How do I prove L'Hôpital's rule variation for complex functions












1












$begingroup$


I need to prove to following version of L'Hôpital's rule for complex functions:



Let $f$ and $g$ be analytical at $z_0$ such that $f(z_0)=g(z_0)=0$ and $g$ is not identically zero: so the following limits exist $($whether $infty$ or final$)$ and equal:
$displaystylelim_{nrightarrow infty}frac{f(z)}{g(z)}=lim_{nrightarrow infty}frac{f'(z)}{g'(z)}$ .



I looked up online and found people using it but not a proof. I've found this, which I'm not sure why doesn't work here:



$$lim_{zrightarrow z_{0}}frac{f(z)}{g(z)} =lim_{zrightarrow z_{0}}frac{f(z)-f(z_{0})}{z-z_{0}}cdotfrac{z-z_{0}}{g(z)-g(z_{0})}
=frac{f'(z)}{g'(z)}$$



My guess it's something with power series.



Thank you.






calculus limits derivatives







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I need to prove to following version of L'Hôpital's rule for complex functions:



    Let $f$ and $g$ be analytical at $z_0$ such that $f(z_0)=g(z_0)=0$ and $g$ is not identically zero: so the following limits exist $($whether $infty$ or final$)$ and equal:
    $displaystylelim_{nrightarrow infty}frac{f(z)}{g(z)}=lim_{nrightarrow infty}frac{f'(z)}{g'(z)}$ .



    I looked up online and found people using it but not a proof. I've found this, which I'm not sure why doesn't work here:



    $$lim_{zrightarrow z_{0}}frac{f(z)}{g(z)} =lim_{zrightarrow z_{0}}frac{f(z)-f(z_{0})}{z-z_{0}}cdotfrac{z-z_{0}}{g(z)-g(z_{0})}
    =frac{f'(z)}{g'(z)}$$



    My guess it's something with power series.



    Thank you.






    calculus limits derivatives







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I need to prove to following version of L'Hôpital's rule for complex functions:



      Let $f$ and $g$ be analytical at $z_0$ such that $f(z_0)=g(z_0)=0$ and $g$ is not identically zero: so the following limits exist $($whether $infty$ or final$)$ and equal:
      $displaystylelim_{nrightarrow infty}frac{f(z)}{g(z)}=lim_{nrightarrow infty}frac{f'(z)}{g'(z)}$ .



      I looked up online and found people using it but not a proof. I've found this, which I'm not sure why doesn't work here:



      $$lim_{zrightarrow z_{0}}frac{f(z)}{g(z)} =lim_{zrightarrow z_{0}}frac{f(z)-f(z_{0})}{z-z_{0}}cdotfrac{z-z_{0}}{g(z)-g(z_{0})}
      =frac{f'(z)}{g'(z)}$$



      My guess it's something with power series.



      Thank you.






      calculus limits derivatives







      share|cite|improve this question











      $endgroup$




      I need to prove to following version of L'Hôpital's rule for complex functions:



      Let $f$ and $g$ be analytical at $z_0$ such that $f(z_0)=g(z_0)=0$ and $g$ is not identically zero: so the following limits exist $($whether $infty$ or final$)$ and equal:
      $displaystylelim_{nrightarrow infty}frac{f(z)}{g(z)}=lim_{nrightarrow infty}frac{f'(z)}{g'(z)}$ .



      I looked up online and found people using it but not a proof. I've found this, which I'm not sure why doesn't work here:



      $$lim_{zrightarrow z_{0}}frac{f(z)}{g(z)} =lim_{zrightarrow z_{0}}frac{f(z)-f(z_{0})}{z-z_{0}}cdotfrac{z-z_{0}}{g(z)-g(z_{0})}
      =frac{f'(z)}{g'(z)}$$



      My guess it's something with power series.



