Finding a complex number












0












$begingroup$


So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.



Here is the illustrative picture that I have:



enter image description here



I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$



Are my calculations right?






complex-numbers







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$endgroup$












  • $begingroup$
    Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
    $endgroup$
    – EdOverflow
    Jan 7 at 21:45










  • $begingroup$
    Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
    $endgroup$
    – Emilio Novati
    Jan 7 at 21:51












  • $begingroup$
    no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
    $endgroup$
    – Student123
    Jan 7 at 21:59


















0












$begingroup$


So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.



Here is the illustrative picture that I have:



enter image description here



I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$



Are my calculations right?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
    $endgroup$
    – EdOverflow
    Jan 7 at 21:45










  • $begingroup$
    Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
    $endgroup$
    – Emilio Novati
    Jan 7 at 21:51












  • $begingroup$
    no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
    $endgroup$
    – Student123
    Jan 7 at 21:59
















0












0








0





$begingroup$


So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.



Here is the illustrative picture that I have:



enter image description here



I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$



Are my calculations right?






complex-numbers







share|cite|improve this question











$endgroup$




So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.



Here is the illustrative picture that I have:



enter image description here



I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$



Are my calculations right?






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 Jan 7 at 22:27







Student123

















asked Jan 7 at 21:19









Student123Student123

536




536












  • $begingroup$
    Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
    $endgroup$
    – EdOverflow
    Jan 7 at 21:45










  • $begingroup$
    Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
    $endgroup$
    – Emilio Novati
    Jan 7 at 21:51












  • $begingroup$
    no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
    $endgroup$
    – Student123
    Jan 7 at 21:59




















  • $begingroup$
    Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
    $endgroup$
    – EdOverflow
    Jan 7 at 21:45










  • $begingroup$
    Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
    $endgroup$
    – Emilio Novati
    Jan 7 at 21:51












  • $begingroup$
    no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
    $endgroup$
    – Student123
    Jan 7 at 21:59


















$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45




$begingroup$
Hint: If you refer to your diagram, $text{Im(}z)$ is negative, but your solution concludes that the imaginary part is positive.
$endgroup$
– EdOverflow
Jan 7 at 21:45












$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51






$begingroup$
Your question is not clear. Are you searching the number $z$ such that $w^6=z$ with $w=1.7e^{-ifrac{61pi}{180}}$?
$endgroup$
– Emilio Novati
Jan 7 at 21:51














$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59






$begingroup$
no, actually $ w^x=z $ this is just illustrative picture that was given to help you understand the problem
$endgroup$
– Student123
Jan 7 at 21:59












3 Answers
3






active

oldest

votes


















1












$begingroup$

Assuming I understand the question correctly, we have:
$$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
$$therefore z = w^6 = 24.00534-2.52306i$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I think you're close.



    If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.



    However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).



    You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.



    Do you see how to fix it?



    That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I assume z=0.824-1.487i maybe?
      $endgroup$
      – Student123
      Jan 7 at 21:44










    • $begingroup$
      @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
      $endgroup$
      – John Doe
      Jan 7 at 22:05












    • $begingroup$
      is it $z = w^6 = 24.00534-2.52306i$ ?
      $endgroup$
      – Student123
      Jan 7 at 22:39





















    -1












    $begingroup$

    Change the $+$ to a $-$ and you're golden.



    Presumably you have calculated
    $$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Assuming I understand the question correctly, we have:
      $$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
      Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
      $$therefore z = w^6 = 24.00534-2.52306i$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Assuming I understand the question correctly, we have:
        $$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
        Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
        $$therefore z = w^6 = 24.00534-2.52306i$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Assuming I understand the question correctly, we have:
          $$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
          Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
          $$therefore z = w^6 = 24.00534-2.52306i$$






          share|cite|improve this answer









          $endgroup$



          Assuming I understand the question correctly, we have:
          $$w=z^{frac{1}{6}}=frac{3.4}{2} cdot (cos{(-61)}+i cdot sin{(-61)}) = 0.82418-1.48685i$$
          Which is what you have. But this is $w=z^{frac{1}{6}}$ ;
          $$therefore z = w^6 = 24.00534-2.52306i$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 21:35









          Peter ForemanPeter Foreman

          7,1841318




          7,1841318























              0












              $begingroup$

              I think you're close.



