$f$ has a pole at $z=a$ implies $1/f$ has a removable singularity at $z=a$
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In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.
For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.
complex-analysis analytic-functions singularity
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In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.
For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.
complex-analysis analytic-functions singularity
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add a comment |
$begingroup$
In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.
For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.
complex-analysis analytic-functions singularity
$endgroup$
In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.
For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.
complex-analysis analytic-functions singularity
complex-analysis analytic-functions singularity
asked Jan 6 at 10:01
Ajay Kumar NairAjay Kumar Nair
1109
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Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.
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1 Answer
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Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.
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Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.
$endgroup$
add a comment |
$begingroup$
Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.
$endgroup$
Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.
answered Jan 6 at 10:04
jmerryjmerry
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