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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                                          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                                          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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