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Intuition contradicts fact: $fracz1+z^2$ has a pole at $infty$?
Clash Royale CLAN TAG#URR8PPP
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Consider the function $$f(z)=fracz1+z^2$$
Clearly, $$lim_ztoinftyf(z)=0$$ from any directions.
However, it still has a pole at infinity as the residue there is non zero:
$$operatorname*Res_z=0frac1z^2frac1/z1+1/z^2ne 0$$
The function does not blow up to infinity but still has a pole there: how to reconcile? Is there an intuitive explanation for this phenomenon?
complex-analysis
asked Aug 2 at 9:09
Szeto
3,8431421
add a comment |Â
up vote
3
down vote
favorite
Consider the function $$f(z)=fracz1+z^2$$
Clearly, $$lim_ztoinftyf(z)=0$$ from any directions.
However, it still has a pole at infinity as the residue there is non zero:
$$operatorname*Res_z=0frac1z^2frac1/z1+1/z^2ne 0$$
The function does not blow up to infinity but still has a pole there: how to reconcile? Is there an intuitive explanation for this phenomenon?
complex-analysis
asked Aug 2 at 9:09
Szeto
3,8431421
Why is there a $1/z$ in the numerator?
– Mattos
Aug 2 at 9:12
2
$f(1/z)=(frac1z)/(1+frac1z^2)=fraczz^2+1$ and it is analytic at $0$.
– daruma
Aug 2 at 9:13
I think a pole at $infty$ is hard to grasp "intuitively"
– Peter
Aug 2 at 9:16
Where are you getting that formula from?
– Lorenzo
Aug 2 at 9:25
I think a function can have a nonzero residue at infinity without having a pole at infinity. Perhaps this previous answer sheds some light.
– Rahul
Aug 2 at 9:25
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider the function $$f(z)=fracz1+z^2$$
Clearly, $$lim_ztoinftyf(z)=0$$ from any directions.
However, it still has a pole at infinity as the residue there is non zero:
$$operatorname*Res_z=0frac1z^2frac1/z1+1/z^2ne 0$$
The function does not blow up to infinity but still has a pole there: how to reconcile? Is there an intuitive explanation for this phenomenon?
complex-analysis
asked Aug 2 at 9:09
Szeto
3,8431421
Consider the function $$f(z)=fracz1+z^2$$
Clearly, $$lim_ztoinftyf(z)=0$$ from any directions.
However, it still has a pole at infinity as the residue there is non zero:
$$operatorname*Res_z=0frac1z^2frac1/z1+1/z^2ne 0$$
The function does not blow up to infinity but still has a pole there: how to reconcile? Is there an intuitive explanation for this phenomenon?
complex-analysis
asked Aug 2 at 9:09
Szeto
3,8431421
asked Aug 2 at 9:09
Szeto
3,8431421
asked Aug 2 at 9:09
Szeto
3,8431421
asked Aug 2 at 9:09
Szeto
3,8431421
3,8431421
Why is there a $1/z$ in the numerator?
– Mattos
Aug 2 at 9:12
2
$f(1/z)=(frac1z)/(1+frac1z^2)=fraczz^2+1$ and it is analytic at $0$.
– daruma
Aug 2 at 9:13
I think a pole at $infty$ is hard to grasp "intuitively"
– Peter
Aug 2 at 9:16
Where are you getting that formula from?
– Lorenzo
Aug 2 at 9:25
I think a function can have a nonzero residue at infinity without having a pole at infinity. Perhaps this previous answer sheds some light.
– Rahul
Aug 2 at 9:25
add a comment |Â
Why is there a $1/z$ in the numerator?
– Mattos
Aug 2 at 9:12
2
$f(1/z)=(frac1z)/(1+frac1z^2)=fraczz^2+1$ and it is analytic at $0$.
– daruma
Aug 2 at 9:13
I think a pole at $infty$ is hard to grasp "intuitively"
– Peter
Aug 2 at 9:16
Where are you getting that formula from?
– Lorenzo
Aug 2 at 9:25
I think a function can have a nonzero residue at infinity without having a pole at infinity. Perhaps this previous answer sheds some light.
– Rahul
Aug 2 at 9:25
Why is there a $1/z$ in the numerator?
– Mattos
Aug 2 at 9:12
Why is there a $1/z$ in the numerator?
– Mattos
Aug 2 at 9:12
2
2
$f(1/z)=(frac1z)/(1+frac1z^2)=fraczz^2+1$ and it is analytic at $0$.
– daruma
Aug 2 at 9:13
$f(1/z)=(frac1z)/(1+frac1z^2)=fraczz^2+1$ and it is analytic at $0$.
