Does $z^4+2z^2+z=0$ have complex roots?












4














Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3itheta}+2re^{itheta}=e^{i(pi+2kpi)}$, $kin mathbb Z$. How to find the complex roots?






calculus complex-numbers







share|cite|improve this question


















  • 3




    By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $mathbb{C}$. As for finding yours, though, I'm not sure.
    – Eevee Trainer
    Nov 29 at 3:01








  • 4




    The polynomial is $zP$ where $P = z^3+2z+1$. Since $frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root.
    – Théophile
    Nov 29 at 3:17












  • Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| le 2$ (since on the circle of radius two, $|2z^2 + z| le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk.
    – Brevan Ellefsen
    Nov 29 at 9:26
















4














Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3itheta}+2re^{itheta}=e^{i(pi+2kpi)}$, $kin mathbb Z$. How to find the complex roots?






calculus complex-numbers







share|cite|improve this question


















  • 3




    By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $mathbb{C}$. As for finding yours, though, I'm not sure.
    – Eevee Trainer
    Nov 29 at 3:01








  • 4




    The polynomial is $zP$ where $P = z^3+2z+1$. Since $frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root.
    – Théophile
    Nov 29 at 3:17












  • Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| le 2$ (since on the circle of radius two, $|2z^2 + z| le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk.
    – Brevan Ellefsen
    Nov 29 at 9:26














4












4








4


3





Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3itheta}+2re^{itheta}=e^{i(pi+2kpi)}$, $kin mathbb Z$. How to find the complex roots?






calculus complex-numbers







share|cite|improve this question













Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3itheta}+2re^{itheta}=e^{i(pi+2kpi)}$, $kin mathbb Z$. How to find the complex roots?






calculus complex-numbers




calculus complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 at 3:00









user398843

598215




598215








  • 3




    By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $mathbb{C}$. As for finding yours, though, I'm not sure.
    – Eevee Trainer
    Nov 29 at 3:01








  • 4




    The polynomial is $zP$ where $P = z^3+2z+1$. Since $frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root.
    – Théophile
    Nov 29 at 3:17












  • Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| le 2$ (since on the circle of radius two, $|2z^2 + z| le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk.
    – Brevan Ellefsen
    Nov 29 at 9:26














  • 3




    By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $mathbb{C}$. As for finding yours, though, I'm not sure.
    – Eevee Trainer
    Nov 29 at 3:01








  • 4




    The polynomial is $zP$ where $P = z^3+2z+1$. Since $frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root.
    – Théophile
    Nov 29 at 3:17












  • Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| le 2$ (since on the circle of radius two, $|2z^2 + z| le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk.
    – Brevan Ellefsen
    Nov 29 at 9:26








3




3




By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $mathbb{C}$. As for finding yours, though, I'm not sure.
– Eevee Trainer
Nov 29 at 3:01






By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $mathbb{C}$. As for finding yours, though, I'm not sure.
– Eevee Trainer
Nov 29 at 3:01






4




4




The polynomial is $zP$ where $P = z^3+2z+1$. Since $frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root.
– Théophile
Nov 29 at 3:17






The polynomial is $zP$ where $P = z^3+2z+1$. Since $frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root.
– Théophile
Nov 29 at 3:17














Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| le 2$ (since on the circle of radius two, $|2z^2 + z| le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk.
– Brevan Ellefsen
Nov 29 at 9:26




Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| le 2$ (since on the circle of radius two, $|2z^2 + z| le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk.
– Brevan Ellefsen
Nov 29 at 9:26










2 Answers
2






active

oldest

votes


















3














Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
$$
Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
$$
$$
Delta_3=begin{cases} >0 & text{3 distinct real roots}\
<0 & text{1 real, 2 conjugate complex roots}\
=0 & text{3 real roots with duplicates}\
end{cases}
$$
In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
$$
x_1=frac{sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{3^{2/3}}-2 sqrt[3]{frac{2}{3
left(sqrt{177}-9right)}}
$$
$$
x_2=left(1+i sqrt{3}right)
sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1-i sqrt{3}right)
sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2 3^{2/3}}
$$
$$
x_3=left(1-i
sqrt{3}right) sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1+i
sqrt{3}right) sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2
3^{2/3}}
$$






share|cite|improve this answer





























    0














    According to Wolfy,
    the roots are
    $0, ≈-0.453397651516404,
    ≈0.22669882575820 pm 1.46771150871022 i
    $
    so the answer is yes.



