Find all complex numbers $z$ such that $frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$ is a real number.












0












$begingroup$


Find all complex numbers $z$ such that:



$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$



is a real number. I'm preparing for a math competition, but I can't solve this one. Can anyone help me? Thanks in advance. :)






complex-numbers







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$endgroup$












  • $begingroup$
    What have you tried? Can you make up a simpler problem of this form that you can solve?
    $endgroup$
    – Hans Engler
    Dec 24 '18 at 14:26






  • 1




    $begingroup$
    for it to be a real number, doesn't both the numerator and denominator need to be reals? (just asking)
    $endgroup$
    – user531476
    Dec 24 '18 at 14:35








  • 1




    $begingroup$
    notice that for purely imaginary numbers, the quotient is real
    $endgroup$
    – user531476
    Dec 24 '18 at 14:37






  • 1




    $begingroup$
    Notice the palindromic coefficients in numerator and denominator.
    $endgroup$
    – Mark Bennet
    Dec 24 '18 at 14:42






  • 3




    $begingroup$
    @archaic Not at all. $frac{1+i}{1+i}$ is real
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 14:42
















0












$begingroup$


Find all complex numbers $z$ such that:



$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$



is a real number. I'm preparing for a math competition, but I can't solve this one. Can anyone help me? Thanks in advance. :)






complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    What have you tried? Can you make up a simpler problem of this form that you can solve?
    $endgroup$
    – Hans Engler
    Dec 24 '18 at 14:26






  • 1




    $begingroup$
    for it to be a real number, doesn't both the numerator and denominator need to be reals? (just asking)
    $endgroup$
    – user531476
    Dec 24 '18 at 14:35








  • 1




    $begingroup$
    notice that for purely imaginary numbers, the quotient is real
    $endgroup$
    – user531476
    Dec 24 '18 at 14:37






  • 1




    $begingroup$
    Notice the palindromic coefficients in numerator and denominator.
    $endgroup$
    – Mark Bennet
    Dec 24 '18 at 14:42






  • 3




    $begingroup$
    @archaic Not at all. $frac{1+i}{1+i}$ is real
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 14:42














0












0








0


0



$begingroup$


Find all complex numbers $z$ such that:



$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$



is a real number. I'm preparing for a math competition, but I can't solve this one. Can anyone help me? Thanks in advance. :)






complex-numbers







share|cite|improve this question









$endgroup$




Find all complex numbers $z$ such that:



$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}$



is a real number. I'm preparing for a math competition, but I can't solve this one. Can anyone help me? Thanks in advance. :)






complex-numbers




complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 14:24









Wolf M.Wolf M.

1097




1097












  • $begingroup$
    What have you tried? Can you make up a simpler problem of this form that you can solve?
    $endgroup$
    – Hans Engler
    Dec 24 '18 at 14:26






  • 1




    $begingroup$
    for it to be a real number, doesn't both the numerator and denominator need to be reals? (just asking)
    $endgroup$
    – user531476
    Dec 24 '18 at 14:35








  • 1




    $begingroup$
    notice that for purely imaginary numbers, the quotient is real
    $endgroup$
    – user531476
    Dec 24 '18 at 14:37






  • 1




    $begingroup$
    Notice the palindromic coefficients in numerator and denominator.
    $endgroup$
    – Mark Bennet
    Dec 24 '18 at 14:42






  • 3




    $begingroup$
    @archaic Not at all. $frac{1+i}{1+i}$ is real
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 14:42


















  • $begingroup$
    What have you tried? Can you make up a simpler problem of this form that you can solve?
    $endgroup$
    – Hans Engler
    Dec 24 '18 at 14:26






  • 1




    $begingroup$
    for it to be a real number, doesn't both the numerator and denominator need to be reals? (just asking)
    $endgroup$
    – user531476
    Dec 24 '18 at 14:35








  • 1




    $begingroup$
    notice that for purely imaginary numbers, the quotient is real
    $endgroup$
    – user531476
    Dec 24 '18 at 14:37






