Calculate the argument of $ (-2{sqrt 3} - 2i)^5$












0












$begingroup$


So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.



Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$



where is my mistake?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
    $endgroup$
    – D.B.
    Jan 13 at 23:28










  • $begingroup$
    More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
    $endgroup$
    – obscurans
    Jan 13 at 23:38
















0












$begingroup$


So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.



Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$



where is my mistake?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
    $endgroup$
    – D.B.
    Jan 13 at 23:28










  • $begingroup$
    More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
    $endgroup$
    – obscurans
    Jan 13 at 23:38














0












0








0





$begingroup$


So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.



Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$



where is my mistake?






complex-numbers







share|cite|improve this question











$endgroup$




So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.



Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$



where is my mistake?






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 13 at 23:24









user376343

3,9834829




3,9834829










asked Jan 13 at 23:22









ythhtrgythhtrg

383




383












  • $begingroup$
    Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
    $endgroup$
    – D.B.
    Jan 13 at 23:28










  • $begingroup$
    More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
    $endgroup$
    – obscurans
    Jan 13 at 23:38


















  • $begingroup$
    Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
    $endgroup$
    – D.B.
    Jan 13 at 23:28










  • $begingroup$
    More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
    $endgroup$
    – obscurans
    Jan 13 at 23:38
















$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28




$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28












$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38




$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38










2 Answers
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$begingroup$

$$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
$$=arg(-1)+arg((2+2sqrt{3})^5)$$
$$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
    $$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
    $$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
    $$330^{circ}.$$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      $$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
      $$=arg(-1)+arg((2+2sqrt{3})^5)$$
      $$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        $$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
        $$=arg(-1)+arg((2+2sqrt{3})^5)$$
        $$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          $$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
          $$=arg(-1)+arg((2+2sqrt{3})^5)$$
          $$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$






          share|cite|improve this answer











          $endgroup$



          $$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
          $$=arg(-1)+arg((2+2sqrt{3})^5)$$
          $$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 13 at 23:44

























          answered Jan 13 at 23:34









          D.B.D.B.

          1,61029




          1,61029























              0












              $begingroup$

              $$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
              $$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
              $$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
              $$330^{circ}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
                $$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
                $$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
                $$330^{circ}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
                  $$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
                  $$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
                  $$330^{circ}.$$






                  share|cite|improve this answer









                  $endgroup$



                  $$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
                  $$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
                  $$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
                  $$330^{circ}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 23:30









                  Michael RozenbergMichael Rozenberg

                  111k1897201




                  111k1897201






























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