$|z|=z^5$ How many solutions does this equation have?












1












$begingroup$


Multiple choice question : How many solutions does the equation $|z|=z^5$ have?



A - $1$ solution



B - $2$ solutions



C - $5$ solutions



D - $6$ solutions



These were the $4$ possible answers.



I started by allowing $z^5in Bbb R$.



$|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.



I found $2$ obvious solutions, $z=1$ and $z=0$
However I couldn't find any other complex solutions; and the B answer was not correct.



I have also tried to expand with $z=a+ib$ but couldn't find anything, still.



The correct answer was D but I couldn't understand why.



Excuse my English, I am not used to doing maths in English.






complex-analysis complex-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Multiple choice question : How many solutions does the equation $|z|=z^5$ have?



    A - $1$ solution



    B - $2$ solutions



    C - $5$ solutions



    D - $6$ solutions



    These were the $4$ possible answers.



    I started by allowing $z^5in Bbb R$.



    $|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.



    I found $2$ obvious solutions, $z=1$ and $z=0$
    However I couldn't find any other complex solutions; and the B answer was not correct.



    I have also tried to expand with $z=a+ib$ but couldn't find anything, still.



    The correct answer was D but I couldn't understand why.



    Excuse my English, I am not used to doing maths in English.






    complex-analysis complex-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Multiple choice question : How many solutions does the equation $|z|=z^5$ have?



      A - $1$ solution



      B - $2$ solutions



      C - $5$ solutions



      D - $6$ solutions



      These were the $4$ possible answers.



      I started by allowing $z^5in Bbb R$.



      $|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.



      I found $2$ obvious solutions, $z=1$ and $z=0$
      However I couldn't find any other complex solutions; and the B answer was not correct.



      I have also tried to expand with $z=a+ib$ but couldn't find anything, still.



      The correct answer was D but I couldn't understand why.



      Excuse my English, I am not used to doing maths in English.






      complex-analysis complex-numbers







      share|cite|improve this question











      $endgroup$




      Multiple choice question : How many solutions does the equation $|z|=z^5$ have?



      A - $1$ solution



      B - $2$ solutions



      C - $5$ solutions



      D - $6$ solutions



      These were the $4$ possible answers.



      I started by allowing $z^5in Bbb R$.



      $|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.



      I found $2$ obvious solutions, $z=1$ and $z=0$
      However I couldn't find any other complex solutions; and the B answer was not correct.



      I have also tried to expand with $z=a+ib$ but couldn't find anything, still.



      The correct answer was D but I couldn't understand why.



      Excuse my English, I am not used to doing maths in English.






      complex-analysis complex-numbers




      complex-analysis complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 17:39









      José Carlos Santos

      173k23133242




      173k23133242










      asked Jan 6 at 17:21









      CaioCaio

      61




      61






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
            $endgroup$
            – NewMath
            Jan 6 at 17:29










          • $begingroup$
            @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
            $endgroup$
            – mathcounterexamples.net
            Jan 6 at 17:31



















          2












          $begingroup$

          It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
            $endgroup$
            – NewMath
            Jan 6 at 17:31












          • $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 17:35










          • $begingroup$
            @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
            $endgroup$
            – Hagen von Eitzen
            Jan 6 at 17:48



















          0












          $begingroup$

          $$
          eqalign{
          & left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
          & Rightarrow quad left{ {matrix{
          {A = A^{,5} } cr
          {e^{,i,5alpha } = 1} cr
          } } right.quad Rightarrow quad left{ {matrix{
          {A = 0,1} cr
          {alpha = 2kpi /5} cr
          } } right.quad Rightarrow cr
          & Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you all this was very helpful
            $endgroup$
            – Caio
            Jan 6 at 18:01










          • $begingroup$
            @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
            $endgroup$
            – G Cab
            Jan 6 at 18:22












          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
            $endgroup$
            – NewMath
            Jan 6 at 17:29










          • $begingroup$
            @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
            $endgroup$
            – mathcounterexamples.net
            Jan 6 at 17:31
















          4












          $begingroup$

          Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
            $endgroup$
            – NewMath
            Jan 6 at 17:29










          • $begingroup$
            @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
            $endgroup$
            – mathcounterexamples.net
            Jan 6 at 17:31














          4












          4








          4





          $begingroup$

          Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.






          share|cite|improve this answer









          $endgroup$



          Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 17:25









          mathcounterexamples.netmathcounterexamples.net

          26.9k22158




          26.9k22158












          • $begingroup$
            You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
            $endgroup$
            – NewMath
            Jan 6 at 17:29










          • $begingroup$
            @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
            $endgroup$
            – mathcounterexamples.net
            Jan 6 at 17:31


















          • $begingroup$
            You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
            $endgroup$
            – NewMath
            Jan 6 at 17:29










          • $begingroup$
            @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
            $endgroup$
            – mathcounterexamples.net
            Jan 6 at 17:31
















          $begingroup$
          You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
          $endgroup$
          – NewMath
          Jan 6 at 17:29




          $begingroup$
          You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
          $endgroup$
          – NewMath
          Jan 6 at 17:29












