Solving $z^3=1+i sqrt3$












1












$begingroup$


How can I solve the complex equation $$z^3=1+i sqrt3$$



Splitting $z$ into $z=a+bi$ gives me the mess



$$a^3+3a^2b^2i-3ab^2-b^3i-i sqrt3-1=0$$



where I dont know how to continue. I have never really worked with complex numbers, so I am not familiar with the "tricks" or steps that have to be done.






complex-numbers roots







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  • $begingroup$
    maybe read your friendly text book first.
    $endgroup$
    – Arjang
    Jan 15 at 12:14
















1












$begingroup$


How can I solve the complex equation $$z^3=1+i sqrt3$$



Splitting $z$ into $z=a+bi$ gives me the mess



$$a^3+3a^2b^2i-3ab^2-b^3i-i sqrt3-1=0$$



where I dont know how to continue. I have never really worked with complex numbers, so I am not familiar with the "tricks" or steps that have to be done.






complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    maybe read your friendly text book first.
    $endgroup$
    – Arjang
    Jan 15 at 12:14














1












1








1





$begingroup$


How can I solve the complex equation $$z^3=1+i sqrt3$$



Splitting $z$ into $z=a+bi$ gives me the mess



$$a^3+3a^2b^2i-3ab^2-b^3i-i sqrt3-1=0$$



where I dont know how to continue. I have never really worked with complex numbers, so I am not familiar with the "tricks" or steps that have to be done.






complex-numbers roots







share|cite|improve this question











$endgroup$




How can I solve the complex equation $$z^3=1+i sqrt3$$



Splitting $z$ into $z=a+bi$ gives me the mess



$$a^3+3a^2b^2i-3ab^2-b^3i-i sqrt3-1=0$$



where I dont know how to continue. I have never really worked with complex numbers, so I am not familiar with the "tricks" or steps that have to be done.






complex-numbers roots




complex-numbers roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 12:24









José Carlos Santos

162k22130233




162k22130233










asked Jan 15 at 12:13









TeslaTesla

885426




885426












  • $begingroup$
    maybe read your friendly text book first.
    $endgroup$
    – Arjang
    Jan 15 at 12:14


















  • $begingroup$
    maybe read your friendly text book first.
    $endgroup$
    – Arjang
    Jan 15 at 12:14
















$begingroup$
maybe read your friendly text book first.
$endgroup$
– Arjang
Jan 15 at 12:14




$begingroup$
maybe read your friendly text book first.
$endgroup$
– Arjang
Jan 15 at 12:14










4 Answers
4






active

oldest

votes


















5












$begingroup$

You approach this using the fact that$$1+isqrt3=2left(cosleft(fracpi3right)+isinleft(fracpi3right)right)$$and using de Moivre's formula.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
    $endgroup$
    – Tesla
    Jan 15 at 13:37










  • $begingroup$
    That's one of three solutions, yes.
    $endgroup$
    – José Carlos Santos
    Jan 15 at 13:38



















4












$begingroup$

Note that $$a^3-3ab^2-1=0$$ and $$3a^2b^2-b^3-sqrt{3}=0$$ must hold. So we obtain $$b=3,{frac {a left( {a}^{4}-sqrt {3}-a right) }{{a}^{3}-1}}$$ and then you can compute $$a$$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is true, but it's not so easy to solve.
    $endgroup$
    – Botond
    Jan 15 at 13:19










  • $begingroup$
    Should i post the solution for you?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 15 at 14:01










  • $begingroup$
    But how do you solve it easily?
    $endgroup$
    – Botond
    Jan 15 at 14:26



















3












$begingroup$

For products of complex numbers it's often easiest to use the polar form, $z = r e^{i phi}$, which yields $z^3 = r^3 e^{3 i phi}$. Converting the right-hand side of the equation to this form yields $1 + i sqrt{3} = 2 e^{i frac{pi}{3}}$ (cf. the hint of José Carlos Santos). Now you compare both sides to obtain two easily solvable equations for $r$ and $phi$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    More generally, by using the exponential form (instead of the cartesian one)
    $$1+i sqrt3=2exp(ipi/3),$$
    it is easier to solve the equation $z^n=1+i sqrt3$ where $n$ is a positive integer. There are $n$ distinct solutions given by
    $$z_k=2^{1/n}expleft(ifrac{pi/3+2kpi}{n}right)quad text{for $k=0,1,2,dots,n-1$}.$$
    As a reference take a look at $n$-roots of a complex number.