  • 1




    $begingroup$
    Notice the palindromic coefficients in numerator and denominator.
    $endgroup$
    – Mark Bennet
    Dec 24 '18 at 14:42






  • 3




    $begingroup$
    @archaic Not at all. $frac{1+i}{1+i}$ is real
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 14:42
















$begingroup$
What have you tried? Can you make up a simpler problem of this form that you can solve?
$endgroup$
– Hans Engler
Dec 24 '18 at 14:26




$begingroup$
What have you tried? Can you make up a simpler problem of this form that you can solve?
$endgroup$
– Hans Engler
Dec 24 '18 at 14:26




1




1




$begingroup$
for it to be a real number, doesn't both the numerator and denominator need to be reals? (just asking)
$endgroup$
– user531476
Dec 24 '18 at 14:35






$begingroup$
for it to be a real number, doesn't both the numerator and denominator need to be reals? (just asking)
$endgroup$
– user531476
Dec 24 '18 at 14:35






1




1




$begingroup$
notice that for purely imaginary numbers, the quotient is real
$endgroup$
– user531476
Dec 24 '18 at 14:37




$begingroup$
notice that for purely imaginary numbers, the quotient is real
$endgroup$
– user531476
Dec 24 '18 at 14:37




1




1




$begingroup$
Notice the palindromic coefficients in numerator and denominator.
$endgroup$
– Mark Bennet
Dec 24 '18 at 14:42




$begingroup$
Notice the palindromic coefficients in numerator and denominator.
$endgroup$
– Mark Bennet
Dec 24 '18 at 14:42




3




3




$begingroup$
@archaic Not at all. $frac{1+i}{1+i}$ is real
$endgroup$
– Ankit Kumar
Dec 24 '18 at 14:42




$begingroup$
@archaic Not at all. $frac{1+i}{1+i}$ is real
$endgroup$
– Ankit Kumar
Dec 24 '18 at 14:42










4 Answers
4






active

oldest

votes


















3












$begingroup$

For a complex number to be purely real, $z=bar{z}$. So, taking $x=z^2,$
$$Y=frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}$$
$$Y=bar{Y}implies frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}=frac{6bar{x^2} + 5bar{x} + 6}{3bar{x^2} + 10bar{x} + 3}$$
Cross multiply, and cancel lot of terms to get
$$(xbar{x}-1)(x-bar{x})=0$$
$$x=bar{x} implies z=pm bar{z}$$
$$xbar{x}=1implies |z|=1.$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You can express:
    $$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=2-frac{15z^2}{3z^4 + 10z^2 + 3}=2-frac{15}{3z^2+10+frac3{x^2}}=2-frac{15}{3(z+frac1z)^2+4} Rightarrow \ left(z+frac1zright)^2=left(a+bi+frac1{a+bi}right)^2=left(a+bi+frac{a-bi}{a^2+b^2}right)^2 Rightarrow \
    1) a^3+ab^2+a=0 text{or} 2) a^2b+b^3-b=0 Rightarrow \
    1) a=0 text{or} 2) b=0 text{or} a^2+b^2=1.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nicely done! I like it.
      $endgroup$
      – Mark Bennet
      Dec 24 '18 at 17:21










    • $begingroup$
      I’m glad to hear that. Thank you.
      $endgroup$
      – farruhota
      Dec 24 '18 at 18:50



















    1












    $begingroup$

    Hint: Plug in your formula $$z=x+iy$$ then you will get$$24x^3y-24xy^3+10xy=0$$ and $$12x^3y-12xy^3+20xy=0$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      This is probably too much of a hint, but is based on my comment.
      I suggest dividing numerator and denominator by $z^2$ and setting $y=z^2+frac 1{z^2}$ when your fraction becomes $$frac {6y+5}{3y+10}$$ and it is easy now to show that $y$ has to be real, and if $y$ is real so is the fraction. So set $y=r$, an arbitrary real number, whence $z^4-rz^2+1=0$.



