Finding roots using Euler's formula












2












$begingroup$


I'm going through past exam papers and have come accross the following question:



find all solutions to $z^5=2-2i $



for this question I was going to use Euler's formula:



$ e^{i(2k+1)pi}=-1 $



$ -2e^{i(2k+1)pi}=2 $



$ -2e^{i(2k+1)pi}-2i=2-2i $



$ z^5 =-2e^{i(2k+1)pi}-2i$



$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$



I'll then sub in k=0,1,2,3,4



am I on the right track here?
thanks!






complex-analysis complex-numbers







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$endgroup$












  • $begingroup$
    Try writing $2-2i$ in the $e^{x}$ form first
    $endgroup$
    – TheMathsGeek
    Jan 5 '17 at 12:47






  • 2




    $begingroup$
    $(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
    $endgroup$
    – ajotatxe
    Jan 5 '17 at 12:47
















2












$begingroup$


I'm going through past exam papers and have come accross the following question:



find all solutions to $z^5=2-2i $



for this question I was going to use Euler's formula:



$ e^{i(2k+1)pi}=-1 $



$ -2e^{i(2k+1)pi}=2 $



$ -2e^{i(2k+1)pi}-2i=2-2i $



$ z^5 =-2e^{i(2k+1)pi}-2i$



$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$



I'll then sub in k=0,1,2,3,4



am I on the right track here?
thanks!






complex-analysis complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try writing $2-2i$ in the $e^{x}$ form first
    $endgroup$
    – TheMathsGeek
    Jan 5 '17 at 12:47






  • 2




    $begingroup$
    $(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
    $endgroup$
    – ajotatxe
    Jan 5 '17 at 12:47














2












2








2





$begingroup$


I'm going through past exam papers and have come accross the following question:



find all solutions to $z^5=2-2i $



for this question I was going to use Euler's formula:



$ e^{i(2k+1)pi}=-1 $



$ -2e^{i(2k+1)pi}=2 $



$ -2e^{i(2k+1)pi}-2i=2-2i $



$ z^5 =-2e^{i(2k+1)pi}-2i$



$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$



I'll then sub in k=0,1,2,3,4



am I on the right track here?
thanks!






complex-analysis complex-numbers







share|cite|improve this question









$endgroup$




I'm going through past exam papers and have come accross the following question:



find all solutions to $z^5=2-2i $



for this question I was going to use Euler's formula:



$ e^{i(2k+1)pi}=-1 $



$ -2e^{i(2k+1)pi}=2 $



$ -2e^{i(2k+1)pi}-2i=2-2i $



$ z^5 =-2e^{i(2k+1)pi}-2i$



$ z = -2^{frac 1 5}e^{frac {i(2k+1)pi}{5}}-2i^{frac 1 5}$



I'll then sub in k=0,1,2,3,4



am I on the right track here?
thanks!






complex-analysis complex-numbers




complex-analysis complex-numbers






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share|cite|improve this question











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asked Jan 5 '17 at 12:42









user6122081user6122081

758




758












  • $begingroup$
    Try writing $2-2i$ in the $e^{x}$ form first
    $endgroup$
    – TheMathsGeek
    Jan 5 '17 at 12:47






  • 2




    $begingroup$
    $(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
    $endgroup$
    – ajotatxe
    Jan 5 '17 at 12:47


















  • $begingroup$
    Try writing $2-2i$ in the $e^{x}$ form first
    $endgroup$
    – TheMathsGeek
    Jan 5 '17 at 12:47






  • 2




    $begingroup$
    $(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
    $endgroup$
    – ajotatxe
    Jan 5 '17 at 12:47
















$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47




$begingroup$
Try writing $2-2i$ in the $e^{x}$ form first
$endgroup$
– TheMathsGeek
Jan 5 '17 at 12:47




2




2




$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47




$begingroup$
$(a+b)^{1/5}neq a^{1/5}+b^{1/5}$
$endgroup$
– ajotatxe
Jan 5 '17 at 12:47










4 Answers
4






active

oldest

votes


















0












$begingroup$

Hint:



$$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$



Therefore



$$2-2i=2sqrt 2e^{7pi/4}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Using the Euler's Formula is a good approach but I would use:



    $$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$



    $$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$



    Now compare both:



    $$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$



    Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:



    begin{cases}
    r^5=2sqrt{2}\
    e^{5xi}=e^{(7pi/4+2kpi)i}
    end{cases}



    P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.



