Limit at infinity for complex functions












3












$begingroup$


Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.



First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?






complex-analysis







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.



    First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.



      First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?






      complex-analysis







      share|cite|improve this question









      $endgroup$




      Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.



      First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?






      complex-analysis




      complex-analysis






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 42 mins ago









      Fluffy SkyeFluffy Skye

      1479




      1479






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          The niceties of the one-point compactification of $Bbb C$ aside, consider:



          $f(z) ne w, ; forall z in Bbb C; tag 1$



          $f(z) - w ne 0, ; forall z in Bbb C; tag 2$



          $(f(z) - w)^{-1}$ is then entire; since



          $f(z) to infty ; text{as} ; z to infty tag 3$



          $(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Nice answer! +1 vote.
            $endgroup$
            – Kavi Rama Murthy
            22 mins ago










          • $begingroup$
            @KaviRamaMurthy: quite a complement, considering the source! Cheers!
            $endgroup$
            – Robert Lewis
            21 mins ago



















          2












          $begingroup$

          The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
          $
          g(z)=f(frac1{z})
          $           for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
          $$
          g(z)=sum_{k=-n}^infty a_kz^k.
          $$           (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
            $endgroup$
            – nicomezi
            16 mins ago










          • $begingroup$
            Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
            $endgroup$
            – nicomezi
            11 mins ago












          • $begingroup$
            @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
            $endgroup$
            – Song
            12 secs ago





















          0












          $begingroup$

          Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.



          Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by



          $g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.



          Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            The niceties of the one-point compactification of $Bbb C$ aside, consider:



            $f(z) ne w, ; forall z in Bbb C; tag 1$



            $f(z) - w ne 0, ; forall z in Bbb C; tag 2$



            $(f(z) - w)^{-1}$ is then entire; since



            $f(z) to infty ; text{as} ; z to infty tag 3$



            $(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Nice answer! +1 vote.
              $endgroup$
              – Kavi Rama Murthy
              22 mins ago










            • $begingroup$
              @KaviRamaMurthy: quite a complement, considering the source! Cheers!
              $endgroup$
              – Robert Lewis
              21 mins ago
















            5












            $begingroup$

            The niceties of the one-point compactification of $Bbb C$ aside, consider:



            $f(z) ne w, ; forall z in Bbb C; tag 1$



            $f(z) - w ne 0, ; forall z in Bbb C; tag 2$



            $(f(z) - w)^{-1}$ is then entire; since



            $f(z) to infty ; text{as} ; z to infty tag 3$



            $(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Nice answer! +1 vote.
              $endgroup$
              – Kavi Rama Murthy
              22 mins ago










            • $begingroup$
              @KaviRamaMurthy: quite a complement, considering the source! Cheers!
              $endgroup$
              – Robert Lewis
              21 mins ago














            5












            5








            5





            $begingroup$

            The niceties of the one-point compactification of $Bbb C$ aside, consider:



            $f(z) ne w, ; forall z in Bbb C; tag 1$



            $f(z) - w ne 0, ; forall z in Bbb C; tag 2$



            $(f(z) - w)^{-1}$ is then entire; since



            $f(z) to infty ; text{as} ; z to infty tag 3$



            $(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.






            share|cite|improve this answer









            $endgroup$



            The niceties of the one-point compactification of $Bbb C$ aside, consider:



            $f(z) ne w, ; forall z in Bbb C; tag 1$



            $f(z) - w ne 0, ; forall z in Bbb C; tag 2$



            $(f(z) - w)^{-1}$ is then entire; since



            $f(z) to infty ; text{as} ; z to infty tag 3$



            $(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 23 mins ago









            Robert LewisRobert Lewis

            47.1k23067




            47.1k23067








            • 1




              $begingroup$
              Nice answer! +1 vote.
              $endgroup$
              – Kavi Rama Murthy
              22 mins ago










            • $begingroup$
              @KaviRamaMurthy: quite a complement, considering the source! Cheers!
              $endgroup$
              – Robert Lewis
              21 mins ago














            • 1




              $begingroup$
              Nice answer! +1 vote.
              $endgroup$
              – Kavi Rama Murthy
              22 mins ago










            • $begingroup$
              @KaviRamaMurthy: quite a complement, considering the source! Cheers!
              $endgroup$
              – Robert Lewis
              21 mins ago








            1




            1




            $begingroup$
            Nice answer! +1 vote.
            $endgroup$
            – Kavi Rama Murthy
            22 mins ago




            $begingroup$
            Nice answer! +1 vote.
            $endgroup$
            – Kavi Rama Murthy
            22 mins ago












            $begingroup$
            @KaviRamaMurthy: quite a complement, considering the source! Cheers!
            $endgroup$
            – Robert Lewis
            21 mins ago




            $begingroup$
            @KaviRamaMurthy: quite a complement, considering the source! Cheers!
            $endgroup$
            – Robert Lewis
            21 mins ago











            2












            $begingroup$

            The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
            $
            g(z)=f(frac1{z})
            $             for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
            $$
            g(z)=sum_{k=-n}^infty a_kz^k.
            $$             (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
              $endgroup$
              – nicomezi
              16 mins ago










            • $begingroup$
              Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
              $endgroup$
              – nicomezi
              11 mins ago












            • $begingroup$
              @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
              $endgroup$
              – Song
              12 secs ago


















            2












            $begingroup$

            The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
            $
            g(z)=f(frac1{z})
            $             for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
            $$
            g(z)=sum_{k=-n}^infty a_kz^k.
            $$             (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
              $endgroup$
              – nicomezi
              16 mins ago










            • $begingroup$
              Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
              $endgroup$
              – nicomezi
              11 mins ago












            • $begingroup$
              @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
              $endgroup$
              – Song
              12 secs ago
















            2












            2








            2





            $begingroup$

            The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
            $
            g(z)=f(frac1{z})
            $             for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
            $$
            g(z)=sum_{k=-n}^infty a_kz^k.
            $$             (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$






            share|cite|improve this answer











            $endgroup$



            The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
            $
            g(z)=f(frac1{z})
            $             for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
            $$
            g(z)=sum_{k=-n}^infty a_kz^k.
            $$             (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 mins ago

























            answered 30 mins ago









            SongSong

            14.9k1635




            14.9k1635












            • $begingroup$
              The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
              $endgroup$
              – nicomezi
              16 mins ago










            • $begingroup$
              Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
              $endgroup$
              – nicomezi
              11 mins ago












            • $begingroup$
              @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
              $endgroup$
              – Song
              12 secs ago




















            • $begingroup$
              The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
              $endgroup$
              – nicomezi
              16 mins ago










            • $begingroup$
              Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
              $endgroup$
              – nicomezi
              11 mins ago












            • $begingroup$
              @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
              $endgroup$
              – Song
              12 secs ago


















            $begingroup$
            The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
            $endgroup$
            – nicomezi
            16 mins ago




            $begingroup$
            The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
            $endgroup$
            – nicomezi
            16 mins ago












            $begingroup$
            Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
            $endgroup$
            – nicomezi
            11 mins ago






            $begingroup$
            Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
            $endgroup$
            – nicomezi
            11 mins ago














            $begingroup$
            @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
            $endgroup$
            – Song
            12 secs ago






            $begingroup$
            @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.)
            $endgroup$
            – Song
            12 secs ago













            0












            $begingroup$

            Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.



            Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by



            $g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.



            Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.



              Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by



              $g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.



              Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.



                Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by



                $g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.



                Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.






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                $endgroup$



                Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.



                Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by



                $g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.



                Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 14 mins ago









                FredFred

                46.9k1848




                46.9k1848






























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