$f : BbbC to BbbC$ is an entire function s.t. $|f(z)| to infty$ as $|z| to infty$. Prove $f$ is a polynomial.If $F$ is entire with removable singularity at $infty$, then $F$ is constant?bounded real part then removable singularityAt $z=0$ the function $f(z)=exp(zover 1-cos z)$ hasMust an entire function be bounded in any open disk around the origin?Extending an analytic function to an entire functionNot understanding how to use the following information: $f$ is entire, and $lim _ fracf(z)z^2=2i$Prove that $fequiv 0$ where $f$ is entireFind all entire function $f$ such that $lim_zto inftyleft|fracf(z)zright|=0$problem on existence of entire functionShouldn't a entire function be strictly a polynomial?
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$f : BbbC to BbbC$ is an entire function s.t. $|f(z)| to infty$ as $|z| to infty$. Prove $f$ is a polynomial.
If $F$ is entire with removable singularity at $infty$, then $F$ is constant?bounded real part then removable singularityAt $z=0$ the function $f(z)=exp(zover 1-cos z)$ hasMust an entire function be bounded in any open disk around the origin?Extending an analytic function to an entire functionNot understanding how to use the following information: $f$ is entire, and $lim _z fracf(z)z^2=2i$Prove that $fequiv 0$ where $f$ is entireFind all entire function $f$ such that $lim_zto inftyleft|fracf(z)zright|=0$problem on existence of entire functionShouldn't a entire function be strictly a polynomial?
$begingroup$
Let $f : mathbbC rightarrow mathbbC$ be an entire function such that $|f(z)| rightarrow infty$ as $|z| rightarrow infty$. Prove that $f$ is a polynomial by following the steps below.
(a) Observe that the function $f(1/z)$ defined in $C setminus0$ has a pole at the origin. Let $S$ be the
singular part of its Laurent series around $0$. Argue that $g(z) = f(z) − S(1/z)$ approaches
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$.
(b) Prove that $g$ extends to a bounded entire function and is therefore constant.
(c) Deduce that $f$ is a polynomial.
I know there are solutions to this but not with these steps. I have been given this as an assignment and really need help to figure out how to write the solution.
For point (a) I wrote
In order for $|f(z)| rightarrow infty$ as $|z| rightarrow infty$, is equavalent saying $|f(frac1z)| rightarrow infty$ as $z rightarrow 0$. So $f(z)$ can be written in Laurent series around the point $0$ such that $sum_n=-infty^infty a_nz^n = g(z) + s(frac1z)$ given $g(z)$ is the power series $sum_n=0^infty a_nz^n$ and $S$ is the singular part of its Laurent series around $0$.
Hence,
$ g(z)= f(z) - s(frac1z)$ and $g(z)$ tends to a
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$ given the power series.
For (b) I wrote
Only removable singularities are left in $g(z)$. So
with Riemann's removable singularity theorem, with $p$ being some removable singularity. $g(z)$ can be extended continuously and holomorphically at $p$ and since $g(z)$ is bonded on some disc around $p$ excluding the point $p$ with a radius bigger than $0$. Hence the extended function is bounded as well and by Liouville's theorem, it is a constant function.
And I have no clue how to answer (c), sorry for the long post and thanks in advance!!!
complex-analysis
$endgroup$
add a comment |
$begingroup$
Let $f : mathbbC rightarrow mathbbC$ be an entire function such that $|f(z)| rightarrow infty$ as $|z| rightarrow infty$. Prove that $f$ is a polynomial by following the steps below.
(a) Observe that the function $f(1/z)$ defined in $C setminus0$ has a pole at the origin. Let $S$ be the
singular part of its Laurent series around $0$. Argue that $g(z) = f(z) − S(1/z)$ approaches
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$.
(b) Prove that $g$ extends to a bounded entire function and is therefore constant.
(c) Deduce that $f$ is a polynomial.
I know there are solutions to this but not with these steps. I have been given this as an assignment and really need help to figure out how to write the solution.
For point (a) I wrote
In order for $|f(z)| rightarrow infty$ as $|z| rightarrow infty$, is equavalent saying $|f(frac1z)| rightarrow infty$ as $z rightarrow 0$. So $f(z)$ can be written in Laurent series around the point $0$ such that $sum_n=-infty^infty a_nz^n = g(z) + s(frac1z)$ given $g(z)$ is the power series $sum_n=0^infty a_nz^n$ and $S$ is the singular part of its Laurent series around $0$.
Hence,
$ g(z)= f(z) - s(frac1z)$ and $g(z)$ tends to a
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$ given the power series.
For (b) I wrote
Only removable singularities are left in $g(z)$. So
with Riemann's removable singularity theorem, with $p$ being some removable singularity. $g(z)$ can be extended continuously and holomorphically at $p$ and since $g(z)$ is bonded on some disc around $p$ excluding the point $p$ with a radius bigger than $0$. Hence the extended function is bounded as well and by Liouville's theorem, it is a constant function.