      This can be solved in various ways, which include $$z^4-2z^2+2-(r-2)z^2=(z^2+zsqrt {r-2}+1)(z^2-zsqrt {r-2}+1)=0$$



      If we now put $r-2=4R$ with $R$ a real number, this has solutions $$z=pm sqrt Rpm sqrt {R-1}$$and though this is a clunky way through in the end, we get $z$ real, or $z$ pure imaginary, or $|z|=1$ when $Rin [0,1]$.



      Others have found easier ways through to these conditions.






      share|cite|improve this answer











      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        For a complex number to be purely real, $z=bar{z}$. So, taking $x=z^2,$
        $$Y=frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}$$
        $$Y=bar{Y}implies frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}=frac{6bar{x^2} + 5bar{x} + 6}{3bar{x^2} + 10bar{x} + 3}$$
        Cross multiply, and cancel lot of terms to get
        $$(xbar{x}-1)(x-bar{x})=0$$
        $$x=bar{x} implies z=pm bar{z}$$
        $$xbar{x}=1implies |z|=1.$$






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          For a complex number to be purely real, $z=bar{z}$. So, taking $x=z^2,$
          $$Y=frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}$$
          $$Y=bar{Y}implies frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}=frac{6bar{x^2} + 5bar{x} + 6}{3bar{x^2} + 10bar{x} + 3}$$
          Cross multiply, and cancel lot of terms to get
          $$(xbar{x}-1)(x-bar{x})=0$$
          $$x=bar{x} implies z=pm bar{z}$$
          $$xbar{x}=1implies |z|=1.$$






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            For a complex number to be purely real, $z=bar{z}$. So, taking $x=z^2,$
            $$Y=frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}$$
            $$Y=bar{Y}implies frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}=frac{6bar{x^2} + 5bar{x} + 6}{3bar{x^2} + 10bar{x} + 3}$$
            Cross multiply, and cancel lot of terms to get
            $$(xbar{x}-1)(x-bar{x})=0$$
            $$x=bar{x} implies z=pm bar{z}$$
            $$xbar{x}=1implies |z|=1.$$






            share|cite|improve this answer









            $endgroup$



            For a complex number to be purely real, $z=bar{z}$. So, taking $x=z^2,$
            $$Y=frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}$$
            $$Y=bar{Y}implies frac{6x^2 + 5x + 6}{3x^2 + 10x + 3}=frac{6bar{x^2} + 5bar{x} + 6}{3bar{x^2} + 10bar{x} + 3}$$
            Cross multiply, and cancel lot of terms to get
            $$(xbar{x}-1)(x-bar{x})=0$$
            $$x=bar{x} implies z=pm bar{z}$$
            $$xbar{x}=1implies |z|=1.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 14:42









            Ankit KumarAnkit Kumar

            1,516221




            1,516221























                2












                $begingroup$

                You can express:
                $$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=2-frac{15z^2}{3z^4 + 10z^2 + 3}=2-frac{15}{3z^2+10+frac3{x^2}}=2-frac{15}{3(z+frac1z)^2+4} Rightarrow \ left(z+frac1zright)^2=left(a+bi+frac1{a+bi}right)^2=left(a+bi+frac{a-bi}{a^2+b^2}right)^2 Rightarrow \
                1) a^3+ab^2+a=0 text{or} 2) a^2b+b^3-b=0 Rightarrow \
                1) a=0 text{or} 2) b=0 text{or} a^2+b^2=1.$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Nicely done! I like it.
                  $endgroup$
                  – Mark Bennet
                  Dec 24 '18 at 17:21










                • $begingroup$
                  I’m glad to hear that. Thank you.
                  $endgroup$
                  – farruhota
                  Dec 24 '18 at 18:50
















