Mixing
Removable singularity at infinity implies the Laurent series has no positive powers
$begingroup$
Suppose that f(z) is analytic on the complex plane minus a single point $z_0$. Suppose further that f has a simple pole at $z_0$ and a removable singularity at infinity. Prove that $f(z) = frac{A}{z − z_0} + B$
Since $z_0$ is a simple pole, we know the Laurent series around $z_0$ is of the form $f(z) = frac{A}{z − z_0} + B + sum_{n=1}^{infty} a_n (z - z_0 )^n $
where $a_n = frac{1}{2 pi i}int_C frac{f(z)}{(z - z_0)^{n+1}} dz$.
And $ a_n = - Res_{w=0} frac{1}{w^2}frac{f(frac{1}{w})}{(frac{1}{w} -z_0)^{n+1}}$ by the change of variable $z = frac{1}{w}$. Since z = ∞ is a removable singularity of f(z), we also know $lim_{x to infty} zf(frac{1}{z}) = 0$. However, I'm not sure if I am missing something or how to finish the argument that $a_n = 0$ for all $n > 0$
complex-analysis
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
$endgroup$
add a comment |
$begingroup$
Suppose that f(z) is analytic on the complex plane minus a single point $z_0$. Suppose further that f has a simple pole at $z_0$ and a removable singularity at infinity. Prove that $f(z) = frac{A}{z − z_0} + B$
Since $z_0$ is a simple pole, we know the Laurent series around $z_0$ is of the form $f(z) = frac{A}{z − z_0} + B + sum_{n=1}^{infty} a_n (z - z_0 )^n $
where $a_n = frac{1}{2 pi i}int_C frac{f(z)}{(z - z_0)^{n+1}} dz$.
And $ a_n = - Res_{w=0} frac{1}{w^2}frac{f(frac{1}{w})}{(frac{1}{w} -z_0)^{n+1}}$ by the change of variable $z = frac{1}{w}$. Since z = ∞ is a removable singularity of f(z), we also know $lim_{x to infty} zf(frac{1}{z}) = 0$. However, I'm not sure if I am missing something or how to finish the argument that $a_n = 0$ for all $n > 0$
complex-analysis
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
$endgroup$
2
$begingroup$
You almost have it. Render $w=1/(color{blue}{z-z_0})$.
$endgroup$
– Oscar Lanzi
Jan 3 at 18:52
2
$begingroup$
Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $infty$, and take $r to infty$.
$endgroup$
– Robert Israel
Jan 3 at 19:03
add a comment |
$begingroup$
Suppose that f(z) is analytic on the complex plane minus a single point $z_0$. Suppose further that f has a simple pole at $z_0$ and a removable singularity at infinity. Prove that $f(z) = frac{A}{z − z_0} + B$
Since $z_0$ is a simple pole, we know the Laurent series around $z_0$ is of the form $f(z) = frac{A}{z − z_0} + B + sum_{n=1}^{infty} a_n (z - z_0 )^n $
where $a_n = frac{1}{2 pi i}int_C frac{f(z)}{(z - z_0)^{n+1}} dz$.
And $ a_n = - Res_{w=0} frac{1}{w^2}frac{f(frac{1}{w})}{(frac{1}{w} -z_0)^{n+1}}$ by the change of variable $z = frac{1}{w}$. Since z = ∞ is a removable singularity of f(z), we also know $lim_{x to infty} zf(frac{1}{z}) = 0$. However, I'm not sure if I am missing something or how to finish the argument that $a_n = 0$ for all $n > 0$
complex-analysis
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
$endgroup$
Suppose that f(z) is analytic on the complex plane minus a single point $z_0$. Suppose further that f has a simple pole at $z_0$ and a removable singularity at infinity. Prove that $f(z) = frac{A}{z − z_0} + B$
Since $z_0$ is a simple pole, we know the Laurent series around $z_0$ is of the form $f(z) = frac{A}{z − z_0} + B + sum_{n=1}^{infty} a_n (z - z_0 )^n $
where $a_n = frac{1}{2 pi i}int_C frac{f(z)}{(z - z_0)^{n+1}} dz$.
And $ a_n = - Res_{w=0} frac{1}{w^2}frac{f(frac{1}{w})}{(frac{1}{w} -z_0)^{n+1}}$ by the change of variable $z = frac{1}{w}$. Since z = ∞ is a removable singularity of f(z), we also know $lim_{x to infty} zf(frac{1}{z}) = 0$. However, I'm not sure if I am missing something or how to finish the argument that $a_n = 0$ for all $n > 0$
complex-analysis
complex-analysis
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
asked Jan 3 at 18:43
David WarrenDavid Warren
586313
586313
2
$begingroup$
You almost have it. Render $w=1/(color{blue}{z-z_0})$.
$endgroup$
– Oscar Lanzi
Jan 3 at 18:52
2
$begingroup$
Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $infty$, and take $r to infty$.
$endgroup$
– Robert Israel
Jan 3 at 19:03
add a comment |
2
$begingroup$
You almost have it. Render $w=1/(color{blue}{z-z_0})$.
$endgroup$
– Oscar Lanzi
Jan 3 at 18:52
2
$begingroup$
Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $infty$, and take $r to infty$.
$endgroup$
– Robert Israel
Jan 3 at 19:03
2
2
$begingroup$
You almost have it. Render $w=1/(color{blue}{z-z_0})$.
$endgroup$
– Oscar Lanzi
Jan 3 at 18:52
$begingroup$
You almost have it. Render $w=1/(color{blue}{z-z_0})$.
$endgroup$
– Oscar Lanzi
Jan 3 at 18:52
2
2
$begingroup$
Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $infty$, and take $r to infty$.
$endgroup$
– Robert Israel
Jan 3 at 19:03
$begingroup$
Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $infty$, and take $r to infty$.
$endgroup$
– Robert Israel
Jan 3 at 19:03
add a comment |
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$begingroup$
You almost have it. Render $w=1/(color{blue}{z-z_0})$.
$endgroup$
– Oscar Lanzi
Jan 3 at 18:52
2
$begingroup$
Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $infty$, and take $r to infty$.
$endgroup$
– Robert Israel
Jan 3 at 19:03