    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









      5












      $begingroup$

      You approach this using the fact that$$1+isqrt3=2left(cosleft(fracpi3right)+isinleft(fracpi3right)right)$$and using de Moivre's formula.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
        $endgroup$
        – Tesla
        Jan 15 at 13:37










      • $begingroup$
        That's one of three solutions, yes.
        $endgroup$
        – José Carlos Santos
        Jan 15 at 13:38
















      5












      $begingroup$

      You approach this using the fact that$$1+isqrt3=2left(cosleft(fracpi3right)+isinleft(fracpi3right)right)$$and using de Moivre's formula.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
        $endgroup$
        – Tesla
        Jan 15 at 13:37










      • $begingroup$
        That's one of three solutions, yes.
        $endgroup$
        – José Carlos Santos
        Jan 15 at 13:38














      5












      5








      5





      $begingroup$

      You approach this using the fact that$$1+isqrt3=2left(cosleft(fracpi3right)+isinleft(fracpi3right)right)$$and using de Moivre's formula.






      share|cite|improve this answer









      $endgroup$



      You approach this using the fact that$$1+isqrt3=2left(cosleft(fracpi3right)+isinleft(fracpi3right)right)$$and using de Moivre's formula.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 15 at 12:15









      José Carlos SantosJosé Carlos Santos

      162k22130233




      162k22130233












      • $begingroup$
        Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
        $endgroup$
        – Tesla
        Jan 15 at 13:37










      • $begingroup$
        That's one of three solutions, yes.
        $endgroup$
        – José Carlos Santos
        Jan 15 at 13:38


















      • $begingroup$
        Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
        $endgroup$
        – Tesla
        Jan 15 at 13:37










      • $begingroup$
        That's one of three solutions, yes.
        $endgroup$
        – José Carlos Santos
        Jan 15 at 13:38
















      $begingroup$
      Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
      $endgroup$
      – Tesla
      Jan 15 at 13:37




      $begingroup$
      Thanks. So what I get is $a=2^{frac{1}{3}}cos({pi over 9}), b=2^{frac{1}{3}} sin({frac{pi}{9}})$?
      $endgroup$
      – Tesla
      Jan 15 at 13:37












      $begingroup$
      That's one of three solutions, yes.
      $endgroup$
      – José Carlos Santos
      Jan 15 at 13:38




      $begingroup$
      That's one of three solutions, yes.
      $endgroup$
      – José Carlos Santos
      Jan 15 at 13:38











      4












      $begingroup$

      Note that $$a^3-3ab^2-1=0$$ and $$3a^2b^2-b^3-sqrt{3}=0$$ must hold. So we obtain $$b=3,{frac {a left( {a}^{4}-sqrt {3}-a right) }{{a}^{3}-1}}$$ and then you can compute $$a$$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This is true, but it's not so easy to solve.
        $endgroup$
        – Botond
        Jan 15 at 13:19










      • $begingroup$
        Should i post the solution for you?
        $endgroup$
        – Dr. Sonnhard Graubner
        Jan 15 at 14:01










      • $begingroup$
        But how do you solve it easily?
        $endgroup$
        – Botond
        Jan 15 at 14:26
















      4












      $begingroup$

      Note that $$a^3-3ab^2-1=0$$ and $$3a^2b^2-b^3-sqrt{3}=0$$ must hold. So we obtain $$b=3,{frac {a left( {a}^{4}-sqrt {3}-a right) }{{a}^{3}-1}}$$ and then you can compute $$a$$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This is true, but it's not so easy to solve.
        $endgroup$
        – Botond
        Jan 15 at 13:19










      • $begingroup$
        Should i post the solution for you?
        $endgroup$
        – Dr. Sonnhard Graubner
        Jan 15 at 14:01










      • $begingroup$
        But how do you solve it easily?
        $endgroup$
        – Botond
        Jan 15 at 14:26














      4












      4








      4





      $begingroup$

      Note that $$a^3-3ab^2-1=0$$ and $$3a^2b^2-b^3-sqrt{3}=0$$ must hold. So we obtain $$b=3,{frac {a left( {a}^{4}-sqrt {3}-a right) }{{a}^{3}-1}}$$ and then you can compute $$a$$.






      share|cite|improve this answer











      $endgroup$



      Note that $$a^3-3ab^2-1=0$$ and $$3a^2b^2-b^3-sqrt{3}=0$$ must hold. So we obtain $$b=3,{frac {a left( {a}^{4}-sqrt {3}-a right) }{{a}^{3}-1}}$$ and then you can compute $$a$$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 15 at 14:10

























      answered Jan 15 at 12:16









      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

      75.7k42866




      75.7k42866








      • 1




        $begingroup$
        This is true, but it's not so easy to solve.
        $endgroup$
        – Botond
        Jan 15 at 13:19










      • $begingroup$
        Should i post the solution for you?
        $endgroup$
        – Dr. Sonnhard Graubner
        Jan 15 at 14:01










      • $begingroup$
        But how do you solve it easily?
        $endgroup$
        – Botond
        Jan 15 at 14:26














      • 1




        $begingroup$
        This is true, but it's not so easy to solve.
        $endgroup$
        – Botond
        Jan 15 at 13:19










