If $z_1,z_2$ satisfy $z+bar{z}=2|z-1|$, $arg(z_1-z_2)=dfrac{pi}{4}$, then find $Im(z_1+z_2)$












3












$begingroup$



If $z_1$ and $z_2$ both satisfy $z+bar{z}=2|z-1|$, $arg(z_1-z_2)=dfrac{pi}{4}$, then find $Im(z_1+z_2)$




My Attempt
$$
z_1+bar{z}_1=2|z_1-1|quad&quad z_2+bar{z}_2=2|z_2-1|quad&quad arg(z_1-z_2)=dfrac{pi}{4}\
z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}Big(1+iBig)\
Re(z_1+z_2)=frac{z_1+z_2+bar{z}_1+bar{z}_2}{2}=|z_1-1|+|z_2-1|\
Im(z_1+z_2)=frac{z_1+z_2-bar{z}_1-bar{z}_2}{2i}\
Re(z_1-z_2)=frac{z_1-z_2+bar{z}_1-bar{z}_2}{2}=frac{|z_1-z_2|}{sqrt{2}}={|z_1-1|-|z_2-1|}\
Im(z_1-z_2)=frac{z_1-z_2-bar{z}_1+bar{z}_2}{2i}=frac{|z_1-z_2|}{sqrt{2}}
$$
My reference gives the solution $2$, but how do I find the solution without actually representing $z=a+ib$ ?






complex-numbers







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    If $z_1$ and $z_2$ both satisfy $z+bar{z}=2|z-1|$, $arg(z_1-z_2)=dfrac{pi}{4}$, then find $Im(z_1+z_2)$




    My Attempt
    $$
    z_1+bar{z}_1=2|z_1-1|quad&quad z_2+bar{z}_2=2|z_2-1|quad&quad arg(z_1-z_2)=dfrac{pi}{4}\
    z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}Big(1+iBig)\
    Re(z_1+z_2)=frac{z_1+z_2+bar{z}_1+bar{z}_2}{2}=|z_1-1|+|z_2-1|\
    Im(z_1+z_2)=frac{z_1+z_2-bar{z}_1-bar{z}_2}{2i}\
    Re(z_1-z_2)=frac{z_1-z_2+bar{z}_1-bar{z}_2}{2}=frac{|z_1-z_2|}{sqrt{2}}={|z_1-1|-|z_2-1|}\
    Im(z_1-z_2)=frac{z_1-z_2-bar{z}_1+bar{z}_2}{2i}=frac{|z_1-z_2|}{sqrt{2}}
    $$
    My reference gives the solution $2$, but how do I find the solution without actually representing $z=a+ib$ ?






    complex-numbers







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$



      If $z_1$ and $z_2$ both satisfy $z+bar{z}=2|z-1|$, $arg(z_1-z_2)=dfrac{pi}{4}$, then find $Im(z_1+z_2)$




      My Attempt
      $$
      z_1+bar{z}_1=2|z_1-1|quad&quad z_2+bar{z}_2=2|z_2-1|quad&quad arg(z_1-z_2)=dfrac{pi}{4}\
      z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}Big(1+iBig)\
      Re(z_1+z_2)=frac{z_1+z_2+bar{z}_1+bar{z}_2}{2}=|z_1-1|+|z_2-1|\
      Im(z_1+z_2)=frac{z_1+z_2-bar{z}_1-bar{z}_2}{2i}\
      Re(z_1-z_2)=frac{z_1-z_2+bar{z}_1-bar{z}_2}{2}=frac{|z_1-z_2|}{sqrt{2}}={|z_1-1|-|z_2-1|}\
      Im(z_1-z_2)=frac{z_1-z_2-bar{z}_1+bar{z}_2}{2i}=frac{|z_1-z_2|}{sqrt{2}}
      $$
      My reference gives the solution $2$, but how do I find the solution without actually representing $z=a+ib$ ?






      complex-numbers







      share|cite|improve this question











      $endgroup$





      If $z_1$ and $z_2$ both satisfy $z+bar{z}=2|z-1|$, $arg(z_1-z_2)=dfrac{pi}{4}$, then find $Im(z_1+z_2)$




      My Attempt
      $$
      z_1+bar{z}_1=2|z_1-1|quad&quad z_2+bar{z}_2=2|z_2-1|quad&quad arg(z_1-z_2)=dfrac{pi}{4}\
      z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}Big(1+iBig)\
      Re(z_1+z_2)=frac{z_1+z_2+bar{z}_1+bar{z}_2}{2}=|z_1-1|+|z_2-1|\
      Im(z_1+z_2)=frac{z_1+z_2-bar{z}_1-bar{z}_2}{2i}\
      Re(z_1-z_2)=frac{z_1-z_2+bar{z}_1-bar{z}_2}{2}=frac{|z_1-z_2|}{sqrt{2}}={|z_1-1|-|z_2-1|}\
      Im(z_1-z_2)=frac{z_1-z_2-bar{z}_1+bar{z}_2}{2i}=frac{|z_1-z_2|}{sqrt{2}}
      $$
      My reference gives the solution $2$, but how do I find the solution without actually representing $z=a+ib$ ?