And I have no clue how to answer (c), sorry for the long post and thanks in advance!!!
complex-analysis
$endgroup$
add a comment |
$begingroup$
Let $f : mathbbC rightarrow mathbbC$ be an entire function such that $|f(z)| rightarrow infty$ as $|z| rightarrow infty$. Prove that $f$ is a polynomial by following the steps below.
(a) Observe that the function $f(1/z)$ defined in $C setminus0$ has a pole at the origin. Let $S$ be the
singular part of its Laurent series around $0$. Argue that $g(z) = f(z) − S(1/z)$ approaches
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$.
(b) Prove that $g$ extends to a bounded entire function and is therefore constant.
(c) Deduce that $f$ is a polynomial.
I know there are solutions to this but not with these steps. I have been given this as an assignment and really need help to figure out how to write the solution.
For point (a) I wrote
In order for $|f(z)| rightarrow infty$ as $|z| rightarrow infty$, is equavalent saying $|f(frac1z)| rightarrow infty$ as $z rightarrow 0$. So $f(z)$ can be written in Laurent series around the point $0$ such that $sum_n=-infty^infty a_nz^n = g(z) + s(frac1z)$ given $g(z)$ is the power series $sum_n=0^infty a_nz^n$ and $S$ is the singular part of its Laurent series around $0$.
Hence,
$ g(z)= f(z) - s(frac1z)$ and $g(z)$ tends to a
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$ given the power series.
For (b) I wrote
Only removable singularities are left in $g(z)$. So
with Riemann's removable singularity theorem, with $p$ being some removable singularity. $g(z)$ can be extended continuously and holomorphically at $p$ and since $g(z)$ is bonded on some disc around $p$ excluding the point $p$ with a radius bigger than $0$. Hence the extended function is bounded as well and by Liouville's theorem, it is a constant function.
And I have no clue how to answer (c), sorry for the long post and thanks in advance!!!
complex-analysis
$endgroup$
Let $f : mathbbC rightarrow mathbbC$ be an entire function such that $|f(z)| rightarrow infty$ as $|z| rightarrow infty$. Prove that $f$ is a polynomial by following the steps below.
(a) Observe that the function $f(1/z)$ defined in $C setminus0$ has a pole at the origin. Let $S$ be the
singular part of its Laurent series around $0$. Argue that $g(z) = f(z) − S(1/z)$ approaches
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$.
(b) Prove that $g$ extends to a bounded entire function and is therefore constant.
(c) Deduce that $f$ is a polynomial.
I know there are solutions to this but not with these steps. I have been given this as an assignment and really need help to figure out how to write the solution.
For point (a) I wrote
In order for $|f(z)| rightarrow infty$ as $|z| rightarrow infty$, is equavalent saying $|f(frac1z)| rightarrow infty$ as $z rightarrow 0$. So $f(z)$ can be written in Laurent series around the point $0$ such that $sum_n=-infty^infty a_nz^n = g(z) + s(frac1z)$ given $g(z)$ is the power series $sum_n=0^infty a_nz^n$ and $S$ is the singular part of its Laurent series around $0$.
Hence,
$ g(z)= f(z) - s(frac1z)$ and $g(z)$ tends to a
finite limits as $z rightarrow 0$ and as $|z| rightarrow infty$ given the power series.
For (b) I wrote
Only removable singularities are left in $g(z)$. So
with Riemann's removable singularity theorem, with $p$ being some removable singularity. $g(z)$ can be extended continuously and holomorphically at $p$ and since $g(z)$ is bonded on some disc around $p$ excluding the point $p$ with a radius bigger than $0$. Hence the extended function is bounded as well and by Liouville's theorem, it is a constant function.
And I have no clue how to answer (c), sorry for the long post and thanks in advance!!!
complex-analysis
complex-analysis
edited Mar 22 at 2:45
Andrews
1,2812422
1,2812422
asked Mar 22 at 0:37
ChengCheng
233
233
add a comment |
add a comment |
1 Answer
1
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$begingroup$
In your solution to part (a), notice that the Laurent series you have written down is for $f(frac 1z)$, so the correct formula is $f(frac 1z)=sum_n=0^infty a_n z^n +S(z)$. So by the change of coordinates $zmapsto frac 1z$, we get $f(z)=sum_n=0^infty a_nfrac 1z^n +S(frac 1z)$. So the $g(z)$ in the question is actually $sum_n=0^infty a_n frac 1z^n$. As $|z|toinfty$, this series indeed tends to $a_0$, which is finite. As for $zto 0$. Now by looking that the Laurent series again, we see that $S(frac 1z)$ has removable singularity at $z=0$. So $g(z)=f(z)-S(frac 1z)$ also has removable singularity at $z=0$, in other words the limit as $zto 0$ exists.
For part (b), holomorphicity of $g(z)=sum_n=0^infty a_n frac 1z^n$ follows directly from $g$ having removable singularity at $0$. By comparing the power series and Laurent series, we see that the only term appearing in the series must be the constant $a_0$ only.
