Find number of roots of a polynomial inside $ |z| < 2 $












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The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?










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  • $begingroup$
    A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
    $endgroup$
    – LutzL
    Jan 6 at 11:59










  • $begingroup$
    So maybe this is not Rouché's theorem?
    $endgroup$
    – calm
    Jan 6 at 12:02
















2












$begingroup$


The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?










share|cite|improve this question











$endgroup$












  • $begingroup$
    A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
    $endgroup$
    – LutzL
    Jan 6 at 11:59










  • $begingroup$
    So maybe this is not Rouché's theorem?
    $endgroup$
    – calm
    Jan 6 at 12:02














2












2








2


1



$begingroup$


The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?










share|cite|improve this question











$endgroup$




The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?







complex-analysis






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share|cite|improve this question













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edited Jan 6 at 12:01







calm

















asked Jan 6 at 9:29









calmcalm

1387




1387












  • $begingroup$
    A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
    $endgroup$
    – LutzL
    Jan 6 at 11:59










  • $begingroup$
    So maybe this is not Rouché's theorem?
    $endgroup$
    – calm
    Jan 6 at 12:02


















  • $begingroup$
    A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
    $endgroup$
    – LutzL
    Jan 6 at 11:59










  • $begingroup$
    So maybe this is not Rouché's theorem?
    $endgroup$
    – calm
    Jan 6 at 12:02
















$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59




$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59












$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02




$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02










1 Answer
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$begingroup$

A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.



winding plot



plot of the function values on the circle of radius $2$





Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.



enter image description here





The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:



enter image description here






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    $begingroup$

    A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.



    winding plot



    plot of the function values on the circle of radius $2$





    Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.



    enter image description here





    The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.



      winding plot



      plot of the function values on the circle of radius $2$





      Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.



      enter image description here





      The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:



      enter image description here






      share|cite|improve this answer









      $endgroup$
















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        $begingroup$

        A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.



        winding plot



        plot of the function values on the circle of radius $2$





        Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.



        enter image description here





        The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:



        enter image description here






        share|cite|improve this answer









        $endgroup$



        A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.



        winding plot



        plot of the function values on the circle of radius $2$





        Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.



        enter image description here





        The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 13:38









        LutzLLutzL

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