How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| <...












3












$begingroup$


How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?



I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.






real-analysis abstract-algebra complex-analysis polynomials roots







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    3












    $begingroup$


    How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?



    I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.






    real-analysis abstract-algebra complex-analysis polynomials roots







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      2



      $begingroup$


      How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?



      I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.






      real-analysis abstract-algebra complex-analysis polynomials roots







      share|cite|improve this question









      $endgroup$




      How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?



      I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.






      real-analysis abstract-algebra complex-analysis polynomials roots




      real-analysis abstract-algebra complex-analysis polynomials roots






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 23 '16 at 18:32









      HappyHappy

      561412




      561412






















          2 Answers
          2






          active

          oldest

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          6












          $begingroup$

          General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.



          How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.



          I will let you do the same with the circle of radius $2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
            $endgroup$
            – Happy
            Apr 25 '16 at 16:52










          • $begingroup$
            Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:18










          • $begingroup$
            That makes perfect sense! How does one know which $f$ to choose?
            $endgroup$
            – Happy
            Apr 25 '16 at 17:24










          • $begingroup$
            Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:28










          • $begingroup$
            I understand now. Thank you : )
            $endgroup$
            – Happy
            Apr 25 '16 at 17:30



















          -3












          $begingroup$

          I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
            $endgroup$
            – hardmath
            Dec 21 '18 at 5:34













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          2 Answers
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          2 Answers
          2






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          active

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          active

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          6












          $begingroup$

          General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.



          How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.



          I will let you do the same with the circle of radius $2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
            $endgroup$
            – Happy
            Apr 25 '16 at 16:52










          • $begingroup$
            Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:18










          • $begingroup$
            That makes perfect sense! How does one know which $f$ to choose?
            $endgroup$
            – Happy
            Apr 25 '16 at 17:24










          • $begingroup$
            Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:28










          • $begingroup$
            I understand now. Thank you : )
            $endgroup$
            – Happy
            Apr 25 '16 at 17:30
















          6












          $begingroup$

          General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.



          How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.



          I will let you do the same with the circle of radius $2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
            $endgroup$
            – Happy
            Apr 25 '16 at 16:52










          • $begingroup$
            Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:18










          • $begingroup$
            That makes perfect sense! How does one know which $f$ to choose?
            $endgroup$
            – Happy
            Apr 25 '16 at 17:24










          • $begingroup$
            Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:28










          • $begingroup$
            I understand now. Thank you : )
            $endgroup$
            – Happy
            Apr 25 '16 at 17:30














          6












          6








          6





          $begingroup$

          General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.



          How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.



          I will let you do the same with the circle of radius $2$.






          share|cite|improve this answer









          $endgroup$



          General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.



          How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.



          I will let you do the same with the circle of radius $2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 23 '16 at 18:44









          JenniferJennifer

          8,44721837




          8,44721837












          • $begingroup$
            I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
            $endgroup$
            – Happy
            Apr 25 '16 at 16:52










          • $begingroup$
            Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:18










          • $begingroup$
            That makes perfect sense! How does one know which $f$ to choose?
            $endgroup$
            – Happy
            Apr 25 '16 at 17:24










          • $begingroup$
            Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:28










          • $begingroup$
            I understand now. Thank you : )
            $endgroup$
            – Happy
            Apr 25 '16 at 17:30


















          • $begingroup$
            I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
            $endgroup$
            – Happy
            Apr 25 '16 at 16:52










          • $begingroup$
            Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:18










          • $begingroup$
            That makes perfect sense! How does one know which $f$ to choose?
            $endgroup$
            – Happy
            Apr 25 '16 at 17:24










          • $begingroup$
            Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
            $endgroup$
            – Jennifer
            Apr 25 '16 at 17:28










          • $begingroup$
            I understand now. Thank you : )
            $endgroup$
            – Happy
            Apr 25 '16 at 17:30
















          $begingroup$
          I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
          $endgroup$
          – Happy
          Apr 25 '16 at 16:52




          $begingroup$
          I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
          $endgroup$
          – Happy
          Apr 25 '16 at 16:52












          $begingroup$
          Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
          $endgroup$
          – Jennifer
          Apr 25 '16 at 17:18




          $begingroup$
          Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
          $endgroup$
          – Jennifer
          Apr 25 '16 at 17:18












          $begingroup$
          That makes perfect sense! How does one know which $f$ to choose?
          $endgroup$
          – Happy
          Apr 25 '16 at 17:24




          $begingroup$
          That makes perfect sense! How does one know which $f$ to choose?
          $endgroup$
          – Happy
          Apr 25 '16 at 17:24












          $begingroup$
          Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
          $endgroup$
          – Jennifer
          Apr 25 '16 at 17:28




          $begingroup$
          Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
          $endgroup$
          – Jennifer
          Apr 25 '16 at 17:28












          $begingroup$
          I understand now. Thank you : )
          $endgroup$
          – Happy
          Apr 25 '16 at 17:30




          $begingroup$
          I understand now. Thank you : )
          $endgroup$
          – Happy
          Apr 25 '16 at 17:30











          -3












          $begingroup$

          I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
            $endgroup$
            – hardmath
            Dec 21 '18 at 5:34


















          -3












          $begingroup$

          I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
            $endgroup$
            – hardmath
            Dec 21 '18 at 5:34
















          -3












          -3








          -3





          $begingroup$

          I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.






          share|cite|improve this answer









          $endgroup$



          I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 '18 at 5:02









          AMRAMR

          1




          1












          • $begingroup$
            Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
            $endgroup$
            – hardmath
            Dec 21 '18 at 5:34




















          • $begingroup$
            Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
            $endgroup$
            – hardmath
            Dec 21 '18 at 5:34


















          $begingroup$
          Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
          $endgroup$
          – hardmath
          Dec 21 '18 at 5:34






          $begingroup$
          Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
          $endgroup$
          – hardmath
          Dec 21 '18 at 5:34




















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