    Another way to see this is that,
    if
    $f(z) = z^4+2z^2+z$,
    then
    $f'(z)
    = 4z^3+4z+1
    $
    and
    $f''(z)
    = 12z^2+4
    gt 0
    $
    so $f(z)$
    can have at most two real roots.



    Since
    $f(0) = 0$
    and $f'(0)=1 ne 0$,
    $f(z)$ has exactly
    two real roots
    so has
    two (conjugate) complex roots
    since all its coefficients
    are real.






    share|cite|improve this answer





















    • I think OP is asking for complex, not real roots.
      – YiFan
      Nov 29 at 13:14










    • I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
      – marty cohen
      Nov 29 at 16:20










    • The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
      – YiFan
      Nov 29 at 20:34











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
    $$
    Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
    $$
    $$
    Delta_3=begin{cases} >0 & text{3 distinct real roots}\
    <0 & text{1 real, 2 conjugate complex roots}\
    =0 & text{3 real roots with duplicates}\
    end{cases}
    $$
    In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
    $$
    x_1=frac{sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{3^{2/3}}-2 sqrt[3]{frac{2}{3
    left(sqrt{177}-9right)}}
    $$
    $$
    x_2=left(1+i sqrt{3}right)
    sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1-i sqrt{3}right)
    sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2 3^{2/3}}
    $$
    $$
    x_3=left(1-i
    sqrt{3}right) sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1+i
    sqrt{3}right) sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2
    3^{2/3}}
    $$






    share|cite|improve this answer


























      3














      Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
      $$
      Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
      $$
      $$
      Delta_3=begin{cases} >0 & text{3 distinct real roots}\
      <0 & text{1 real, 2 conjugate complex roots}\
      =0 & text{3 real roots with duplicates}\
      end{cases}
      $$
      In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
      $$
      x_1=frac{sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{3^{2/3}}-2 sqrt[3]{frac{2}{3
      left(sqrt{177}-9right)}}
      $$
      $$
      x_2=left(1+i sqrt{3}right)
      sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1-i sqrt{3}right)
      sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2 3^{2/3}}
      $$
      $$
      x_3=left(1-i
      sqrt{3}right) sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1+i
      sqrt{3}right) sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2
      3^{2/3}}
      $$






      share|cite|improve this answer
























        3












        3








        3






        Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
        $$
        Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
        $$
        $$
        Delta_3=begin{cases} >0 & text{3 distinct real roots}\
        <0 & text{1 real, 2 conjugate complex roots}\
        =0 & text{3 real roots with duplicates}\
        end{cases}
        $$
        In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
        $$
        x_1=frac{sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{3^{2/3}}-2 sqrt[3]{frac{2}{3
        left(sqrt{177}-9right)}}
        $$
        $$
        x_2=left(1+i sqrt{3}right)
        sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1-i sqrt{3}right)
        sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2 3^{2/3}}
        $$
        $$
        x_3=left(1-i
        sqrt{3}right) sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1+i
        sqrt{3}right) sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2
        3^{2/3}}
        $$






        share|cite|improve this answer












        Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
        $$
        Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
        $$
        $$
        Delta_3=begin{cases} >0 & text{3 distinct real roots}\
        <0 & text{1 real, 2 conjugate complex roots}\
        =0 & text{3 real roots with duplicates}\
        end{cases}
        $$
        In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
        $$
        x_1=frac{sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{3^{2/3}}-2 sqrt[3]{frac{2}{3
        left(sqrt{177}-9right)}}
        $$
        $$
        x_2=left(1+i sqrt{3}right)
        sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1-i sqrt{3}right)
        sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2 3^{2/3}}
        $$
        $$
        x_3=left(1-i
        sqrt{3}right) sqrt[3]{frac{2}{3 left(sqrt{177}-9right)}}-frac{left(1+i
        sqrt{3}right) sqrt[3]{frac{1}{2} left(sqrt{177}-9right)}}{2
        3^{2/3}}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 3:11









        Ákos Somogyi

        1,547417




        1,547417























            0














            According to Wolfy,
            the roots are
            $0, ≈-0.453397651516404,
            ≈0.22669882575820 pm 1.46771150871022 i
            $
            so the answer is yes.