                2












                $begingroup$

                You can express:
                $$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=2-frac{15z^2}{3z^4 + 10z^2 + 3}=2-frac{15}{3z^2+10+frac3{x^2}}=2-frac{15}{3(z+frac1z)^2+4} Rightarrow \ left(z+frac1zright)^2=left(a+bi+frac1{a+bi}right)^2=left(a+bi+frac{a-bi}{a^2+b^2}right)^2 Rightarrow \
                1) a^3+ab^2+a=0 text{or} 2) a^2b+b^3-b=0 Rightarrow \
                1) a=0 text{or} 2) b=0 text{or} a^2+b^2=1.$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Nicely done! I like it.
                  $endgroup$
                  – Mark Bennet
                  Dec 24 '18 at 17:21










                • $begingroup$
                  I’m glad to hear that. Thank you.
                  $endgroup$
                  – farruhota
                  Dec 24 '18 at 18:50














                2












                2








                2





                $begingroup$

                You can express:
                $$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=2-frac{15z^2}{3z^4 + 10z^2 + 3}=2-frac{15}{3z^2+10+frac3{x^2}}=2-frac{15}{3(z+frac1z)^2+4} Rightarrow \ left(z+frac1zright)^2=left(a+bi+frac1{a+bi}right)^2=left(a+bi+frac{a-bi}{a^2+b^2}right)^2 Rightarrow \
                1) a^3+ab^2+a=0 text{or} 2) a^2b+b^3-b=0 Rightarrow \
                1) a=0 text{or} 2) b=0 text{or} a^2+b^2=1.$$






                share|cite|improve this answer









                $endgroup$



                You can express:
                $$frac{6z^4 + 5z^2 + 6}{3z^4 + 10z^2 + 3}=2-frac{15z^2}{3z^4 + 10z^2 + 3}=2-frac{15}{3z^2+10+frac3{x^2}}=2-frac{15}{3(z+frac1z)^2+4} Rightarrow \ left(z+frac1zright)^2=left(a+bi+frac1{a+bi}right)^2=left(a+bi+frac{a-bi}{a^2+b^2}right)^2 Rightarrow \
                1) a^3+ab^2+a=0 text{or} 2) a^2b+b^3-b=0 Rightarrow \
                1) a=0 text{or} 2) b=0 text{or} a^2+b^2=1.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 15:09









                farruhotafarruhota

                20.7k2740




                20.7k2740












                • $begingroup$
                  Nicely done! I like it.
                  $endgroup$
                  – Mark Bennet
                  Dec 24 '18 at 17:21










                • $begingroup$
                  I’m glad to hear that. Thank you.
                  $endgroup$
                  – farruhota
                  Dec 24 '18 at 18:50


















                • $begingroup$
                  Nicely done! I like it.
                  $endgroup$
                  – Mark Bennet
                  Dec 24 '18 at 17:21










                • $begingroup$
                  I’m glad to hear that. Thank you.
                  $endgroup$
                  – farruhota
                  Dec 24 '18 at 18:50
















                $begingroup$
                Nicely done! I like it.
                $endgroup$
                – Mark Bennet
                Dec 24 '18 at 17:21




                $begingroup$
                Nicely done! I like it.
                $endgroup$
                – Mark Bennet
                Dec 24 '18 at 17:21












                $begingroup$
                I’m glad to hear that. Thank you.
                $endgroup$
                – farruhota
                Dec 24 '18 at 18:50




                $begingroup$
                I’m glad to hear that. Thank you.
                $endgroup$
                – farruhota
                Dec 24 '18 at 18:50











                1












                $begingroup$

                Hint: Plug in your formula $$z=x+iy$$ then you will get$$24x^3y-24xy^3+10xy=0$$ and $$12x^3y-12xy^3+20xy=0$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Hint: Plug in your formula $$z=x+iy$$ then you will get$$24x^3y-24xy^3+10xy=0$$ and $$12x^3y-12xy^3+20xy=0$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hint: Plug in your formula $$z=x+iy$$ then you will get$$24x^3y-24xy^3+10xy=0$$ and $$12x^3y-12xy^3+20xy=0$$






                    share|cite|improve this answer









                    $endgroup$



                    Hint: Plug in your formula $$z=x+iy$$ then you will get$$24x^3y-24xy^3+10xy=0$$ and $$12x^3y-12xy^3+20xy=0$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 24 '18 at 14:43









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    77k42866




                    77k42866























                        1












                        $begingroup$

                        This is probably too much of a hint, but is based on my comment.
                        I suggest dividing numerator and denominator by $z^2$ and setting $y=z^2+frac 1{z^2}$ when your fraction becomes $$frac {6y+5}{3y+10}$$ and it is easy now to show that $y$ has to be real, and if $y$ is real so is the fraction. So set $y=r$, an arbitrary real number, whence $z^4-rz^2+1=0$.