      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 Jan 9 at 6:56







      ss1729

















      asked Jan 8 at 10:07









      ss1729ss1729

      2,07811124




      2,07811124






















          2 Answers
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          active

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          2












          $begingroup$

          It's easy if you consider



          $z_1=a_1+ib_1$ , $z_2=a_2+ib_2$



          $bar{z_1}=a_1-ib_1$ , $bar{z_2}=a_2-ib_2$



          Given $z+bar{z}=2|z-1|$



          Since, $z_1$ and $z_2$ both satisfy, $z+bar{z}=2|z-1|$, we can say that $$z_1+bar{z_1}=2|z_1-1|$$ and $$z_2+bar{z_2}=2|z_2-1|$$



          Firstly, $$z_1+bar{z_1}=2|z_1-1|$$
          $$a_1+ib_1+a_1-ib_1=2|(a_1-1)+ib_1|$$
          After solving we get, $$b_1^2=2a_1-1tag1$$
          Similarly, we get $$b_1^2=2a_2-1tag3$$



          Now, $$arg(z_1-z_2)=dfrac{pi}{4}$$
          $$arg(a_1+ib_1-a_2-ib_2)=dfrac{pi}{4}$$
          $$dfrac{b_1-b_2}{a_1-a_2}=tandfrac{pi}{4}$$
          $$dfrac{b_1-b_2}{a_1-a_2}=1$$
          $$b_1-b_2=a_1-a_2tag3$$
          Now use the equations $(1),(2)$ and $(3)$ to get the final answer.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Without representing $z=a+bi$.
            begin{align}
            z+bar z&=2|z-1|\
            (z+bar z)^2&=4|z-1|^2=4(z-1)(bar z-1)\
            z^2+2zbar z+bar z^2&=4(zbar z-z-bar z+1)\
            z^2-2zbar z+bar z^2&=4-4(z+bar z)\
            (z-bar z)^2&=4-4(z+bar z)tag{1}
            end{align}
            Now take (1) for $z_2$ minus (1) for $z_1$ and use
            $$
            z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}(1+i)tag{2}
            $$
            to get
            begin{align}
            (z_2-bar z_2)^2-(z_1-bar z_1)^2&=4(z_1+bar z_1-z_2-bar z_2)\
            (underbrace{z_2-bar z_2-z_1+bar z_1}_{-sqrt2i|z_1-z_2|})(underbrace{z_2-bar z_2+z_1-bar z_1}_{2ioperatorname{Im}(z_1+z_2)})&=4(underbrace{z_1+bar z_1-z_2-bar z_2}_{sqrt2|z_1-z_2|})\
            |z_1-z_2|(operatorname{Im}(z_1+z_2)-2)&=0.
            end{align}






            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              2 Answers
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              2












              $begingroup$

              It's easy if you consider



              $z_1=a_1+ib_1$ , $z_2=a_2+ib_2$



              $bar{z_1}=a_1-ib_1$ , $bar{z_2}=a_2-ib_2$



              Given $z+bar{z}=2|z-1|$



              Since, $z_1$ and $z_2$ both satisfy, $z+bar{z}=2|z-1|$, we can say that $$z_1+bar{z_1}=2|z_1-1|$$ and $$z_2+bar{z_2}=2|z_2-1|$$



              Firstly, $$z_1+bar{z_1}=2|z_1-1|$$
              $$a_1+ib_1+a_1-ib_1=2|(a_1-1)+ib_1|$$
              After solving we get, $$b_1^2=2a_1-1tag1$$
              Similarly, we get $$b_1^2=2a_2-1tag3$$



              Now, $$arg(z_1-z_2)=dfrac{pi}{4}$$
              $$arg(a_1+ib_1-a_2-ib_2)=dfrac{pi}{4}$$
              $$dfrac{b_1-b_2}{a_1-a_2}=tandfrac{pi}{4}$$
              $$dfrac{b_1-b_2}{a_1-a_2}=1$$
              $$b_1-b_2=a_1-a_2tag3$$
              Now use the equations $(1),(2)$ and $(3)$ to get the final answer.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                It's easy if you consider



