$endgroup$
– user398843
2 days ago
$begingroup$
The possible reason is that a non-integer power is always multivalued in the domain containing its zero or pole.
$endgroup$
– user
2 days ago
add a comment |
$begingroup$
Ahlfors' Complex Analysis page 128:
For a more detailed discussion of isolated singularities, we consider the conditions (1) $lim_zto a|z-a|^alpha|f(z)|=0$, (2) $lim_zto a|z-a|^alpha|f(z)|=infty$, for real values of $alpha$. If (1) hold for a certain $alpha$, then it holds for all larger $alpha$, and hence for some integer $m$. Then $(z-a)^mf(z)$ has a removable singularity and vanishes for $z=a$. Either $f(z)$ is identically zero, in which case (1) holds for all $alpha$, or $(z-a)^mf(z)$ has a zero of finite order $k$. In the latter case it follows at once that (1) holds for all $alpha>h=m-k$, while (2) holds for all $alpha<h$. Assume now that (2) holds for some $alpha$; then it holds for all smaller $alpha$, and hence for some integer $n$. The function $(z-a)^nf(z)$ has a pole of finite order $l$, and setting $h=n+l$ we find again that (1) holds for $alpha>h$ and (2) for $alpha<h$. The discussion shows that there are three possibilities: (i) condition (1) holds for all $alpha$, and $f(z)$ vanishes identically; (ii) there exists an integer $h$ such that (1) holds for $alpha>h$ and (2) for $alpha<h$; (iii) neither (1) nor (2) holds for any $alpha$.
Case (i) is uninteresting. In case (ii) $h$ may be called the algebraic order of $f(z)$ at $alpha$. It is positive in case of a pole, negative in case of a zero, and zero if $f(z)$ is analytic but $ne0$ at $a$. The remarkable thing is that the order is always an integer; there is no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$.
Why is there no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$?
complex-analysis
edited yesterday
Parcly Taxel
43.7k1375104
asked 2 days ago
user398843user398843
689216
$endgroup$
Ahlfors' Complex Analysis page 128:
For a more detailed discussion of isolated singularities, we consider the conditions (1) $lim_zto a|z-a|^alpha|f(z)|=0$, (2) $lim_zto a|z-a|^alpha|f(z)|=infty$, for real values of $alpha$. If (1) hold for a certain $alpha$, then it holds for all larger $alpha$, and hence for some integer $m$. Then $(z-a)^mf(z)$ has a removable singularity and vanishes for $z=a$. Either $f(z)$ is identically zero, in which case (1) holds for all $alpha$, or $(z-a)^mf(z)$ has a zero of finite order $k$. In the latter case it follows at once that (1) holds for all $alpha>h=m-k$, while (2) holds for all $alpha<h$. Assume now that (2) holds for some $alpha$; then it holds for all smaller $alpha$, and hence for some integer $n$. The function $(z-a)^nf(z)$ has a pole of finite order $l$, and setting $h=n+l$ we find again that (1) holds for $alpha>h$ and (2) for $alpha<h$. The discussion shows that there are three possibilities: (i) condition (1) holds for all $alpha$, and $f(z)$ vanishes identically; (ii) there exists an integer $h$ such that (1) holds for $alpha>h$ and (2) for $alpha<h$; (iii) neither (1) nor (2) holds for any $alpha$.
Case (i) is uninteresting. In case (ii) $h$ may be called the algebraic order of $f(z)$ at $alpha$. It is positive in case of a pole, negative in case of a zero, and zero if $f(z)$ is analytic but $ne0$ at $a$. The remarkable thing is that the order is always an integer; there is no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$.
Why is there no single-valued analytic function which tends to $0$ or $infty$ like a fractional power of $|z-a|$?
complex-analysis
complex-analysis
edited yesterday
Parcly Taxel
43.7k1375104
asked 2 days ago
user398843user398843
689216
edited yesterday
Parcly Taxel
43.7k1375104
asked 2 days ago
user398843user398843
689216
edited yesterday
Parcly Taxel
43.7k1375104
edited yesterday
Parcly Taxel
43.7k1375104
edited yesterday
Parcly Taxel
43.7k1375104
43.7k1375104
asked 2 days ago
user398843user398843
689216
asked 2 days ago
user398843user398843
689216
asked 2 days ago
user398843user398843
689216
689216
put on hold as off-topic by Theo Bendit, José Carlos Santos, mrtaurho, Peter Foreman, Eevee Trainer 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Theo Bendit, José Carlos Santos, Peter Foreman, Eevee Trainer
put on hold as off-topic by Theo Bendit, José Carlos Santos, mrtaurho, Peter Foreman, Eevee Trainer 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Theo Bendit, José Carlos Santos, Peter Foreman, Eevee Trainer
$begingroup$
Does Ahlfors specify exactly what tending to $0$ or $infty$ "like" another function means explicitly? I could take a guess, but a bit more context would be good.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@TheoBendit I upload a new picture that includes a previous paragraph. I am not sure why he points out that.
$endgroup$
– user398843
2 days ago
$begingroup$
The possible reason is that a non-integer power is always multivalued in the domain containing its zero or pole.
$endgroup$
– user
2 days ago
add a comment |
$begingroup$
Does Ahlfors specify exactly what tending to $0$ or $infty$ "like" another function means explicitly? I could take a guess, but a bit more context would be good.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@TheoBendit I upload a new picture that includes a previous paragraph. I am not sure why he points out that.
$endgroup$
– user398843
2 days ago
$begingroup$
The possible reason is that a non-integer power is always multivalued in the domain containing its zero or pole.
$endgroup$
– user
2 days ago
$begingroup$
Does Ahlfors specify exactly what tending to $0$ or $infty$ "like" another function means explicitly? I could take a guess, but a bit more context would be good.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
Does Ahlfors specify exactly what tending to $0$ or $infty$ "like" another function means explicitly? I could take a guess, but a bit more context would be good.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@TheoBendit I upload a new picture that includes a previous paragraph. I am not sure why he points out that.
$endgroup$
– user398843
2 days ago
$begingroup$
@TheoBendit I upload a new picture that includes a previous paragraph. I am not sure why he points out that.
$endgroup$
– user398843
2 days ago
$begingroup$
The possible reason is that a non-integer power is always multivalued in the domain containing its zero or pole.
$endgroup$
– user
2 days ago
$begingroup$
The possible reason is that a non-integer power is always multivalued in the domain containing its zero or pole.
$endgroup$
– user
2 days ago
add a comment |
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$begingroup$
Does Ahlfors specify exactly what tending to $0$ or $infty$ "like" another function means explicitly? I could take a guess, but a bit more context would be good.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@TheoBendit I upload a new picture that includes a previous paragraph. I am not sure why he points out that.
$endgroup$
– user398843
2 days ago
$begingroup$
The possible reason is that a non-integer power is always multivalued in the domain containing its zero or pole.
$endgroup$
– user
2 days ago