                        This can be solved in various ways, which include $$z^4-2z^2+2-(r-2)z^2=(z^2+zsqrt {r-2}+1)(z^2-zsqrt {r-2}+1)=0$$



                        If we now put $r-2=4R$ with $R$ a real number, this has solutions $$z=pm sqrt Rpm sqrt {R-1}$$and though this is a clunky way through in the end, we get $z$ real, or $z$ pure imaginary, or $|z|=1$ when $Rin [0,1]$.



                        Others have found easier ways through to these conditions.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          This is probably too much of a hint, but is based on my comment.
                          I suggest dividing numerator and denominator by $z^2$ and setting $y=z^2+frac 1{z^2}$ when your fraction becomes $$frac {6y+5}{3y+10}$$ and it is easy now to show that $y$ has to be real, and if $y$ is real so is the fraction. So set $y=r$, an arbitrary real number, whence $z^4-rz^2+1=0$.



                          This can be solved in various ways, which include $$z^4-2z^2+2-(r-2)z^2=(z^2+zsqrt {r-2}+1)(z^2-zsqrt {r-2}+1)=0$$



                          If we now put $r-2=4R$ with $R$ a real number, this has solutions $$z=pm sqrt Rpm sqrt {R-1}$$and though this is a clunky way through in the end, we get $z$ real, or $z$ pure imaginary, or $|z|=1$ when $Rin [0,1]$.



                          Others have found easier ways through to these conditions.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            This is probably too much of a hint, but is based on my comment.
                            I suggest dividing numerator and denominator by $z^2$ and setting $y=z^2+frac 1{z^2}$ when your fraction becomes $$frac {6y+5}{3y+10}$$ and it is easy now to show that $y$ has to be real, and if $y$ is real so is the fraction. So set $y=r$, an arbitrary real number, whence $z^4-rz^2+1=0$.



                            This can be solved in various ways, which include $$z^4-2z^2+2-(r-2)z^2=(z^2+zsqrt {r-2}+1)(z^2-zsqrt {r-2}+1)=0$$



                            If we now put $r-2=4R$ with $R$ a real number, this has solutions $$z=pm sqrt Rpm sqrt {R-1}$$and though this is a clunky way through in the end, we get $z$ real, or $z$ pure imaginary, or $|z|=1$ when $Rin [0,1]$.



                            Others have found easier ways through to these conditions.






                            share|cite|improve this answer











                            $endgroup$



                            This is probably too much of a hint, but is based on my comment.
                            I suggest dividing numerator and denominator by $z^2$ and setting $y=z^2+frac 1{z^2}$ when your fraction becomes $$frac {6y+5}{3y+10}$$ and it is easy now to show that $y$ has to be real, and if $y$ is real so is the fraction. So set $y=r$, an arbitrary real number, whence $z^4-rz^2+1=0$.



                            This can be solved in various ways, which include $$z^4-2z^2+2-(r-2)z^2=(z^2+zsqrt {r-2}+1)(z^2-zsqrt {r-2}+1)=0$$



                            If we now put $r-2=4R$ with $R$ a real number, this has solutions $$z=pm sqrt Rpm sqrt {R-1}$$and though this is a clunky way through in the end, we get $z$ real, or $z$ pure imaginary, or $|z|=1$ when $Rin [0,1]$.



                            Others have found easier ways through to these conditions.







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                            edited Dec 24 '18 at 19:00

























                            answered Dec 24 '18 at 15:03









                            Mark BennetMark Bennet

                            81.5k984181




                            81.5k984181






























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