            Another way to see this is that,
            if
            $f(z) = z^4+2z^2+z$,
            then
            $f'(z)
            = 4z^3+4z+1
            $
            and
            $f''(z)
            = 12z^2+4
            gt 0
            $
            so $f(z)$
            can have at most two real roots.



            Since
            $f(0) = 0$
            and $f'(0)=1 ne 0$,
            $f(z)$ has exactly
            two real roots
            so has
            two (conjugate) complex roots
            since all its coefficients
            are real.






            share|cite|improve this answer





















            • I think OP is asking for complex, not real roots.
              – YiFan
              Nov 29 at 13:14










            • I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
              – marty cohen
              Nov 29 at 16:20










            • The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
              – YiFan
              Nov 29 at 20:34
















            0














            According to Wolfy,
            the roots are
            $0, ≈-0.453397651516404,
            ≈0.22669882575820 pm 1.46771150871022 i
            $
            so the answer is yes.



            Another way to see this is that,
            if
            $f(z) = z^4+2z^2+z$,
            then
            $f'(z)
            = 4z^3+4z+1
            $
            and
            $f''(z)
            = 12z^2+4
            gt 0
            $
            so $f(z)$
            can have at most two real roots.



            Since
            $f(0) = 0$
            and $f'(0)=1 ne 0$,
            $f(z)$ has exactly
            two real roots
            so has
            two (conjugate) complex roots
            since all its coefficients
            are real.






            share|cite|improve this answer





















            • I think OP is asking for complex, not real roots.
              – YiFan
              Nov 29 at 13:14










            • I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
              – marty cohen
              Nov 29 at 16:20










            • The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
              – YiFan
              Nov 29 at 20:34














            0












            0








            0






            According to Wolfy,
            the roots are
            $0, ≈-0.453397651516404,
            ≈0.22669882575820 pm 1.46771150871022 i
            $
            so the answer is yes.



            Another way to see this is that,
            if
            $f(z) = z^4+2z^2+z$,
            then
            $f'(z)
            = 4z^3+4z+1
            $
            and
            $f''(z)
            = 12z^2+4
            gt 0
            $
            so $f(z)$
            can have at most two real roots.



            Since
            $f(0) = 0$
            and $f'(0)=1 ne 0$,
            $f(z)$ has exactly
            two real roots
            so has
            two (conjugate) complex roots
            since all its coefficients
            are real.






            share|cite|improve this answer












            According to Wolfy,
            the roots are
            $0, ≈-0.453397651516404,
            ≈0.22669882575820 pm 1.46771150871022 i
            $
            so the answer is yes.



            Another way to see this is that,
            if
            $f(z) = z^4+2z^2+z$,
            then
            $f'(z)
            = 4z^3+4z+1
            $
            and
            $f''(z)
            = 12z^2+4
            gt 0
            $
            so $f(z)$
            can have at most two real roots.



            Since
            $f(0) = 0$
            and $f'(0)=1 ne 0$,
            $f(z)$ has exactly
            two real roots
            so has
            two (conjugate) complex roots
            since all its coefficients
            are real.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 at 6:05









            marty cohen

            72.3k549127




            72.3k549127












            • I think OP is asking for complex, not real roots.
              – YiFan
              Nov 29 at 13:14










            • I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
              – marty cohen
              Nov 29 at 16:20










            • The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
              – YiFan
              Nov 29 at 20:34


















            • I think OP is asking for complex, not real roots.
              – YiFan
              Nov 29 at 13:14










            • I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
              – marty cohen
              Nov 29 at 16:20










            • The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
              – YiFan
              Nov 29 at 20:34
















            I think OP is asking for complex, not real roots.
            – YiFan
            Nov 29 at 13:14




            I think OP is asking for complex, not real roots.
            – YiFan
            Nov 29 at 13:14












            I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
            – marty cohen
            Nov 29 at 16:20




            I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence?
            – marty cohen
            Nov 29 at 16:20












            The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
            – YiFan
            Nov 29 at 20:34




            The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy.
            – YiFan
            Nov 29 at 20:34


















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