– daruma
Aug 2 at 9:13
I think a pole at $infty$ is hard to grasp "intuitively"
– Peter
Aug 2 at 9:16
I think a pole at $infty$ is hard to grasp "intuitively"
– Peter
Aug 2 at 9:16
Where are you getting that formula from?
– Lorenzo
Aug 2 at 9:25
Where are you getting that formula from?
– Lorenzo
Aug 2 at 9:25
I think a function can have a nonzero residue at infinity without having a pole at infinity. Perhaps this previous answer sheds some light.
– Rahul
Aug 2 at 9:25
I think a function can have a nonzero residue at infinity without having a pole at infinity. Perhaps this previous answer sheds some light.
– Rahul
Aug 2 at 9:25
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
Let $w=1/z$ then $$f(z) = g(w)=frac w1+w^2$$
$f$ has a pole at $infty $ if and only if g has a pole at $w=0$ which is not the case.
Thus $f$ does not have a pole at $infty$
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
add a comment |Â
up vote
1
down vote
Functions have poles; differentials have poles.
The function
$$frac z1+z^2$$
has no pole at $infty$ (it has a simple zero there).
The differential
$$frac z,dz1+z^2$$
has a simple pole at $infty$. To see this, set $w=1/z$. Then
$$frac z1+z^2=frac w1+w^2$$
and
$$frac z,dz1+z^2
=-frac dww(1+w^2).$$
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
2
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
add a comment |Â
up vote
0
down vote
According to https://en.wikipedia.org/wiki/Zeros_and_poles the point at infinity is a pole of order n for $f(z)$ if for $n > 0$ $$lim_zto inftyfracf(z)z^n$$ exists and is a non zero complex number. At this case $$dfracf(z)z^n=dfracz^1-n1+z^2=dfrac1z^n-1(1+z^2)$$if $nge 1$ then the fraction tends to $0$ as $z$ tends to $infty$ but we can say that
$infty$ is a 1st-order zero for $dfracz1+z^2$
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
Over the complex plane, every rational function factors into linear factors. Some factors are in the numerator and some in the denominator. Consider any linear factor $ z-c. $ It clearly has a simple zero at $ z=c. $ We consider that it also has a simple pole at infinity to balance the simple zero. That is, the number of zeros and poles are equal. This makes sense for several reasons. If we extend this idea to all the linear factors of any rational function, then the number of zeros and poles of it are requal, up to multiplicity.
In the case of your function $ f(z) = z/(1+z^2) = z/((z+i)(z-i)). $ The $ z $ gives a zero at $0$ and a pole at $infty.$ The two factors in the denominator contribute a double zero at $infty$ along with poles at $i$ and $-i.$ Thus, in total, there are simple zeros at $0$ and $infty$, and simple poles at $i$ and $-i.$ You can see this in another way by using partial fractions. Here
$ f(x) = frac12(frac1x+i+frac1x-i). $ Again, there are simple poles at $i$ and $-i.$ Check that the simple zeros are at $0$ and $infty.$
answered Aug 2 at 11:30
Somos
10.9k1831
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $w=1/z$ then $$f(z) = g(w)=frac w1+w^2$$
$f$ has a pole at $infty $ if and only if g has a pole at $w=0$ which is not the case.
Thus $f$ does not have a pole at $infty$
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
add a comment |Â
up vote
3
down vote
Let $w=1/z$ then $$f(z) = g(w)=frac w1+w^2$$
$f$ has a pole at $infty $ if and only if g has a pole at $w=0$ which is not the case.
Thus $f$ does not have a pole at $infty$
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $w=1/z$ then $$f(z) = g(w)=frac w1+w^2$$
$f$ has a pole at $infty $ if and only if g has a pole at $w=0$ which is not the case.
Thus $f$ does not have a pole at $infty$
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
Let $w=1/z$ then $$f(z) = g(w)=frac w1+w^2$$
$f$ has a pole at $infty $ if and only if g has a pole at $w=0$ which is not the case.
Thus $f$ does not have a pole at $infty$
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
answered Aug 2 at 9:24
Mohammad Riazi-Kermani
27.1k41851
27.1k41851
add a comment |Â
add a comment |Â
up vote
1
down vote
Functions have poles; differentials have poles.
The function
$$frac z1+z^2$$
has no pole at $infty$ (it has a simple zero there).