      • $begingroup$
        Should i post the solution for you?
        $endgroup$
        – Dr. Sonnhard Graubner
        Jan 15 at 14:01










      • $begingroup$
        But how do you solve it easily?
        $endgroup$
        – Botond
        Jan 15 at 14:26








      1




      1




      $begingroup$
      This is true, but it's not so easy to solve.
      $endgroup$
      – Botond
      Jan 15 at 13:19




      $begingroup$
      This is true, but it's not so easy to solve.
      $endgroup$
      – Botond
      Jan 15 at 13:19












      $begingroup$
      Should i post the solution for you?
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 15 at 14:01




      $begingroup$
      Should i post the solution for you?
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 15 at 14:01












      $begingroup$
      But how do you solve it easily?
      $endgroup$
      – Botond
      Jan 15 at 14:26




      $begingroup$
      But how do you solve it easily?
      $endgroup$
      – Botond
      Jan 15 at 14:26











      3












      $begingroup$

      For products of complex numbers it's often easiest to use the polar form, $z = r e^{i phi}$, which yields $z^3 = r^3 e^{3 i phi}$. Converting the right-hand side of the equation to this form yields $1 + i sqrt{3} = 2 e^{i frac{pi}{3}}$ (cf. the hint of José Carlos Santos). Now you compare both sides to obtain two easily solvable equations for $r$ and $phi$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        For products of complex numbers it's often easiest to use the polar form, $z = r e^{i phi}$, which yields $z^3 = r^3 e^{3 i phi}$. Converting the right-hand side of the equation to this form yields $1 + i sqrt{3} = 2 e^{i frac{pi}{3}}$ (cf. the hint of José Carlos Santos). Now you compare both sides to obtain two easily solvable equations for $r$ and $phi$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          For products of complex numbers it's often easiest to use the polar form, $z = r e^{i phi}$, which yields $z^3 = r^3 e^{3 i phi}$. Converting the right-hand side of the equation to this form yields $1 + i sqrt{3} = 2 e^{i frac{pi}{3}}$ (cf. the hint of José Carlos Santos). Now you compare both sides to obtain two easily solvable equations for $r$ and $phi$.






          share|cite|improve this answer









          $endgroup$



          For products of complex numbers it's often easiest to use the polar form, $z = r e^{i phi}$, which yields $z^3 = r^3 e^{3 i phi}$. Converting the right-hand side of the equation to this form yields $1 + i sqrt{3} = 2 e^{i frac{pi}{3}}$ (cf. the hint of José Carlos Santos). Now you compare both sides to obtain two easily solvable equations for $r$ and $phi$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 12:29









          ChristophChristoph

          58116




          58116























              2












              $begingroup$

              More generally, by using the exponential form (instead of the cartesian one)
              $$1+i sqrt3=2exp(ipi/3),$$
              it is easier to solve the equation $z^n=1+i sqrt3$ where $n$ is a positive integer. There are $n$ distinct solutions given by
              $$z_k=2^{1/n}expleft(ifrac{pi/3+2kpi}{n}right)quad text{for $k=0,1,2,dots,n-1$}.$$
              As a reference take a look at $n$-roots of a complex number.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                More generally, by using the exponential form (instead of the cartesian one)
                $$1+i sqrt3=2exp(ipi/3),$$
                it is easier to solve the equation $z^n=1+i sqrt3$ where $n$ is a positive integer. There are $n$ distinct solutions given by
                $$z_k=2^{1/n}expleft(ifrac{pi/3+2kpi}{n}right)quad text{for $k=0,1,2,dots,n-1$}.$$
                As a reference take a look at $n$-roots of a complex number.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  More generally, by using the exponential form (instead of the cartesian one)
                  $$1+i sqrt3=2exp(ipi/3),$$
                  it is easier to solve the equation $z^n=1+i sqrt3$ where $n$ is a positive integer. There are $n$ distinct solutions given by
                  $$z_k=2^{1/n}expleft(ifrac{pi/3+2kpi}{n}right)quad text{for $k=0,1,2,dots,n-1$}.$$
                  As a reference take a look at $n$-roots of a complex number.






                  share|cite|improve this answer











                  $endgroup$



                  More generally, by using the exponential form (instead of the cartesian one)
                  $$1+i sqrt3=2exp(ipi/3),$$
                  it is easier to solve the equation $z^n=1+i sqrt3$ where $n$ is a positive integer. There are $n$ distinct solutions given by
                  $$z_k=2^{1/n}expleft(ifrac{pi/3+2kpi}{n}right)quad text{for $k=0,1,2,dots,n-1$}.$$
                  As a reference take a look at $n$-roots of a complex number.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 15 at 12:35

























                  answered Jan 15 at 12:27









                  Robert ZRobert Z

                  98.3k1067139




                  98.3k1067139






























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