    Can you finish?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
      $endgroup$
      – user6122081
      Jan 5 '17 at 12:54










    • $begingroup$
      I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
      $endgroup$
      – Arnaldo
      Jan 5 '17 at 12:56












    • $begingroup$
      @user6122081: are you ok now?
      $endgroup$
      – Arnaldo
      Jan 5 '17 at 21:11



















    0












    $begingroup$

    We can write $2-2i$ as:



    $z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$



    now we take the root:



    $z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$



    with $kin mathbb Z$, $k=0,1,2,3,4$.



    Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
      $endgroup$
      – Nosrati
      Jan 5 '17 at 14:12



















    0












    $begingroup$

    You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.



    So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).



    Now $z^5=2sqrt2e^{frac{7pi}4i}$.



    So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.



    There are $4$ more.



    Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.



    That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.






    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









      0












      $begingroup$

      Hint:



      $$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$



      Therefore



      $$2-2i=2sqrt 2e^{7pi/4}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Hint:



        $$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$



        Therefore



        $$2-2i=2sqrt 2e^{7pi/4}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint:



          $$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$



          Therefore



          $$2-2i=2sqrt 2e^{7pi/4}$$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$e^{7pi/4}=frac1{sqrt 2}-ifrac1{sqrt 2}$$



          Therefore



          $$2-2i=2sqrt 2e^{7pi/4}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 '17 at 12:50









          ajotatxeajotatxe

          53.8k23890




          53.8k23890























              0












              $begingroup$

              Using the Euler's Formula is a good approach but I would use:



              $$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$



              $$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Now compare both:



              $$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:



              begin{cases}
              r^5=2sqrt{2}\
              e^{5xi}=e^{(7pi/4+2kpi)i}
              end{cases}



              P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.



              Can you finish?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
                $endgroup$
                – user6122081
                Jan 5 '17 at 12:54










              • $begingroup$
                I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 12:56












              • $begingroup$
                @user6122081: are you ok now?
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 21:11
















              0












              $begingroup$

              Using the Euler's Formula is a good approach but I would use:



              $$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$



              $$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Now compare both:



              $$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:



              begin{cases}
              r^5=2sqrt{2}\
              e^{5xi}=e^{(7pi/4+2kpi)i}
              end{cases}



              P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.



              Can you finish?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
                $endgroup$
                – user6122081
                Jan 5 '17 at 12:54










              • $begingroup$
                I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 12:56












              • $begingroup$
                @user6122081: are you ok now?
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 21:11














              0












              0








              0





              $begingroup$

              Using the Euler's Formula is a good approach but I would use:



              $$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$



              $$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Now compare both:



              $$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:



              begin{cases}
              r^5=2sqrt{2}\
              e^{5xi}=e^{(7pi/4+2kpi)i}
              end{cases}



              P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.



              Can you finish?






              share|cite|improve this answer











              $endgroup$



              Using the Euler's Formula is a good approach but I would use:



              $$z=re^{ix} rightarrow z^5=r^5e^{5xi}$$



              $$2-2i=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Now compare both:



              $$r^5e^{5xi}=2sqrt{2}e^{(7pi/4+2kpi)i}$$



              Once both complex numbers are equal then they must have the same magnitude ($r^5=2sqrt{2}$) and also $e^{5xi}=e^{(7pi/4+2kpi)i}$, so:



              begin{cases}
              r^5=2sqrt{2}\
              e^{5xi}=e^{(7pi/4+2kpi)i}
              end{cases}



              P.S: Remember that in general if we have $u=r_1e^{ix_1}$ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.