The differential
$$frac z,dz1+z^2$$
has a simple pole at $infty$. To see this, set $w=1/z$. Then
$$frac z1+z^2=frac w1+w^2$$
and
$$frac z,dz1+z^2
=-frac dww(1+w^2).$$
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
2
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
add a comment |Â
up vote
1
down vote
Functions have poles; differentials have poles.
The function
$$frac z1+z^2$$
has no pole at $infty$ (it has a simple zero there).
The differential
$$frac z,dz1+z^2$$
has a simple pole at $infty$. To see this, set $w=1/z$. Then
$$frac z1+z^2=frac w1+w^2$$
and
$$frac z,dz1+z^2
=-frac dww(1+w^2).$$
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
2
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Functions have poles; differentials have poles.
The function
$$frac z1+z^2$$
has no pole at $infty$ (it has a simple zero there).
The differential
$$frac z,dz1+z^2$$
has a simple pole at $infty$. To see this, set $w=1/z$. Then
$$frac z1+z^2=frac w1+w^2$$
and
$$frac z,dz1+z^2
=-frac dww(1+w^2).$$
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
Functions have poles; differentials have poles.
The function
$$frac z1+z^2$$
has no pole at $infty$ (it has a simple zero there).
The differential
$$frac z,dz1+z^2$$
has a simple pole at $infty$. To see this, set $w=1/z$. Then
$$frac z1+z^2=frac w1+w^2$$
and
$$frac z,dz1+z^2
=-frac dww(1+w^2).$$
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
answered Aug 2 at 9:31
Lord Shark the Unknown
84.2k950111
84.2k950111
2
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
add a comment |Â
2
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
2
2
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
Is there a typo in the first sentence? It looks like it's intended to be a contrast.
– Peter Taylor
Aug 2 at 10:05
add a comment |Â
up vote
0
down vote
According to https://en.wikipedia.org/wiki/Zeros_and_poles the point at infinity is a pole of order n for $f(z)$ if for $n > 0$ $$lim_zto inftyfracf(z)z^n$$ exists and is a non zero complex number. At this case $$dfracf(z)z^n=dfracz^1-n1+z^2=dfrac1z^n-1(1+z^2)$$if $nge 1$ then the fraction tends to $0$ as $z$ tends to $infty$ but we can say that
$infty$ is a 1st-order zero for $dfracz1+z^2$
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
According to https://en.wikipedia.org/wiki/Zeros_and_poles the point at infinity is a pole of order n for $f(z)$ if for $n > 0$ $$lim_zto inftyfracf(z)z^n$$ exists and is a non zero complex number. At this case $$dfracf(z)z^n=dfracz^1-n1+z^2=dfrac1z^n-1(1+z^2)$$if $nge 1$ then the fraction tends to $0$ as $z$ tends to $infty$ but we can say that
$infty$ is a 1st-order zero for $dfracz1+z^2$
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
According to https://en.wikipedia.org/wiki/Zeros_and_poles the point at infinity is a pole of order n for $f(z)$ if for $n > 0$ $$lim_zto inftyfracf(z)z^n$$ exists and is a non zero complex number. At this case $$dfracf(z)z^n=dfracz^1-n1+z^2=dfrac1z^n-1(1+z^2)$$if $nge 1$ then the fraction tends to $0$ as $z$ tends to $infty$ but we can say that
$infty$ is a 1st-order zero for $dfracz1+z^2$
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
According to https://en.wikipedia.org/wiki/Zeros_and_poles the point at infinity is a pole of order n for $f(z)$ if for $n > 0$ $$lim_zto inftyfracf(z)z^n$$ exists and is a non zero complex number. At this case $$dfracf(z)z^n=dfracz^1-n1+z^2=dfrac1z^n-1(1+z^2)$$if $nge 1$ then the fraction tends to $0$ as $z$ tends to $infty$ but we can say that
$infty$ is a 1st-order zero for $dfracz1+z^2$
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
answered Aug 2 at 9:29
Mostafa Ayaz
8,5183630
8,5183630
add a comment |Â
add a comment |Â
up vote
0
down vote
Over the complex plane, every rational function factors into linear factors. Some factors are in the numerator and some in the denominator. Consider any linear factor $ z-c. $ It clearly has a simple zero at $ z=c. $ We consider that it also has a simple pole at infinity to balance the simple zero. That is, the number of zeros and poles are equal. This makes sense for several reasons. If we extend this idea to all the linear factors of any rational function, then the number of zeros and poles of it are requal, up to multiplicity.
