How many roots does $g(z)=z^7-2z^5+6z^3-z+1$ have inside the unit disk - Rouche's Theorem Application...












1












$begingroup$


$g(z)=z^7-2z^5+6z^3-z+1$



and choose $f(z)=2z^5-6z^3$.



On $mid zmid =1$, we have



$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$



so $$mid f(z)-g(z)mid leq mid f(z)mid$$



and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.



Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.



Is this argument correct?






complex-analysis proof-verification complex-numbers roots rouches-theorem







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$endgroup$












  • $begingroup$
    why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
    $endgroup$
    – MoonKnight
    Dec 18 '18 at 22:07
















1












$begingroup$


$g(z)=z^7-2z^5+6z^3-z+1$



and choose $f(z)=2z^5-6z^3$.



On $mid zmid =1$, we have



$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$



so $$mid f(z)-g(z)mid leq mid f(z)mid$$



and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.



Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.



Is this argument correct?






complex-analysis proof-verification complex-numbers roots rouches-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
    $endgroup$
    – MoonKnight
    Dec 18 '18 at 22:07














1












1








1





$begingroup$


$g(z)=z^7-2z^5+6z^3-z+1$



and choose $f(z)=2z^5-6z^3$.



On $mid zmid =1$, we have



$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$



so $$mid f(z)-g(z)mid leq mid f(z)mid$$



and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.



Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.



Is this argument correct?






complex-analysis proof-verification complex-numbers roots rouches-theorem







share|cite|improve this question









$endgroup$




$g(z)=z^7-2z^5+6z^3-z+1$



and choose $f(z)=2z^5-6z^3$.



On $mid zmid =1$, we have



$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$



so $$mid f(z)-g(z)mid leq mid f(z)mid$$



and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.



Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.



Is this argument correct?






complex-analysis proof-verification complex-numbers roots rouches-theorem




complex-analysis proof-verification complex-numbers roots rouches-theorem






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asked Dec 18 '18 at 21:49









MikeMike

769415




769415












  • $begingroup$
    why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
    $endgroup$
    – MoonKnight
    Dec 18 '18 at 22:07


















  • $begingroup$
    why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
    $endgroup$
    – MoonKnight
    Dec 18 '18 at 22:07
















$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07




$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07










1 Answer
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$begingroup$

I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.



Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
$$lvert f(z)rvert= lvert 6z^3rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.






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    $begingroup$

    I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.



    Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
    $$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
    $$lvert f(z)rvert= lvert 6z^3rvert=6.$$
    So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.






    share|cite|improve this answer









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      1












      $begingroup$

      I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.



      Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
      $$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
      $$lvert f(z)rvert= lvert 6z^3rvert=6.$$
      So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.



        Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
        $$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
        $$lvert f(z)rvert= lvert 6z^3rvert=6.$$
        So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.






        share|cite|improve this answer









        $endgroup$



        I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.



        Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
        $$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
        $$lvert f(z)rvert= lvert 6z^3rvert=6.$$
        So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.







        share|cite|improve this answer












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        answered Dec 18 '18 at 22:11









        Antonios-Alexandros RobotisAntonios-Alexandros Robotis

        10.5k41741




        10.5k41741






























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