              If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.



              However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).



              You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.



              Do you see how to fix it?



              That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I assume z=0.824-1.487i maybe?
                $endgroup$
                – Student123
                Jan 7 at 21:44










              • $begingroup$
                @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
                $endgroup$
                – John Doe
                Jan 7 at 22:05












              • $begingroup$
                is it $z = w^6 = 24.00534-2.52306i$ ?
                $endgroup$
                – Student123
                Jan 7 at 22:39


















              0












              $begingroup$

              I think you're close.



              If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.



              However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).



              You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.



              Do you see how to fix it?



              That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I assume z=0.824-1.487i maybe?
                $endgroup$
                – Student123
                Jan 7 at 21:44










              • $begingroup$
                @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
                $endgroup$
                – John Doe
                Jan 7 at 22:05












              • $begingroup$
                is it $z = w^6 = 24.00534-2.52306i$ ?
                $endgroup$
                – Student123
                Jan 7 at 22:39
















              0












              0








              0





              $begingroup$

              I think you're close.



              If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.



              However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).



              You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.



              Do you see how to fix it?



              That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.






              share|cite|improve this answer









              $endgroup$



              I think you're close.



              If I take $sqrt{0.824^2+1.487^2}$, this equals $1.7$, so that checks.



              However, if your angle is $-61^{circ}$, I'd expect the imaginary part to be negative (the angle lies in the fourth quadrant).



              You have $tan ^{-1} 1.487/0.824 = 61^{circ}$, so the magnitudes are right.



              Do you see how to fix it?



              That's one of the roots. Now for the other five roots, you add $60^{circ}, 120^{circ}, 180^{circ}, 240^{circ}, 300^{circ}$ to the angle.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 7 at 21:35









              JohnJohn

              22.8k32550




              22.8k32550












              • $begingroup$
                I assume z=0.824-1.487i maybe?
                $endgroup$
                – Student123
                Jan 7 at 21:44










              • $begingroup$
                @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
                $endgroup$
                – John Doe
                Jan 7 at 22:05












              • $begingroup$
                is it $z = w^6 = 24.00534-2.52306i$ ?
                $endgroup$
                – Student123
                Jan 7 at 22:39




















              • $begingroup$
                I assume z=0.824-1.487i maybe?
                $endgroup$
                – Student123
                Jan 7 at 21:44










              • $begingroup$
                @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
                $endgroup$
                – John Doe
                Jan 7 at 22:05












              • $begingroup$
                is it $z = w^6 = 24.00534-2.52306i$ ?
                $endgroup$
                – Student123
                Jan 7 at 22:39


















              $begingroup$
              I assume z=0.824-1.487i maybe?
              $endgroup$
              – Student123
              Jan 7 at 21:44




              $begingroup$
              I assume z=0.824-1.487i maybe?
              $endgroup$
              – Student123
              Jan 7 at 21:44












              $begingroup$
              @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
              $endgroup$
              – John Doe
              Jan 7 at 22:05






              $begingroup$
              @Student123 you have the right complex number, but that is not $z$ - it is in fact $w$. The points plotted on the graph are the roots of $z$. To get $z$ back you need to raise $w$ to a power. But yes, your answer is correct for $w$.
              $endgroup$
              – John Doe
              Jan 7 at 22:05














              $begingroup$
              is it $z = w^6 = 24.00534-2.52306i$ ?
              $endgroup$
              – Student123
              Jan 7 at 22:39






              $begingroup$
              is it $z = w^6 = 24.00534-2.52306i$ ?
              $endgroup$
              – Student123
              Jan 7 at 22:39













              -1












              $begingroup$

              Change the $+$ to a $-$ and you're golden.



              Presumably you have calculated
              $$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                Change the $+$ to a $-$ and you're golden.



                Presumably you have calculated
                $$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  Change the $+$ to a $-$ and you're golden.



                  Presumably you have calculated
                  $$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.






                  share|cite|improve this answer









                  $endgroup$



                  Change the $+$ to a $-$ and you're golden.



                  Presumably you have calculated
                  $$x = 1.7cos(-61^circ), y = 1.7sin(-61^circ),$$ which does indeed give the answer you suggest.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 21:34









                  bouncebackbounceback

                  446212




                  446212






























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