Resolution exercises on complex numbers












6














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







share|cite|improve this question




















  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 '18 at 11:20
















6














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







share|cite|improve this question




















  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 '18 at 11:20














6












6








6







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







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




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 7:41

























asked Nov 21 '18 at 7:35









Vincenzo Iannucci

343




343








  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 '18 at 11:20














  • 1




    $-1$ is not a solution as you would get $1+1-1not =0$
    – Henry
    Nov 21 '18 at 11:20








1




1




$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20




$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20










5 Answers
5






active

oldest

votes


















10














$$z=dfrac{|z|^2}{|z|-1}$$ 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 '18 at 23:29





















6














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





























    2














    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





























      2














      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































        1














        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

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          10














          $$z=dfrac{|z|^2}{|z|-1}$$ 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 '18 at 23:29


















          10














          $$z=dfrac{|z|^2}{|z|-1}$$ 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 '18 at 23:29
















          10












          10








          10






          $$z=dfrac{|z|^2}{|z|-1}$$ 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 '18 at 13:51

























          answered Nov 21 '18 at 7:58









          lab bhattacharjee

          223k15156274




          223k15156274












          • 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 '18 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 '18 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 '18 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 '18 at 23:29













          6














          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


























            6














            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
























              6












              6








              6






              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 '18 at 7:42









              Fred

              44.2k1845




              44.2k1845























                  2














                  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


























                    2














                    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
























                      2












                      2








                      2






                      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 '18 at 7:42









                      lab bhattacharjee

                      223k15156274




                      223k15156274























                          2














                          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




























                            2














                            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


























                              2












                              2








                              2






                              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 '18 at 15:38

























                              answered Nov 21 '18 at 8:29









                              trancelocation

                              9,1051521




                              9,1051521























                                  1














                                  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


























                                    1














                                    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
























                                      1












                                      1








                                      1






                                      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 '18 at 8:58









                                      Deepak

                                      16.8k11436




                                      16.8k11436






























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                                          Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                                          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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