Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$












0















Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.




This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.






complex-analysis inequality







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  • That's wrong if $n le R$.
    – Martin R
    Nov 20 '18 at 5:46
















0















Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.




This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.






complex-analysis inequality







share|cite|improve this question
























  • That's wrong if $n le R$.
    – Martin R
    Nov 20 '18 at 5:46














0












0








0








Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.




This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.






complex-analysis inequality







share|cite|improve this question
















Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.




This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.






complex-analysis inequality




complex-analysis inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 6:15

























asked Nov 20 '18 at 5:34









Cute Brownie

995416




995416












  • That's wrong if $n le R$.
    – Martin R
    Nov 20 '18 at 5:46


















  • That's wrong if $n le R$.
    – Martin R
    Nov 20 '18 at 5:46
















That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46




That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46










1 Answer
1






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oldest

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1














Hint:



begin{align}
|z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
end{align}






share|cite|improve this answer





















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Hint:



    begin{align}
    |z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
    end{align}






    share|cite|improve this answer


























      1














      Hint:



      begin{align}
      |z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
      end{align}






      share|cite|improve this answer
























        1












        1








        1






        Hint:



        begin{align}
        |z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
        end{align}






        share|cite|improve this answer












        Hint:



        begin{align}
        |z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 5:37









        Jacky Chong

        17.7k21128




        17.7k21128






























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