For part (c), the hint is to look at the Laurent expansion of $f(frac 1z)$ again, and recover the series for $f(z)$.
$endgroup$
add a comment |
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$begingroup$
In your solution to part (a), notice that the Laurent series you have written down is for $f(frac 1z)$, so the correct formula is $f(frac 1z)=sum_n=0^infty a_n z^n +S(z)$. So by the change of coordinates $zmapsto frac 1z$, we get $f(z)=sum_n=0^infty a_nfrac 1z^n +S(frac 1z)$. So the $g(z)$ in the question is actually $sum_n=0^infty a_n frac 1z^n$. As $|z|toinfty$, this series indeed tends to $a_0$, which is finite. As for $zto 0$. Now by looking that the Laurent series again, we see that $S(frac 1z)$ has removable singularity at $z=0$. So $g(z)=f(z)-S(frac 1z)$ also has removable singularity at $z=0$, in other words the limit as $zto 0$ exists.
For part (b), holomorphicity of $g(z)=sum_n=0^infty a_n frac 1z^n$ follows directly from $g$ having removable singularity at $0$. By comparing the power series and Laurent series, we see that the only term appearing in the series must be the constant $a_0$ only.
For part (c), the hint is to look at the Laurent expansion of $f(frac 1z)$ again, and recover the series for $f(z)$.
$endgroup$
add a comment |
$begingroup$
In your solution to part (a), notice that the Laurent series you have written down is for $f(frac 1z)$, so the correct formula is $f(frac 1z)=sum_n=0^infty a_n z^n +S(z)$. So by the change of coordinates $zmapsto frac 1z$, we get $f(z)=sum_n=0^infty a_nfrac 1z^n +S(frac 1z)$. So the $g(z)$ in the question is actually $sum_n=0^infty a_n frac 1z^n$. As $|z|toinfty$, this series indeed tends to $a_0$, which is finite. As for $zto 0$. Now by looking that the Laurent series again, we see that $S(frac 1z)$ has removable singularity at $z=0$. So $g(z)=f(z)-S(frac 1z)$ also has removable singularity at $z=0$, in other words the limit as $zto 0$ exists.
For part (b), holomorphicity of $g(z)=sum_n=0^infty a_n frac 1z^n$ follows directly from $g$ having removable singularity at $0$. By comparing the power series and Laurent series, we see that the only term appearing in the series must be the constant $a_0$ only.
For part (c), the hint is to look at the Laurent expansion of $f(frac 1z)$ again, and recover the series for $f(z)$.
$endgroup$
add a comment |
$begingroup$
In your solution to part (a), notice that the Laurent series you have written down is for $f(frac 1z)$, so the correct formula is $f(frac 1z)=sum_n=0^infty a_n z^n +S(z)$. So by the change of coordinates $zmapsto frac 1z$, we get $f(z)=sum_n=0^infty a_nfrac 1z^n +S(frac 1z)$. So the $g(z)$ in the question is actually $sum_n=0^infty a_n frac 1z^n$. As $|z|toinfty$, this series indeed tends to $a_0$, which is finite. As for $zto 0$. Now by looking that the Laurent series again, we see that $S(frac 1z)$ has removable singularity at $z=0$. So $g(z)=f(z)-S(frac 1z)$ also has removable singularity at $z=0$, in other words the limit as $zto 0$ exists.
For part (b), holomorphicity of $g(z)=sum_n=0^infty a_n frac 1z^n$ follows directly from $g$ having removable singularity at $0$. By comparing the power series and Laurent series, we see that the only term appearing in the series must be the constant $a_0$ only.
For part (c), the hint is to look at the Laurent expansion of $f(frac 1z)$ again, and recover the series for $f(z)$.
$endgroup$
In your solution to part (a), notice that the Laurent series you have written down is for $f(frac 1z)$, so the correct formula is $f(frac 1z)=sum_n=0^infty a_n z^n +S(z)$. So by the change of coordinates $zmapsto frac 1z$, we get $f(z)=sum_n=0^infty a_nfrac 1z^n +S(frac 1z)$. So the $g(z)$ in the question is actually $sum_n=0^infty a_n frac 1z^n$. As $|z|toinfty$, this series indeed tends to $a_0$, which is finite. As for $zto 0$. Now by looking that the Laurent series again, we see that $S(frac 1z)$ has removable singularity at $z=0$. So $g(z)=f(z)-S(frac 1z)$ also has removable singularity at $z=0$, in other words the limit as $zto 0$ exists.
For part (b), holomorphicity of $g(z)=sum_n=0^infty a_n frac 1z^n$ follows directly from $g$ having removable singularity at $0$. By comparing the power series and Laurent series, we see that the only term appearing in the series must be the constant $a_0$ only.
For part (c), the hint is to look at the Laurent expansion of $f(frac 1z)$ again, and recover the series for $f(z)$.
answered Mar 22 at 3:53
lEmlEm
3,3871921
3,3871921
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