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Singularity Type Of $f(z^2+z)$
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$f(z)$ has essential singularity at $z=0$, what type of singularity $f(z^2+z)$ has?
$f(z)$ has essential singularity at $z=0$ so it can be written has $sum_n=0^-infty c_nz^n=c_0+fracc_-1z+fracc_-2z^2+...+$
substituting $z=z^2+z$ will leave all the power to be negative e.g $c_0+fracc_-1z^2+z+fracc_-2(z^2+z)^2+...+$
So $f(z^2+z)$ has essential singularity too, moreover can we even "get rid" of essential singularity?
complex-analysis singularity
asked Jul 16 at 21:06
gbox
5,30851841
add a comment |Â
up vote
3
down vote
favorite
$f(z)$ has essential singularity at $z=0$, what type of singularity $f(z^2+z)$ has?
$f(z)$ has essential singularity at $z=0$ so it can be written has $sum_n=0^-infty c_nz^n=c_0+fracc_-1z+fracc_-2z^2+...+$
substituting $z=z^2+z$ will leave all the power to be negative e.g $c_0+fracc_-1z^2+z+fracc_-2(z^2+z)^2+...+$
So $f(z^2+z)$ has essential singularity too, moreover can we even "get rid" of essential singularity?
complex-analysis singularity
asked Jul 16 at 21:06
gbox
5,30851841
1
Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $infty$), but there will always remain an essential singularity "somewhere" in the plane.
– Clayton
Jul 16 at 21:34
3
Possible duplicate of Singularities of Composition of Functions
– Clayton
Jul 16 at 21:38
Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$.
– GEdgar
Jul 18 at 14:23
@Clayton That question is different: it asks about $g circ f$ where $f$ has a(n essential) singularity. This question is about $f circ g$ (with $g(z) = z^2 + z$).
– Strants
Jul 23 at 13:51
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$f(z)$ has essential singularity at $z=0$, what type of singularity $f(z^2+z)$ has?
$f(z)$ has essential singularity at $z=0$ so it can be written has $sum_n=0^-infty c_nz^n=c_0+fracc_-1z+fracc_-2z^2+...+$
substituting $z=z^2+z$ will leave all the power to be negative e.g $c_0+fracc_-1z^2+z+fracc_-2(z^2+z)^2+...+$
So $f(z^2+z)$ has essential singularity too, moreover can we even "get rid" of essential singularity?
complex-analysis singularity
asked Jul 16 at 21:06
gbox
5,30851841
$f(z)$ has essential singularity at $z=0$, what type of singularity $f(z^2+z)$ has?
$f(z)$ has essential singularity at $z=0$ so it can be written has $sum_n=0^-infty c_nz^n=c_0+fracc_-1z+fracc_-2z^2+...+$
substituting $z=z^2+z$ will leave all the power to be negative e.g $c_0+fracc_-1z^2+z+fracc_-2(z^2+z)^2+...+$
So $f(z^2+z)$ has essential singularity too, moreover can we even "get rid" of essential singularity?
complex-analysis singularity
asked Jul 16 at 21:06
gbox
5,30851841
asked Jul 16 at 21:06
gbox
5,30851841
asked Jul 16 at 21:06
gbox
5,30851841
asked Jul 16 at 21:06
gbox
5,30851841
5,30851841
1
Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $infty$), but there will always remain an essential singularity "somewhere" in the plane.
– Clayton
Jul 16 at 21:34
3
Possible duplicate of Singularities of Composition of Functions
– Clayton
Jul 16 at 21:38
Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$.
– GEdgar
Jul 18 at 14:23
@Clayton That question is different: it asks about $g circ f$ where $f$ has a(n essential) singularity. This question is about $f circ g$ (with $g(z) = z^2 + z$).
– Strants
Jul 23 at 13:51
add a comment |Â
1
Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $infty$), but there will always remain an essential singularity "somewhere" in the plane.
– Clayton
Jul 16 at 21:34
3
Possible duplicate of Singularities of Composition of Functions
– Clayton
Jul 16 at 21:38
Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$.
– GEdgar
Jul 18 at 14:23
@Clayton That question is different: it asks about $g circ f$ where $f$ has a(n essential) singularity. This question is about $f circ g$ (with $g(z) = z^2 + z$).
– Strants
Jul 23 at 13:51
1
1
Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $infty$), but there will always remain an essential singularity "somewhere" in the plane.
