If $f$ is entire such that $|f(z)|leq C|z|^{5/2}$, $f$ is polynomial of degree two.












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$begingroup$


$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis







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  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54
















0












$begingroup$


$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54














0












0








0


1



$begingroup$


$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis







share|cite|improve this question











$endgroup$




$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis




complex-analysis






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share|cite|improve this question













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share|cite|improve this question








edited Dec 21 '18 at 18:10







Ya G

















asked Dec 21 '18 at 17:55









Ya GYa G

536211




536211








  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54














  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54








1




1




$begingroup$
No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
$endgroup$
– David C. Ullrich
Dec 21 '18 at 18:09




$begingroup$
No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
$endgroup$
– David C. Ullrich
Dec 21 '18 at 18:09




2




2




$begingroup$
Cauchy's Estimate
$endgroup$
– Story123
Dec 21 '18 at 18:54




$begingroup$
Cauchy's Estimate
$endgroup$
– Story123
Dec 21 '18 at 18:54










4 Answers
4






active

oldest

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2












$begingroup$

Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
$$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
for $r>0$. Here, then
$$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
    $endgroup$
    – Ya G
    Dec 21 '18 at 20:24










  • $begingroup$
    @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 5:56



















7












$begingroup$

Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
$$
[f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
$$
The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
$$
|g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
$$
Hence, by Liouville's theorem, we must have $g = 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
    $endgroup$
    – Ya G
    Dec 21 '18 at 18:19






  • 1




    $begingroup$
    Added a line in the proof.
    $endgroup$
    – Rigel
    Dec 21 '18 at 18:24



















1












$begingroup$

It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A lot of complex analysis follows from basic facts about Fourier series.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 18:55



















0












$begingroup$

Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



This can be done as follows:
Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



Then $h$ becomes an entire function and
$$lim_{z to infty} h(z)=0$$
From here it is easy to conclude that $h$ is constant.






share|cite|improve this answer









$endgroup$













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    4 Answers
    4






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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56
















    2












    $begingroup$

    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56














    2












    2








    2





    $begingroup$

    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






    share|cite|improve this answer











    $endgroup$



    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 22 '18 at 5:57

























    answered Dec 21 '18 at 19:03









    Lord Shark the UnknownLord Shark the Unknown

    107k1162135




    107k1162135












    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56


















    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56
















    $begingroup$
    Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
    $endgroup$
    – Ya G
    Dec 21 '18 at 20:24




    $begingroup$
    Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
    $endgroup$
    – Ya G
    Dec 21 '18 at 20:24












    $begingroup$
    @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 5:56




    $begingroup$
    @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 5:56











    7












    $begingroup$

    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24
















    7












    $begingroup$

    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24














    7












    7








    7





    $begingroup$

    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.






    share|cite|improve this answer











    $endgroup$



    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 18:24

























    answered Dec 21 '18 at 18:03









    RigelRigel

    11.4k11320




    11.4k11320












    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24


















    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24
















    $begingroup$
    Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
    $endgroup$
    – Ya G
    Dec 21 '18 at 18:19




    $begingroup$
    Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
    $endgroup$
    – Ya G
    Dec 21 '18 at 18:19




    1




    1




    $begingroup$
    Added a line in the proof.
    $endgroup$
    – Rigel
    Dec 21 '18 at 18:24




    $begingroup$
    Added a line in the proof.
    $endgroup$
    – Rigel
    Dec 21 '18 at 18:24











    1












    $begingroup$

    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55
















    1












    $begingroup$

    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55














    1












    1








    1





    $begingroup$

    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






    share|cite|improve this answer









    $endgroup$



    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 18:16









    David C. UllrichDavid C. Ullrich

    61.6k43995




    61.6k43995












    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55


















    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55
















    $begingroup$
    A lot of complex analysis follows from basic facts about Fourier series.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 18:55




    $begingroup$
    A lot of complex analysis follows from basic facts about Fourier series.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 18:55











    0












    $begingroup$

    Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



    This can be done as follows:
    Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



    Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



    Then $h$ becomes an entire function and
    $$lim_{z to infty} h(z)=0$$
    From here it is easy to conclude that $h$ is constant.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



      This can be done as follows:
      Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



      Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



      Then $h$ becomes an entire function and
      $$lim_{z to infty} h(z)=0$$
      From here it is easy to conclude that $h$ is constant.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



        This can be done as follows:
        Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



        Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



        Then $h$ becomes an entire function and
        $$lim_{z to infty} h(z)=0$$
        From here it is easy to conclude that $h$ is constant.






        share|cite|improve this answer









        $endgroup$



        Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



        This can be done as follows:
        Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



        Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



        Then $h$ becomes an entire function and
        $$lim_{z to infty} h(z)=0$$
        From here it is easy to conclude that $h$ is constant.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 19:14









        N. S.N. S.

        105k7114210




        105k7114210






























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