Complex Analysis: Prove that an entire function with $lim_{ztoinfty}f(z)=infty$ is surjective.












1












$begingroup$



If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.




This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    What about the exponential function?
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 4:51






  • 2




    $begingroup$
    With that hypothesis, prove that $f$ has to be a polynomial.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:53






  • 1




    $begingroup$
    @JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:54










  • $begingroup$
    I see! I'm still asleep.
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 6:23
















1












$begingroup$



If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.




This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    What about the exponential function?
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 4:51






  • 2




    $begingroup$
    With that hypothesis, prove that $f$ has to be a polynomial.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:53






  • 1




    $begingroup$
    @JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:54










  • $begingroup$
    I see! I'm still asleep.
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 6:23














1












1








1





$begingroup$



If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.




This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.






complex-analysis







share|cite|improve this question









$endgroup$





If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.




This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 4:44









MephestophelesMephestopheles

83




83












  • $begingroup$
    What about the exponential function?
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 4:51






  • 2




    $begingroup$
    With that hypothesis, prove that $f$ has to be a polynomial.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:53






  • 1




    $begingroup$
    @JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:54










  • $begingroup$
    I see! I'm still asleep.
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 6:23


















  • $begingroup$
    What about the exponential function?
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 4:51






  • 2




    $begingroup$
    With that hypothesis, prove that $f$ has to be a polynomial.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:53






  • 1




    $begingroup$
    @JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
    $endgroup$
    – Zircht
    Dec 8 '18 at 4:54










  • $begingroup$
    I see! I'm still asleep.
    $endgroup$
    – Jens Schwaiger
    Dec 8 '18 at 6:23
















$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51




$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51




2




2




$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53




$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53




1




1




$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54




$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54












$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23




$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23










2 Answers
2






active

oldest

votes


















3












$begingroup$

Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
$$ f(z)=sum_{k=0}^infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.



Can you conclude using Rouché's Theorem?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
    $endgroup$
    – clark
    Dec 8 '18 at 6:00



















0












$begingroup$

Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    votes






    active

    oldest

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    3












    $begingroup$

    Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
    $$ f(z)=sum_{k=0}^infty a_k z^k.$$
    At infinity, the Laurent series looks like
    $$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
    If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.



    Can you conclude using Rouché's Theorem?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
      $endgroup$
      – clark
      Dec 8 '18 at 6:00
















    3












    $begingroup$

    Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
    $$ f(z)=sum_{k=0}^infty a_k z^k.$$
    At infinity, the Laurent series looks like
    $$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
    If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.



    Can you conclude using Rouché's Theorem?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
      $endgroup$
      – clark
      Dec 8 '18 at 6:00














    3












    3








    3





    $begingroup$

    Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
    $$ f(z)=sum_{k=0}^infty a_k z^k.$$
    At infinity, the Laurent series looks like
    $$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
    If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.



    Can you conclude using Rouché's Theorem?






    share|cite|improve this answer











    $endgroup$



    Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
    $$ f(z)=sum_{k=0}^infty a_k z^k.$$
    At infinity, the Laurent series looks like
    $$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
    If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.



    Can you conclude using Rouché's Theorem?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 8 '18 at 5:51

























    answered Dec 8 '18 at 5:04









    Antonios-Alexandros RobotisAntonios-Alexandros Robotis

    9,81241640




    9,81241640












    • $begingroup$
      Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
      $endgroup$
      – clark
      Dec 8 '18 at 6:00


















    • $begingroup$
      Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
      $endgroup$
      – clark
      Dec 8 '18 at 6:00
















    $begingroup$
    Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
    $endgroup$
    – clark
    Dec 8 '18 at 6:00




    $begingroup$
    Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
    $endgroup$
    – clark
    Dec 8 '18 at 6:00











    0












    $begingroup$

    Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.






        share|cite|improve this answer









        $endgroup$



        Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 5:56









        user10354138user10354138

        7,3772925




        7,3772925






























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