Does $(a + bi)^3 = (x+yi)^3 implies a + bi = x + yi$?












0












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Does $(a + bi)^3 = (x+yi)^3 implies a + bi = x + yi$, where $a,b,x,y in mathbb{R}?$




Please provide a proof/counterexample. I've tried brute force and ended up with:



$$(a - x)(a^2 + ax + x^2) + 3(-ab^2 + xy^2) = 0$$



$$(y - b)(y^2 + by + b^2) + 3(a^2b - x^2y) = 0$$






complex-numbers







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  • 1




    $begingroup$
    what are $c$ and $d$?
    $endgroup$
    – asd
    Feb 2 at 11:12










  • $begingroup$
    @asd Sorry, I meant $x,y$
    $endgroup$
    – user168651
    Feb 2 at 11:13












  • $begingroup$
    you mean $(a+ib)^3 = (x+yi)^3 Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false...
    $endgroup$
    – Matteo
    Feb 2 at 11:16
















0












$begingroup$



Does $(a + bi)^3 = (x+yi)^3 implies a + bi = x + yi$, where $a,b,x,y in mathbb{R}?$




Please provide a proof/counterexample. I've tried brute force and ended up with:



$$(a - x)(a^2 + ax + x^2) + 3(-ab^2 + xy^2) = 0$$



$$(y - b)(y^2 + by + b^2) + 3(a^2b - x^2y) = 0$$






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    what are $c$ and $d$?
    $endgroup$
    – asd
    Feb 2 at 11:12










  • $begingroup$
    @asd Sorry, I meant $x,y$
    $endgroup$
    – user168651
    Feb 2 at 11:13












  • $begingroup$
    you mean $(a+ib)^3 = (x+yi)^3 Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false...
    $endgroup$
    – Matteo
    Feb 2 at 11:16














0












0








0





$begingroup$



Does $(a + bi)^3 = (x+yi)^3 implies a + bi = x + yi$, where $a,b,x,y in mathbb{R}?$




Please provide a proof/counterexample. I've tried brute force and ended up with:



$$(a - x)(a^2 + ax + x^2) + 3(-ab^2 + xy^2) = 0$$



$$(y - b)(y^2 + by + b^2) + 3(a^2b - x^2y) = 0$$






complex-numbers







share|cite|improve this question











$endgroup$





Does $(a + bi)^3 = (x+yi)^3 implies a + bi = x + yi$, where $a,b,x,y in mathbb{R}?$




Please provide a proof/counterexample. I've tried brute force and ended up with:



$$(a - x)(a^2 + ax + x^2) + 3(-ab^2 + xy^2) = 0$$



$$(y - b)(y^2 + by + b^2) + 3(a^2b - x^2y) = 0$$






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 11:58









jvdhooft

5,65961641




5,65961641










asked Feb 2 at 11:11









user168651user168651

62




62








  • 1




    $begingroup$
    what are $c$ and $d$?
    $endgroup$
    – asd
    Feb 2 at 11:12










  • $begingroup$
    @asd Sorry, I meant $x,y$
    $endgroup$
    – user168651
    Feb 2 at 11:13












  • $begingroup$
    you mean $(a+ib)^3 = (x+yi)^3 Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false...
    $endgroup$
    – Matteo
    Feb 2 at 11:16














  • 1




    $begingroup$
    what are $c$ and $d$?
    $endgroup$
    – asd
    Feb 2 at 11:12










  • $begingroup$
    @asd Sorry, I meant $x,y$
    $endgroup$
    – user168651
    Feb 2 at 11:13












  • $begingroup$
    you mean $(a+ib)^3 = (x+yi)^3 Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false...
    $endgroup$
    – Matteo
    Feb 2 at 11:16








1




1




$begingroup$
what are $c$ and $d$?
$endgroup$
– asd
Feb 2 at 11:12




$begingroup$
what are $c$ and $d$?
$endgroup$
– asd
Feb 2 at 11:12












$begingroup$
@asd Sorry, I meant $x,y$
$endgroup$
– user168651
Feb 2 at 11:13






$begingroup$
@asd Sorry, I meant $x,y$
$endgroup$
– user168651
Feb 2 at 11:13














$begingroup$
you mean $(a+ib)^3 = (x+yi)^3 Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false...
$endgroup$
– Matteo
Feb 2 at 11:16




$begingroup$
you mean $(a+ib)^3 = (x+yi)^3 Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false...
$endgroup$
– Matteo
Feb 2 at 11:16










3 Answers
3






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$begingroup$

Take $a+ib=e^{frac{2pi i}{3}}$ and $x+iy=1$ We have
$$
(a+ib)^3=(e^{frac{2pi i}{3}})^3=e^{2pi i}=1=1^3
$$
but $a+ibneq 1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    If $z^3=w^3$, then $$z^3=(we^{2pi ik})^3\z=we^{frac{2pi ik}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:




    • $a+bi=(x+iy)cdot 1$

    • $a+bi=(x+iy)cdot e^{frac{2pi i}{3}}$

    • $a+bi=(x+iy)cdot e^{frac{4pi i}{3}}$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      No. There are $3$ roots of $z^3=z_0$ for any $0not=z_0inBbb C$.