                $z_1=a_1+ib_1$ , $z_2=a_2+ib_2$



                $bar{z_1}=a_1-ib_1$ , $bar{z_2}=a_2-ib_2$



                Given $z+bar{z}=2|z-1|$



                Since, $z_1$ and $z_2$ both satisfy, $z+bar{z}=2|z-1|$, we can say that $$z_1+bar{z_1}=2|z_1-1|$$ and $$z_2+bar{z_2}=2|z_2-1|$$



                Firstly, $$z_1+bar{z_1}=2|z_1-1|$$
                $$a_1+ib_1+a_1-ib_1=2|(a_1-1)+ib_1|$$
                After solving we get, $$b_1^2=2a_1-1tag1$$
                Similarly, we get $$b_1^2=2a_2-1tag3$$



                Now, $$arg(z_1-z_2)=dfrac{pi}{4}$$
                $$arg(a_1+ib_1-a_2-ib_2)=dfrac{pi}{4}$$
                $$dfrac{b_1-b_2}{a_1-a_2}=tandfrac{pi}{4}$$
                $$dfrac{b_1-b_2}{a_1-a_2}=1$$
                $$b_1-b_2=a_1-a_2tag3$$
                Now use the equations $(1),(2)$ and $(3)$ to get the final answer.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  It's easy if you consider



                  $z_1=a_1+ib_1$ , $z_2=a_2+ib_2$



                  $bar{z_1}=a_1-ib_1$ , $bar{z_2}=a_2-ib_2$



                  Given $z+bar{z}=2|z-1|$



                  Since, $z_1$ and $z_2$ both satisfy, $z+bar{z}=2|z-1|$, we can say that $$z_1+bar{z_1}=2|z_1-1|$$ and $$z_2+bar{z_2}=2|z_2-1|$$



                  Firstly, $$z_1+bar{z_1}=2|z_1-1|$$
                  $$a_1+ib_1+a_1-ib_1=2|(a_1-1)+ib_1|$$
                  After solving we get, $$b_1^2=2a_1-1tag1$$
                  Similarly, we get $$b_1^2=2a_2-1tag3$$



                  Now, $$arg(z_1-z_2)=dfrac{pi}{4}$$
                  $$arg(a_1+ib_1-a_2-ib_2)=dfrac{pi}{4}$$
                  $$dfrac{b_1-b_2}{a_1-a_2}=tandfrac{pi}{4}$$
                  $$dfrac{b_1-b_2}{a_1-a_2}=1$$
                  $$b_1-b_2=a_1-a_2tag3$$
                  Now use the equations $(1),(2)$ and $(3)$ to get the final answer.






                  share|cite|improve this answer









                  $endgroup$



                  It's easy if you consider



                  $z_1=a_1+ib_1$ , $z_2=a_2+ib_2$



                  $bar{z_1}=a_1-ib_1$ , $bar{z_2}=a_2-ib_2$



                  Given $z+bar{z}=2|z-1|$



                  Since, $z_1$ and $z_2$ both satisfy, $z+bar{z}=2|z-1|$, we can say that $$z_1+bar{z_1}=2|z_1-1|$$ and $$z_2+bar{z_2}=2|z_2-1|$$



                  Firstly, $$z_1+bar{z_1}=2|z_1-1|$$
                  $$a_1+ib_1+a_1-ib_1=2|(a_1-1)+ib_1|$$
                  After solving we get, $$b_1^2=2a_1-1tag1$$
                  Similarly, we get $$b_1^2=2a_2-1tag3$$



                  Now, $$arg(z_1-z_2)=dfrac{pi}{4}$$
                  $$arg(a_1+ib_1-a_2-ib_2)=dfrac{pi}{4}$$
                  $$dfrac{b_1-b_2}{a_1-a_2}=tandfrac{pi}{4}$$
                  $$dfrac{b_1-b_2}{a_1-a_2}=1$$
                  $$b_1-b_2=a_1-a_2tag3$$
                  Now use the equations $(1),(2)$ and $(3)$ to get the final answer.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 10:33