– Clayton
Jul 16 at 21:34
Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $infty$), but there will always remain an essential singularity "somewhere" in the plane.
– Clayton
Jul 16 at 21:34
3
3
Possible duplicate of Singularities of Composition of Functions
– Clayton
Jul 16 at 21:38
Possible duplicate of Singularities of Composition of Functions
– Clayton
Jul 16 at 21:38
Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$.
– GEdgar
Jul 18 at 14:23
Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$.
– GEdgar
Jul 18 at 14:23
@Clayton That question is different: it asks about $g circ f$ where $f$ has a(n essential) singularity. This question is about $f circ g$ (with $g(z) = z^2 + z$).
– Strants
Jul 23 at 13:51
@Clayton That question is different: it asks about $g circ f$ where $f$ has a(n essential) singularity. This question is about $f circ g$ (with $g(z) = z^2 + z$).
– Strants
Jul 23 at 13:51
add a comment |Â
1 Answer
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Here is a rigorous proof of the fact that $g(z)equiv f(z^2+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<delta $ with $0<delta <1$. If $|w|$ is sufficiently small take $z=frac -1+c 2$ where $c$ is such that $c^2=1+4z$ and $Re c>0$. Then $z^2+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| to infty $ as $z to 0$ and an argument similar to the previous one shows that $|g(z)| to infty $ as $z to 0$, again a contradiction.
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here is a rigorous proof of the fact that $g(z)equiv f(z^2+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<delta $ with $0<delta <1$. If $|w|$ is sufficiently small take $z=frac -1+c 2$ where $c$ is such that $c^2=1+4z$ and $Re c>0$. Then $z^2+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| to infty $ as $z to 0$ and an argument similar to the previous one shows that $|g(z)| to infty $ as $z to 0$, again a contradiction.
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
add a comment |Â
up vote
0
down vote
Here is a rigorous proof of the fact that $g(z)equiv f(z^2+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<delta $ with $0<delta <1$. If $|w|$ is sufficiently small take $z=frac -1+c 2$ where $c$ is such that $c^2=1+4z$ and $Re c>0$. Then $z^2+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| to infty $ as $z to 0$ and an argument similar to the previous one shows that $|g(z)| to infty $ as $z to 0$, again a contradiction.
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a rigorous proof of the fact that $g(z)equiv f(z^2+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<delta $ with $0<delta <1$. If $|w|$ is sufficiently small take $z=frac -1+c 2$ where $c$ is such that $c^2=1+4z$ and $Re c>0$. Then $z^2+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| to infty $ as $z to 0$ and an argument similar to the previous one shows that $|g(z)| to infty $ as $z to 0$, again a contradiction.
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
Here is a rigorous proof of the fact that $g(z)equiv f(z^2+z)$ has an essential singularity without using Laurent series. If $g$ has a removable singularity at $0$ then so it is bounded in a neighborhood of $0$, say $|g(z)| <M$ for $|z|<delta $ with $0<delta <1$. If $|w|$ is sufficiently small take $z=frac -1+c 2$ where $c$ is such that $c^2=1+4z$ and $Re c>0$. Then $z^2+z=w$ so we get $|f(w)|<M$. Thus $f$ is bounded in a neighborhood of $0$ contradicting the fact that $f$ has an essential singularity. If $g$ has a pole at $0$ then $|g(z)| to infty $ as $z to 0$ and an argument similar to the previous one shows that $|g(z)| to infty $ as $z to 0$, again a contradiction.
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
answered Jul 18 at 10:23
Kavi Rama Murthy
21k2829
21k2829
add a comment |Â
add a comment |Â
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1
Do you have anything specific in mind when you ask about getting rid of an essential singularity? You can "move" an essential singularity via compositions of functions (in some cases, you can even move it to $infty$), but there will always remain an essential singularity "somewhere" in the plane.
– Clayton
Jul 16 at 21:34
3
Possible duplicate of Singularities of Composition of Functions
– Clayton
Jul 16 at 21:38
Essential singularity at $z=0$ can have Laurent series with both positive and negative exponents on $z$.
– GEdgar
Jul 18 at 14:23
@Clayton That question is different: it asks about $g circ f$ where $f$ has a(n essential) singularity. This question is about $f circ g$ (with $g(z) = z^2 + z$).
– Strants
Jul 23 at 13:51