Resolution exercises on complex numbers











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6
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How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$










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  • 2




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Nov 21 at 7:37






  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 at 11:20















up vote
6
down vote

favorite












How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$










share|cite|improve this question




















  • 2




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Nov 21 at 7:37






  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 at 11:20













up vote
6
down vote

favorite









up vote
6
down vote

favorite











How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$










share|cite|improve this question















How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$







complex-numbers






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share|cite|improve this question













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share|cite|improve this question








edited Nov 21 at 7:41

























asked Nov 21 at 7:35









Vincenzo Iannucci

343




343








  • 2




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Nov 21 at 7:37






  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 at 11:20














  • 2




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Nov 21 at 7:37






  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 at 11:20








2




2




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 21 at 7:37




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 21 at 7:37




1




1




$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 at 11:20




$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 at 11:20










5 Answers
5






active

oldest

votes

















up vote
10
down vote













$$z=dfrac{|z|^2}{|z|-1}$$ which is real



If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable



If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$






share|cite|improve this answer























  • Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
    – ruakh
    Nov 21 at 23:29




















up vote
6
down vote













Observe that $z=0$ is a solution of the equation



$$|z|^2 - z|z| + z = 0.$$



( $z=-1$ is not a solution !)



Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$



$overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence



$z=|z|-1 in mathbb R$.



If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.






share|cite|improve this answer




























    up vote
    2
    down vote













    WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real



    $$r(r-r(cos t+isin t)+(cos t+isin t))=0$$



    If $rne0,$ equating the real & the imaginary parts



    $r-rcos t+cos t=0=(1-r)sin t$



    Case $#1:$



    If $r=1,1=0$ which is untenable



    If $sin t=0,$



    Case $#2A:cos t=1,r=0$ which is untenable



    Case $#2A:cos t=-1,r+r-1=0iff r=?$






    share|cite|improve this answer




























      up vote
      2
      down vote













      You may also proceed as follows:




      • Rewrite the equation to
        $$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$

      • Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.

      • Noting the solution $boxed{z = 0}$ we get
        $$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$






      share|cite|improve this answer






























        up vote
        1
        down vote













        Let $r = e^{itheta}$.



        We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.



        Factorising, we get $r = 0$, giving $z = 0$ as one solution or:



        $r - re^{itheta} + e^{itheta} = 0$



        $e^{itheta} = frac{r}{r-1}$



        From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.



        $frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.



        Therefore the two solutions are $z = 0$ and $z = -frac 12$.






        share|cite|improve this answer





















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          5 Answers
          5






          active

          oldest

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          5 Answers
          5






          active

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          active

          oldest

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          active

          oldest

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          up vote
          10
          down vote













          $$z=dfrac{|z|^2}{|z|-1}$$ which is real



          If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable



          If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$






          share|cite|improve this answer























          • Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
            – ruakh
            Nov 21 at 23:29

















          up vote
          10
          down vote













          $$z=dfrac{|z|^2}{|z|-1}$$ which is real



          If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable



          If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$






          share|cite|improve this answer























          • Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
            – ruakh
            Nov 21 at 23:29















          up vote
          10
          down vote










          up vote
          10
          down vote









          $$z=dfrac{|z|^2}{|z|-1}$$ which is real



          If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable



          If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$






          share|cite|improve this answer














          $$z=dfrac{|z|^2}{|z|-1}$$ which is real



          If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable



          If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 13:51

























          answered Nov 21 at 7:58









          lab bhattacharjee

          221k15154271




          221k15154271












          • Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
            – ruakh
            Nov 21 at 23:29




















          • Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
            – ruakh
            Nov 21 at 23:29


















          Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
          – ruakh
          Nov 21 at 23:29






          Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
          – ruakh
          Nov 21 at 23:29












          up vote
          6
          down vote













          Observe that $z=0$ is a solution of the equation



          $$|z|^2 - z|z| + z = 0.$$



          ( $z=-1$ is not a solution !)



          Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$



          $overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence



          $z=|z|-1 in mathbb R$.



          If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.






          share|cite|improve this answer

























            up vote
            6
            down vote













            Observe that $z=0$ is a solution of the equation



            $$|z|^2 - z|z| + z = 0.$$



            ( $z=-1$ is not a solution !)



            Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$



            $overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence



            $z=|z|-1 in mathbb R$.



            If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.






            share|cite|improve this answer























              up vote
              6
              down vote










              up vote
              6
              down vote









              Observe that $z=0$ is a solution of the equation



              $$|z|^2 - z|z| + z = 0.$$



              ( $z=-1$ is not a solution !)



              Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$



              $overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence



              $z=|z|-1 in mathbb R$.



              If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.






              share|cite|improve this answer












              Observe that $z=0$ is a solution of the equation



              $$|z|^2 - z|z| + z = 0.$$



              ( $z=-1$ is not a solution !)



              Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$



              $overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence



              $z=|z|-1 in mathbb R$.