In the case of your function $ f(z) = z/(1+z^2) = z/((z+i)(z-i)). $ The $ z $ gives a zero at $0$ and a pole at $infty.$ The two factors in the denominator contribute a double zero at $infty$ along with poles at $i$ and $-i.$ Thus, in total, there are simple zeros at $0$ and $infty$, and simple poles at $i$ and $-i.$ You can see this in another way by using partial fractions. Here
$ f(x) = frac12(frac1x+i+frac1x-i). $ Again, there are simple poles at $i$ and $-i.$ Check that the simple zeros are at $0$ and $infty.$
answered Aug 2 at 11:30
Somos
10.9k1831
add a comment |Â
up vote
0
down vote
Over the complex plane, every rational function factors into linear factors. Some factors are in the numerator and some in the denominator. Consider any linear factor $ z-c. $ It clearly has a simple zero at $ z=c. $ We consider that it also has a simple pole at infinity to balance the simple zero. That is, the number of zeros and poles are equal. This makes sense for several reasons. If we extend this idea to all the linear factors of any rational function, then the number of zeros and poles of it are requal, up to multiplicity.
In the case of your function $ f(z) = z/(1+z^2) = z/((z+i)(z-i)). $ The $ z $ gives a zero at $0$ and a pole at $infty.$ The two factors in the denominator contribute a double zero at $infty$ along with poles at $i$ and $-i.$ Thus, in total, there are simple zeros at $0$ and $infty$, and simple poles at $i$ and $-i.$ You can see this in another way by using partial fractions. Here
$ f(x) = frac12(frac1x+i+frac1x-i). $ Again, there are simple poles at $i$ and $-i.$ Check that the simple zeros are at $0$ and $infty.$
answered Aug 2 at 11:30
Somos
10.9k1831
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Over the complex plane, every rational function factors into linear factors. Some factors are in the numerator and some in the denominator. Consider any linear factor $ z-c. $ It clearly has a simple zero at $ z=c. $ We consider that it also has a simple pole at infinity to balance the simple zero. That is, the number of zeros and poles are equal. This makes sense for several reasons. If we extend this idea to all the linear factors of any rational function, then the number of zeros and poles of it are requal, up to multiplicity.
In the case of your function $ f(z) = z/(1+z^2) = z/((z+i)(z-i)). $ The $ z $ gives a zero at $0$ and a pole at $infty.$ The two factors in the denominator contribute a double zero at $infty$ along with poles at $i$ and $-i.$ Thus, in total, there are simple zeros at $0$ and $infty$, and simple poles at $i$ and $-i.$ You can see this in another way by using partial fractions. Here
$ f(x) = frac12(frac1x+i+frac1x-i). $ Again, there are simple poles at $i$ and $-i.$ Check that the simple zeros are at $0$ and $infty.$
answered Aug 2 at 11:30
Somos
10.9k1831
Over the complex plane, every rational function factors into linear factors. Some factors are in the numerator and some in the denominator. Consider any linear factor $ z-c. $ It clearly has a simple zero at $ z=c. $ We consider that it also has a simple pole at infinity to balance the simple zero. That is, the number of zeros and poles are equal. This makes sense for several reasons. If we extend this idea to all the linear factors of any rational function, then the number of zeros and poles of it are requal, up to multiplicity.
In the case of your function $ f(z) = z/(1+z^2) = z/((z+i)(z-i)). $ The $ z $ gives a zero at $0$ and a pole at $infty.$ The two factors in the denominator contribute a double zero at $infty$ along with poles at $i$ and $-i.$ Thus, in total, there are simple zeros at $0$ and $infty$, and simple poles at $i$ and $-i.$ You can see this in another way by using partial fractions. Here
$ f(x) = frac12(frac1x+i+frac1x-i). $ Again, there are simple poles at $i$ and $-i.$ Check that the simple zeros are at $0$ and $infty.$
answered Aug 2 at 11:30
Somos
10.9k1831
edited Aug 2 at 23:55
answered Aug 2 at 11:30
Somos
10.9k1831
answered Aug 2 at 11:30
Somos
10.9k1831
answered Aug 2 at 11:30
Somos
10.9k1831
10.9k1831
add a comment |Â
add a comment |Â
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Why is there a $1/z$ in the numerator?
– Mattos
Aug 2 at 9:12
2
$f(1/z)=(frac1z)/(1+frac1z^2)=fraczz^2+1$ and it is analytic at $0$.
– daruma
Aug 2 at 9:13
I think a pole at $infty$ is hard to grasp "intuitively"
– Peter
Aug 2 at 9:16
Where are you getting that formula from?
– Lorenzo
Aug 2 at 9:25
I think a function can have a nonzero residue at infinity without having a pole at infinity. Perhaps this previous answer sheds some light.
– Rahul
Aug 2 at 9:25