              Can you finish?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 5 '17 at 13:07

























              answered Jan 5 '17 at 12:49









              ArnaldoArnaldo

              18.1k42246




              18.1k42246












              • $begingroup$
                just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
                $endgroup$
                – user6122081
                Jan 5 '17 at 12:54










              • $begingroup$
                I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 12:56












              • $begingroup$
                @user6122081: are you ok now?
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 21:11


















              • $begingroup$
                just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
                $endgroup$
                – user6122081
                Jan 5 '17 at 12:54










              • $begingroup$
                I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 12:56












              • $begingroup$
                @user6122081: are you ok now?
                $endgroup$
                – Arnaldo
                Jan 5 '17 at 21:11
















              $begingroup$
              just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
              $endgroup$
              – user6122081
              Jan 5 '17 at 12:54




              $begingroup$
              just confused at where you got $ e^{5xi}=e^{(7pi/4+2kpi)i} $ Thanks for your reply!
              $endgroup$
              – user6122081
              Jan 5 '17 at 12:54












              $begingroup$
              I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
              $endgroup$
              – Arnaldo
              Jan 5 '17 at 12:56






              $begingroup$
              I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$.
              $endgroup$
              – Arnaldo
              Jan 5 '17 at 12:56














              $begingroup$
              @user6122081: are you ok now?
              $endgroup$
              – Arnaldo
              Jan 5 '17 at 21:11




              $begingroup$
              @user6122081: are you ok now?
              $endgroup$
              – Arnaldo
              Jan 5 '17 at 21:11











              0












              $begingroup$

              We can write $2-2i$ as:



              $z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$



              now we take the root:



              $z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$



              with $kin mathbb Z$, $k=0,1,2,3,4$.



              Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
                $endgroup$
                – Nosrati
                Jan 5 '17 at 14:12
















              0












              $begingroup$

              We can write $2-2i$ as:



              $z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$



              now we take the root:



              $z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$



              with $kin mathbb Z$, $k=0,1,2,3,4$.



              Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
                $endgroup$
                – Nosrati
                Jan 5 '17 at 14:12














              0












              0








              0





              $begingroup$

              We can write $2-2i$ as:



              $z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$



              now we take the root:



              $z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$



              with $kin mathbb Z$, $k=0,1,2,3,4$.



              Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$






              share|cite|improve this answer











              $endgroup$



              We can write $2-2i$ as:



              $z^5=2-2i=2sqrt2e^{i{-piover4}}=2sqrt2e^{i{-piover4}+2kpi}$



              now we take the root:



              $z=(2sqrt2)^{1over5}e^{i{-piover20}+{2kpiover5}}$



              with $kin mathbb Z$, $k=0,1,2,3,4$.



              Your approach fails because $z=(2-2i)^{1over5}ne(2)^{1over5}-(2i)^{1over5}$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 5 '17 at 17:53

























              answered Jan 5 '17 at 12:49









              MattG88MattG88

              2,4582815




              2,4582815












              • $begingroup$
                $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
                $endgroup$
                – Nosrati
                Jan 5 '17 at 14:12


















              • $begingroup$
                $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
                $endgroup$
                – Nosrati
                Jan 5 '17 at 14:12
















              $begingroup$
              $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
              $endgroup$
              – Nosrati
              Jan 5 '17 at 14:12




              $begingroup$
              $arg(2-2i)$ is $displaystyle7frac{pi}{4}$
              $endgroup$
              – Nosrati
              Jan 5 '17 at 14:12











              0












              $begingroup$

              You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.



              So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).



              Now $z^5=2sqrt2e^{frac{7pi}4i}$.



              So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.



              There are $4$ more.



              Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.



              That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.



                So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).



                Now $z^5=2sqrt2e^{frac{7pi}4i}$.



                So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.



                There are $4$ more.



                Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.



                That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.



                  So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).



                  Now $z^5=2sqrt2e^{frac{7pi}4i}$.



                  So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.



                  There are $4$ more.



                  Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.



                  That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.






                  share|cite|improve this answer











                  $endgroup$



                  You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.



                  So, instead, write $2-2i=2sqrt2e^{frac{7pi}4i}$ (using Euler's formula).



                  Now $z^5=2sqrt2e^{frac{7pi}4i}$.



                  So $z=sqrt[10]{8}e^{frac{7pi}{20}i}$ is one solution.



                  There are $4$ more.



                  Take a primitive $5$th root of unity, $rho=e^{frac{2pi i}5}$. Then the other $4$ are: $rho z,rho^2z,rho^3z$ and $rho^4z$.



                  That is, $sqrt[10]8e^{(frac{7pi+8pi k}{20})i}$, for $k=0,1,2,3$ and $4$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 17:43

























                  answered Jan 2 at 8:57









                  Chris CusterChris Custer

                  11.6k3824




                  11.6k3824






























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