              If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 21 at 7:42









              Fred

              42.9k1643




              42.9k1643






















                  up vote
                  2
                  down vote













                  WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real



                  $$r(r-r(cos t+isin t)+(cos t+isin t))=0$$



                  If $rne0,$ equating the real & the imaginary parts



                  $r-rcos t+cos t=0=(1-r)sin t$



                  Case $#1:$



                  If $r=1,1=0$ which is untenable



                  If $sin t=0,$



                  Case $#2A:cos t=1,r=0$ which is untenable



                  Case $#2A:cos t=-1,r+r-1=0iff r=?$






                  share|cite|improve this answer

























                    up vote
                    2
                    down vote













                    WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real



                    $$r(r-r(cos t+isin t)+(cos t+isin t))=0$$



                    If $rne0,$ equating the real & the imaginary parts



                    $r-rcos t+cos t=0=(1-r)sin t$



                    Case $#1:$



                    If $r=1,1=0$ which is untenable



                    If $sin t=0,$



                    Case $#2A:cos t=1,r=0$ which is untenable



                    Case $#2A:cos t=-1,r+r-1=0iff r=?$






                    share|cite|improve this answer























                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real



                      $$r(r-r(cos t+isin t)+(cos t+isin t))=0$$



                      If $rne0,$ equating the real & the imaginary parts



                      $r-rcos t+cos t=0=(1-r)sin t$



                      Case $#1:$



                      If $r=1,1=0$ which is untenable



                      If $sin t=0,$



                      Case $#2A:cos t=1,r=0$ which is untenable



                      Case $#2A:cos t=-1,r+r-1=0iff r=?$






                      share|cite|improve this answer












                      WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real



                      $$r(r-r(cos t+isin t)+(cos t+isin t))=0$$



                      If $rne0,$ equating the real & the imaginary parts



                      $r-rcos t+cos t=0=(1-r)sin t$



                      Case $#1:$



                      If $r=1,1=0$ which is untenable



                      If $sin t=0,$



                      Case $#2A:cos t=1,r=0$ which is untenable



                      Case $#2A:cos t=-1,r+r-1=0iff r=?$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 21 at 7:42









                      lab bhattacharjee

                      221k15154271




                      221k15154271






















                          up vote
                          2
                          down vote













                          You may also proceed as follows:




                          • Rewrite the equation to
                            $$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$

                          • Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.

                          • Noting the solution $boxed{z = 0}$ we get
                            $$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$






                          share|cite|improve this answer



























                            up vote
                            2
                            down vote













                            You may also proceed as follows:




                            • Rewrite the equation to
                              $$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$

                            • Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.

                            • Noting the solution $boxed{z = 0}$ we get
                              $$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$






                            share|cite|improve this answer

























                              up vote
                              2
                              down vote










                              up vote
                              2
                              down vote









                              You may also proceed as follows:




                              • Rewrite the equation to
                                $$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$

                              • Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.

                              • Noting the solution $boxed{z = 0}$ we get
                                $$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$






                              share|cite|improve this answer














                              You may also proceed as follows:




                              • Rewrite the equation to
                                $$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$

                              • Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.

                              • Noting the solution $boxed{z = 0}$ we get
                                $$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 21 at 15:38

























                              answered Nov 21 at 8:29









                              trancelocation

                              8,4571520




                              8,4571520






















                                  up vote
                                  1
                                  down vote













                                  Let $r = e^{itheta}$.



                                  We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.



                                  Factorising, we get $r = 0$, giving $z = 0$ as one solution or:



                                  $r - re^{itheta} + e^{itheta} = 0$



                                  $e^{itheta} = frac{r}{r-1}$



                                  From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.



                                  $frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.



                                  Therefore the two solutions are $z = 0$ and $z = -frac 12$.






                                  share|cite|improve this answer

























                                    up vote
                                    1
                                    down vote













                                    Let $r = e^{itheta}$.



                                    We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.



                                    Factorising, we get $r = 0$, giving $z = 0$ as one solution or:



                                    $r - re^{itheta} + e^{itheta} = 0$



                                    $e^{itheta} = frac{r}{r-1}$



                                    From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.



                                    $frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.



                                    Therefore the two solutions are $z = 0$ and $z = -frac 12$.






                                    share|cite|improve this answer























                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      Let $r = e^{itheta}$.



                                      We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.



                                      Factorising, we get $r = 0$, giving $z = 0$ as one solution or:



                                      $r - re^{itheta} + e^{itheta} = 0$



                                      $e^{itheta} = frac{r}{r-1}$



                                      From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.



                                      $frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.



                                      Therefore the two solutions are $z = 0$ and $z = -frac 12$.






                                      share|cite|improve this answer












                                      Let $r = e^{itheta}$.



                                      We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.



                                      Factorising, we get $r = 0$, giving $z = 0$ as one solution or:



                                      $r - re^{itheta} + e^{itheta} = 0$



                                      $e^{itheta} = frac{r}{r-1}$



                                      From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.



                                      $frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.



                                      Therefore the two solutions are $z = 0$ and $z = -frac 12$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 21 at 8:58









                                      Deepak

                                      16.5k11436




                                      16.5k11436






























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                                          QoS: MAC-Priority for clients behind a repeater