      Thank you.






      calculus limits derivatives




      calculus limits derivatives






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 23 '18 at 15:43









      mrtaurho

      4,53121235




      4,53121235










      asked Dec 23 '18 at 15:40









      RoeeRoee

      374




      374






















          1 Answer
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          $begingroup$

          Assum that $z_0$ is a simple zero of both $f$ and $g$. Then, if$$f(z)=a_1(z-z_0)+a_2(z-z_0)^2+cdotstext{ and }g(z)=b_1(z-z_0)+b_2(z-z_0)^2+cdots,$$we havebegin{align}lim_{zto z_0}frac{f(z)}{g(z)}&=lim_{zto z_0}frac{a_1(z-z_0)+a_2(z-z_0)^2+cdots}{b_1(z-z_0)+b_2(z-z_0)^2+cdots}\&=lim_{zto z_0}frac{a_1+a_2(z-z_0)+cdots}{b_1+b_2(z-z_0)+cdots}\&=frac{a_1}{b_1}\&=lim_{zto z_0}frac{f'(z)}{g'(z)}.end{align}Can you do the general case now?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
            $endgroup$
            – Roee
            Dec 23 '18 at 16:26








          • 1




            $begingroup$
            I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:29












          • $begingroup$
            I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
            $endgroup$
            – Roee
            Dec 23 '18 at 16:49






          • 1




            $begingroup$
            It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:56











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          1












          $begingroup$

          Assum that $z_0$ is a simple zero of both $f$ and $g$. Then, if$$f(z)=a_1(z-z_0)+a_2(z-z_0)^2+cdotstext{ and }g(z)=b_1(z-z_0)+b_2(z-z_0)^2+cdots,$$we havebegin{align}lim_{zto z_0}frac{f(z)}{g(z)}&=lim_{zto z_0}frac{a_1(z-z_0)+a_2(z-z_0)^2+cdots}{b_1(z-z_0)+b_2(z-z_0)^2+cdots}\&=lim_{zto z_0}frac{a_1+a_2(z-z_0)+cdots}{b_1+b_2(z-z_0)+cdots}\&=frac{a_1}{b_1}\&=lim_{zto z_0}frac{f'(z)}{g'(z)}.end{align}Can you do the general case now?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
            $endgroup$
            – Roee
            Dec 23 '18 at 16:26








          • 1




            $begingroup$
            I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:29












          • $begingroup$
            I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
            $endgroup$
            – Roee
            Dec 23 '18 at 16:49






          • 1




            $begingroup$
            It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:56
















          1












          $begingroup$

          Assum that $z_0$ is a simple zero of both $f$ and $g$. Then, if$$f(z)=a_1(z-z_0)+a_2(z-z_0)^2+cdotstext{ and }g(z)=b_1(z-z_0)+b_2(z-z_0)^2+cdots,$$we havebegin{align}lim_{zto z_0}frac{f(z)}{g(z)}&=lim_{zto z_0}frac{a_1(z-z_0)+a_2(z-z_0)^2+cdots}{b_1(z-z_0)+b_2(z-z_0)^2+cdots}\&=lim_{zto z_0}frac{a_1+a_2(z-z_0)+cdots}{b_1+b_2(z-z_0)+cdots}\&=frac{a_1}{b_1}\&=lim_{zto z_0}frac{f'(z)}{g'(z)}.end{align}Can you do the general case now?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
            $endgroup$
            – Roee
            Dec 23 '18 at 16:26








          • 1




            $begingroup$
            I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:29












          • $begingroup$
            I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
            $endgroup$
            – Roee
            Dec 23 '18 at 16:49






          • 1




            $begingroup$
            It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:56














          1












          1








          1





          $begingroup$

          Assum that $z_0$ is a simple zero of both $f$ and $g$. Then, if$$f(z)=a_1(z-z_0)+a_2(z-z_0)^2+cdotstext{ and }g(z)=b_1(z-z_0)+b_2(z-z_0)^2+cdots,$$we havebegin{align}lim_{zto z_0}frac{f(z)}{g(z)}&=lim_{zto z_0}frac{a_1(z-z_0)+a_2(z-z_0)^2+cdots}{b_1(z-z_0)+b_2(z-z_0)^2+cdots}\&=lim_{zto z_0}frac{a_1+a_2(z-z_0)+cdots}{b_1+b_2(z-z_0)+cdots}\&=frac{a_1}{b_1}\&=lim_{zto z_0}frac{f'(z)}{g'(z)}.end{align}Can you do the general case now?






