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 $bar{mathbb{C}}$?










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 $bar{mathbb{C}}$?










    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 $bar{mathbb{C}}$?










      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






      share|cite|improve this question













      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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3125734%2flimit-at-infinity-for-complex-functions%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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.






                share|cite|improve this answer









                $endgroup$



                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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3125734%2flimit-at-infinity-for-complex-functions%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Liste der Baudenkmale in Friedland (Mecklenburg)

                    Single-Malt-Whisky

                    Czorneboh