      If $gamma$ is one, then $sigmagamma,sigma ^2gamma $ are the other two, where $sigma =e^{frac{2pi i}3}$.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Take $a+ib=e^{frac{2pi i}{3}}$ and $x+iy=1$ We have
        $$
        (a+ib)^3=(e^{frac{2pi i}{3}})^3=e^{2pi i}=1=1^3
        $$
        but $a+ibneq 1$.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Take $a+ib=e^{frac{2pi i}{3}}$ and $x+iy=1$ We have
          $$
          (a+ib)^3=(e^{frac{2pi i}{3}})^3=e^{2pi i}=1=1^3
          $$
          but $a+ibneq 1$.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Take $a+ib=e^{frac{2pi i}{3}}$ and $x+iy=1$ We have
            $$
            (a+ib)^3=(e^{frac{2pi i}{3}})^3=e^{2pi i}=1=1^3
            $$
            but $a+ibneq 1$.






            share|cite|improve this answer









            $endgroup$



            Take $a+ib=e^{frac{2pi i}{3}}$ and $x+iy=1$ We have
            $$
            (a+ib)^3=(e^{frac{2pi i}{3}})^3=e^{2pi i}=1=1^3
            $$
            but $a+ibneq 1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 2 at 11:17









            asdasd

            30729




            30729























                2












                $begingroup$

                If $z^3=w^3$, then $$z^3=(we^{2pi ik})^3\z=we^{frac{2pi ik}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:




                • $a+bi=(x+iy)cdot 1$

                • $a+bi=(x+iy)cdot e^{frac{2pi i}{3}}$

                • $a+bi=(x+iy)cdot e^{frac{4pi i}{3}}$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  If $z^3=w^3$, then $$z^3=(we^{2pi ik})^3\z=we^{frac{2pi ik}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:




                  • $a+bi=(x+iy)cdot 1$

                  • $a+bi=(x+iy)cdot e^{frac{2pi i}{3}}$

                  • $a+bi=(x+iy)cdot e^{frac{4pi i}{3}}$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    If $z^3=w^3$, then $$z^3=(we^{2pi ik})^3\z=we^{frac{2pi ik}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:




                    • $a+bi=(x+iy)cdot 1$

                    • $a+bi=(x+iy)cdot e^{frac{2pi i}{3}}$

                    • $a+bi=(x+iy)cdot e^{frac{4pi i}{3}}$






                    share|cite|improve this answer









                    $endgroup$



                    If $z^3=w^3$, then $$z^3=(we^{2pi ik})^3\z=we^{frac{2pi ik}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:




                    • $a+bi=(x+iy)cdot 1$

                    • $a+bi=(x+iy)cdot e^{frac{2pi i}{3}}$

                    • $a+bi=(x+iy)cdot e^{frac{4pi i}{3}}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 2 at 11:23









                    cansomeonehelpmeoutcansomeonehelpmeout

                    7,3323935




                    7,3323935























                        0












                        $begingroup$

                        No. There are $3$ roots of $z^3=z_0$ for any $0not=z_0inBbb C$.



                        If $gamma$ is one, then $sigmagamma,sigma ^2gamma $ are the other two, where $sigma =e^{frac{2pi i}3}$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          No. There are $3$ roots of $z^3=z_0$ for any $0not=z_0inBbb C$.



                          If $gamma$ is one, then $sigmagamma,sigma ^2gamma $ are the other two, where $sigma =e^{frac{2pi i}3}$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            No. There are $3$ roots of $z^3=z_0$ for any $0not=z_0inBbb C$.



                            If $gamma$ is one, then $sigmagamma,sigma ^2gamma $ are the other two, where $sigma =e^{frac{2pi i}3}$.






                            share|cite|improve this answer









                            $endgroup$



                            No. There are $3$ roots of $z^3=z_0$ for any $0not=z_0inBbb C$.



                            If $gamma$ is one, then $sigmagamma,sigma ^2gamma $ are the other two, where $sigma =e^{frac{2pi i}3}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 2 at 11:41









                            Chris CusterChris Custer

                            14.4k3827




                            14.4k3827






























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