          share|cite|improve this answer









          $endgroup$



          Assum that $z_0$ is a simple zero of both $f$ and $g$. Then, if$$f(z)=a_1(z-z_0)+a_2(z-z_0)^2+cdotstext{ and }g(z)=b_1(z-z_0)+b_2(z-z_0)^2+cdots,$$we havebegin{align}lim_{zto z_0}frac{f(z)}{g(z)}&=lim_{zto z_0}frac{a_1(z-z_0)+a_2(z-z_0)^2+cdots}{b_1(z-z_0)+b_2(z-z_0)^2+cdots}\&=lim_{zto z_0}frac{a_1+a_2(z-z_0)+cdots}{b_1+b_2(z-z_0)+cdots}\&=frac{a_1}{b_1}\&=lim_{zto z_0}frac{f'(z)}{g'(z)}.end{align}Can you do the general case now?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 15:45









          José Carlos SantosJosé Carlos Santos

          159k22126231




          159k22126231












          • $begingroup$
            Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
            $endgroup$
            – Roee
            Dec 23 '18 at 16:26








          • 1




            $begingroup$
            I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:29












          • $begingroup$
            I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
            $endgroup$
            – Roee
            Dec 23 '18 at 16:49






          • 1




            $begingroup$
            It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:56


















          • $begingroup$
            Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
            $endgroup$
            – Roee
            Dec 23 '18 at 16:26








          • 1




            $begingroup$
            I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:29












          • $begingroup$
            I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
            $endgroup$
            – Roee
            Dec 23 '18 at 16:49






          • 1




            $begingroup$
            It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 16:56
















          $begingroup$
          Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
          $endgroup$
          – Roee
          Dec 23 '18 at 16:26






          $begingroup$
          Forgive me but I'm going to need a little bit of elaboration. All I know for $f$ and $g$ is that $z_0$ is their zero of some value, and maybe even different values. Why do you assume that they both have a simple zero? and could you please explain where is your usage of it? I'm not fluent in zeros and poles. Lastly, what do you mean by the "general case"? Thanks again and sorry.
          $endgroup$
          – Roee
          Dec 23 '18 at 16:26






          1




          1




          $begingroup$
          I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
          $endgroup$
          – José Carlos Santos
          Dec 23 '18 at 16:29






          $begingroup$
          I proved the statmens assuming explicitely that $z_0$ is a simple zero of both $f$ and $g$. The general case is when there is som $ninmathbb N$ such that$$f(z)=a_n(z-z_0)^n+a_{n+1}(z-z_0)^{n+1}+cdots$$and there is a $minmathbb N$ such that$$g(z)=b_m(z-z_0)^m+b_{m+1}(z-z_0)^{m+1}+cdots$$but you can deal with it the same way that I dealt with the case in which $n=m=1$.
          $endgroup$
          – José Carlos Santos
          Dec 23 '18 at 16:29














          $begingroup$
          I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
          $endgroup$
          – Roee
          Dec 23 '18 at 16:49




          $begingroup$
          I now understand what you mean about the zeros, but why can you expand $f$ and $g$ into power series if they're only analytical at $z_0$? Is it Taylor series or Laurent series? Thanks again
          $endgroup$
          – Roee
          Dec 23 '18 at 16:49




          1




          1




          $begingroup$
          It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
          $endgroup$
          – José Carlos Santos
          Dec 23 '18 at 16:56




          $begingroup$
          It's the Taylor series. If it didn't exist or if it didn't converge to $f(z)$ and to $g(z)$ near $z_0$, then $f$ and $g$ would no be analytic there.
          $endgroup$
          – José Carlos Santos
          Dec 23 '18 at 16:56


















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          Cabo Verde

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            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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          Gyllenstierna

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