Prove $ lim_z rightarrow z_o f(z) in mathbbC$ if $lim_z rightarrow z_o (z-z_0) f(z) = 0$

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Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)




Question



From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?



Update: Attempt is now moved to answer.





Some intuition:



$(2) Leftarrow (1)$



  • If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$

$(2) Rightarrow (1)$



  • If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.


  • If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.


Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.







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    Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)




    Question



    From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?



    Update: Attempt is now moved to answer.





    Some intuition:



    $(2) Leftarrow (1)$



    • If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$

    $(2) Rightarrow (1)$



    • If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.


    • If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.


    Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.







    share|cite|improve this question
























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)




      Question



      From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?



      Update: Attempt is now moved to answer.





      Some intuition:



      $(2) Leftarrow (1)$



      • If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$

      $(2) Rightarrow (1)$



      • If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.


      • If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.


      Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.







      share|cite|improve this question














      Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)




      Question



      From (here): How do we have that $$lim_z rightarrow z_o f(z) in mathbbC textfor removable singularity z_0 tag1$$ under the characterisation that $f$ has a removable singularity if $$lim_z rightarrow z_o (z-z_0) f(z) = 0 tag2$$?



      Update: Attempt is now moved to answer.





      Some intuition:



      $(2) Leftarrow (1)$



      • If $L in mathbbC$, then $lim_z rightarrow z_o (z-z_0) f(z) = lim_z rightarrow z_o (z-z_0) lim_z rightarrow z_o f(z) = (0)(L) = 0$

      $(2) Rightarrow (1)$



      • If $L = infty$, then $lim_z rightarrow z_o (z-z_0) f(z) = 0 cdot infty$, indeterminate.


      • If $L$ dne, then $lim_z rightarrow z_o (z-z_0) f(z)$ dne.


      Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 15 at 8:58

























      asked Aug 8 at 7:00









      BCLC

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          1 Answer
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          Pf $(2) iff (1)$:



          Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.



          Now for the forward implication:




          $(2) Leftarrow (1)$:




          • Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


          • TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$


          • Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$


          $$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$



          $$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$



          (*) $textwhenever 0 < |z-z_0| < delta_1$



          QED $(2) Leftarrow (1)$



          Now for the backward implication:




          $(2) Rightarrow (1)$




          • Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$


          • TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


          • Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$


          $$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$



          (**) $textwhenever 0 < |z-z_0| < delta$



          QED $(2) Rightarrow (1)$



          Now both the forward and backward implications have been proven.



          QED $(2) iff (1)$






          share|cite|improve this answer




















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            1 Answer
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            active

            oldest

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            up vote
            1
            down vote



            accepted










            Pf $(2) iff (1)$:



            Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.



            Now for the forward implication:




            $(2) Leftarrow (1)$:




            • Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


            • TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$


            • Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$


            $$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$



            $$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$



            (*) $textwhenever 0 < |z-z_0| < delta_1$



            QED $(2) Leftarrow (1)$



            Now for the backward implication:




            $(2) Rightarrow (1)$




            • Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$


            • TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


            • Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$


            $$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$



            (**) $textwhenever 0 < |z-z_0| < delta$



            QED $(2) Rightarrow (1)$



            Now both the forward and backward implications have been proven.



            QED $(2) iff (1)$






            share|cite|improve this answer
























              up vote
              1
              down vote



              accepted










              Pf $(2) iff (1)$:



              Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.



              Now for the forward implication:




              $(2) Leftarrow (1)$:




              • Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


              • TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$


              • Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$


              $$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$



              $$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$



              (*) $textwhenever 0 < |z-z_0| < delta_1$



              QED $(2) Leftarrow (1)$



              Now for the backward implication:




              $(2) Rightarrow (1)$




              • Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$


              • TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


              • Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$


              $$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$



              (**) $textwhenever 0 < |z-z_0| < delta$



              QED $(2) Rightarrow (1)$



              Now both the forward and backward implications have been proven.



              QED $(2) iff (1)$






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Pf $(2) iff (1)$:



                Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.



                Now for the forward implication:




                $(2) Leftarrow (1)$:




                • Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


                • TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$


                • Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$


                $$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$



                $$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$



                (*) $textwhenever 0 < |z-z_0| < delta_1$



                QED $(2) Leftarrow (1)$



                Now for the backward implication:




                $(2) Rightarrow (1)$




                • Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$


                • TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


                • Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$


                $$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$



                (**) $textwhenever 0 < |z-z_0| < delta$



                QED $(2) Rightarrow (1)$



                Now both the forward and backward implications have been proven.



                QED $(2) iff (1)$






                share|cite|improve this answer












                Pf $(2) iff (1)$:



                Firstly, let $lim_z rightarrow z_o f(z) =: L$, a letter which could be a finite complex number, $infty$ or 'dne'.



                Now for the forward implication:




                $(2) Leftarrow (1)$:




                • Given: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|<varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


                • TS: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|<varepsilon textwhenever 0<|z-z_0|<delta$


                • Pf: Let $varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$


                $$= |(z-z_0)| |f(z)-L+L| le |z-z_0| (|f(z)-L|+|L|)$$



                $$stackrel(*)le |z-z_0| varepsilon_1+|L| < varepsilon textwhenever 0 < |z-z_0| < delta := minfracvarepsilonvarepsilon_1+,delta_1$$



                (*) $textwhenever 0 < |z-z_0| < delta_1$



                QED $(2) Leftarrow (1)$



                Now for the backward implication:




                $(2) Rightarrow (1)$




                • Given: $forall varepsilon > 0, exists delta > 0: |(z-z_0)f(z)|< varepsilon textwhenever 0<|z-z_0|<delta$


                • TS: $forall varepsilon_1 > 0, exists delta_1 > 0: |f(z)-L|< varepsilon_1 textwhenever 0<|z-z_0|<delta_1$


                • Pf: Let $varepsilon_1 > 0$. $$|f(z)-L| = |frac(z-z_0)[f(z)]z-z_0-L| le |frac(z-z_0)[f(z)]z-z_0| + |L|$$


                $$stackrel(**)le |fracvarepsilonz-z_0| + |L| le varepsilon_1 textwhenever 0<|z-z_0|<delta_1 := mindelta,fracvarepsilonL$$



                (**) $textwhenever 0 < |z-z_0| < delta$



                QED $(2) Rightarrow (1)$



                Now both the forward and backward implications have been proven.



                QED $(2) iff (1)$







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                answered Aug 15 at 11:29









                BCLC

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