          $begingroup$
          @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
          $endgroup$
          – mathcounterexamples.net
          Jan 6 at 17:31




          $begingroup$
          @NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
          $endgroup$
          – mathcounterexamples.net
          Jan 6 at 17:31











          2












          $begingroup$

          It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
            $endgroup$
            – NewMath
            Jan 6 at 17:31












          • $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 17:35










          • $begingroup$
            @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
            $endgroup$
            – Hagen von Eitzen
            Jan 6 at 17:48
















          2












          $begingroup$

          It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
            $endgroup$
            – NewMath
            Jan 6 at 17:31












          • $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 17:35










          • $begingroup$
            @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
            $endgroup$
            – Hagen von Eitzen
            Jan 6 at 17:48














          2












          2








          2





          $begingroup$

          It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.






          share|cite|improve this answer











          $endgroup$



          It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 17:34

























          answered Jan 6 at 17:28









          José Carlos SantosJosé Carlos Santos

          173k23133242




          173k23133242












          • $begingroup$
            Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
            $endgroup$
            – NewMath
            Jan 6 at 17:31












          • $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 17:35










          • $begingroup$
            @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
            $endgroup$
            – Hagen von Eitzen
            Jan 6 at 17:48


















          • $begingroup$
            Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
            $endgroup$
            – NewMath
            Jan 6 at 17:31












          • $begingroup$
            I've edited my answer. I hope that everything is clear now.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 17:35










          • $begingroup$
            @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
            $endgroup$
            – Hagen von Eitzen
            Jan 6 at 17:48
















          $begingroup$
          Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
          $endgroup$
          – NewMath
          Jan 6 at 17:31






          $begingroup$
          Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
          $endgroup$
          – NewMath
          Jan 6 at 17:31














          $begingroup$
          I've edited my answer. I hope that everything is clear now.
          $endgroup$
          – José Carlos Santos
          Jan 6 at 17:35




          $begingroup$
          I've edited my answer. I hope that everything is clear now.
          $endgroup$
          – José Carlos Santos
          Jan 6 at 17:35












          $begingroup$
          @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
          $endgroup$
          – Hagen von Eitzen
          Jan 6 at 17:48




          $begingroup$
          @NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
          $endgroup$
          – Hagen von Eitzen
          Jan 6 at 17:48











          0












          $begingroup$

          $$
          eqalign{
          & left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
          & Rightarrow quad left{ {matrix{
          {A = A^{,5} } cr
          {e^{,i,5alpha } = 1} cr
          } } right.quad Rightarrow quad left{ {matrix{
          {A = 0,1} cr
          {alpha = 2kpi /5} cr
          } } right.quad Rightarrow cr
          & Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you all this was very helpful
            $endgroup$
            – Caio
            Jan 6 at 18:01










          • $begingroup$
            @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
            $endgroup$
            – G Cab
            Jan 6 at 18:22
















          0












          $begingroup$

          $$
          eqalign{
          & left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
          & Rightarrow quad left{ {matrix{
          {A = A^{,5} } cr
          {e^{,i,5alpha } = 1} cr
          } } right.quad Rightarrow quad left{ {matrix{
          {A = 0,1} cr
          {alpha = 2kpi /5} cr
          } } right.quad Rightarrow cr
          & Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you all this was very helpful
            $endgroup$
            – Caio
            Jan 6 at 18:01










          • $begingroup$
            @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
            $endgroup$
            – G Cab
            Jan 6 at 18:22














          0












          0








          0





          $begingroup$

          $$
          eqalign{
          & left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
          & Rightarrow quad left{ {matrix{
          {A = A^{,5} } cr
          {e^{,i,5alpha } = 1} cr
          } } right.quad Rightarrow quad left{ {matrix{
          {A = 0,1} cr
          {alpha = 2kpi /5} cr
          } } right.quad Rightarrow cr
          & Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
          $$






          share|cite|improve this answer









          $endgroup$



          $$
          eqalign{
          & left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
          & Rightarrow quad left{ {matrix{
          {A = A^{,5} } cr
          {e^{,i,5alpha } = 1} cr
          } } right.quad Rightarrow quad left{ {matrix{
          {A = 0,1} cr
          {alpha = 2kpi /5} cr
          } } right.quad Rightarrow cr
          & Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 17:36









          G CabG Cab

          20.4k31341




          20.4k31341












          • $begingroup$
            thank you all this was very helpful
            $endgroup$
            – Caio
            Jan 6 at 18:01










          • $begingroup$
            @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
            $endgroup$
            – G Cab
            Jan 6 at 18:22


















          • $begingroup$
            thank you all this was very helpful
            $endgroup$
            – Caio
            Jan 6 at 18:01










          • $begingroup$
            @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
            $endgroup$
            – G Cab
            Jan 6 at 18:22
















          $begingroup$
          thank you all this was very helpful
          $endgroup$
          – Caio
          Jan 6 at 18:01




          $begingroup$
          thank you all this was very helpful
          $endgroup$
          – Caio
          Jan 6 at 18:01












          $begingroup$
          @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
          $endgroup$
          – G Cab
          Jan 6 at 18:22




          $begingroup$
          @Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
          $endgroup$
          – G Cab
          Jan 6 at 18:22


















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