                  Key FlexKey Flex

                  8,56171233




                  8,56171233























                      2












                      $begingroup$

                      Without representing $z=a+bi$.
                      begin{align}
                      z+bar z&=2|z-1|\
                      (z+bar z)^2&=4|z-1|^2=4(z-1)(bar z-1)\
                      z^2+2zbar z+bar z^2&=4(zbar z-z-bar z+1)\
                      z^2-2zbar z+bar z^2&=4-4(z+bar z)\
                      (z-bar z)^2&=4-4(z+bar z)tag{1}
                      end{align}
                      Now take (1) for $z_2$ minus (1) for $z_1$ and use
                      $$
                      z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}(1+i)tag{2}
                      $$
                      to get
                      begin{align}
                      (z_2-bar z_2)^2-(z_1-bar z_1)^2&=4(z_1+bar z_1-z_2-bar z_2)\
                      (underbrace{z_2-bar z_2-z_1+bar z_1}_{-sqrt2i|z_1-z_2|})(underbrace{z_2-bar z_2+z_1-bar z_1}_{2ioperatorname{Im}(z_1+z_2)})&=4(underbrace{z_1+bar z_1-z_2-bar z_2}_{sqrt2|z_1-z_2|})\
                      |z_1-z_2|(operatorname{Im}(z_1+z_2)-2)&=0.
                      end{align}






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Without representing $z=a+bi$.
                        begin{align}
                        z+bar z&=2|z-1|\
                        (z+bar z)^2&=4|z-1|^2=4(z-1)(bar z-1)\
                        z^2+2zbar z+bar z^2&=4(zbar z-z-bar z+1)\
                        z^2-2zbar z+bar z^2&=4-4(z+bar z)\
                        (z-bar z)^2&=4-4(z+bar z)tag{1}
                        end{align}
                        Now take (1) for $z_2$ minus (1) for $z_1$ and use
                        $$
                        z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}(1+i)tag{2}
                        $$
                        to get
                        begin{align}
                        (z_2-bar z_2)^2-(z_1-bar z_1)^2&=4(z_1+bar z_1-z_2-bar z_2)\
                        (underbrace{z_2-bar z_2-z_1+bar z_1}_{-sqrt2i|z_1-z_2|})(underbrace{z_2-bar z_2+z_1-bar z_1}_{2ioperatorname{Im}(z_1+z_2)})&=4(underbrace{z_1+bar z_1-z_2-bar z_2}_{sqrt2|z_1-z_2|})\
                        |z_1-z_2|(operatorname{Im}(z_1+z_2)-2)&=0.
                        end{align}






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Without representing $z=a+bi$.
                          begin{align}
                          z+bar z&=2|z-1|\
                          (z+bar z)^2&=4|z-1|^2=4(z-1)(bar z-1)\
                          z^2+2zbar z+bar z^2&=4(zbar z-z-bar z+1)\
                          z^2-2zbar z+bar z^2&=4-4(z+bar z)\
                          (z-bar z)^2&=4-4(z+bar z)tag{1}
                          end{align}
                          Now take (1) for $z_2$ minus (1) for $z_1$ and use
                          $$
                          z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}(1+i)tag{2}
                          $$
                          to get
                          begin{align}
                          (z_2-bar z_2)^2-(z_1-bar z_1)^2&=4(z_1+bar z_1-z_2-bar z_2)\
                          (underbrace{z_2-bar z_2-z_1+bar z_1}_{-sqrt2i|z_1-z_2|})(underbrace{z_2-bar z_2+z_1-bar z_1}_{2ioperatorname{Im}(z_1+z_2)})&=4(underbrace{z_1+bar z_1-z_2-bar z_2}_{sqrt2|z_1-z_2|})\
                          |z_1-z_2|(operatorname{Im}(z_1+z_2)-2)&=0.
                          end{align}






                          share|cite|improve this answer









                          $endgroup$



                          Without representing $z=a+bi$.
                          begin{align}
                          z+bar z&=2|z-1|\
                          (z+bar z)^2&=4|z-1|^2=4(z-1)(bar z-1)\
                          z^2+2zbar z+bar z^2&=4(zbar z-z-bar z+1)\
                          z^2-2zbar z+bar z^2&=4-4(z+bar z)\
                          (z-bar z)^2&=4-4(z+bar z)tag{1}
                          end{align}
                          Now take (1) for $z_2$ minus (1) for $z_1$ and use
                          $$
                          z_1-z_2=frac{|z_1-z_2|}{sqrt{2}}(1+i)tag{2}
                          $$
                          to get
                          begin{align}
                          (z_2-bar z_2)^2-(z_1-bar z_1)^2&=4(z_1+bar z_1-z_2-bar z_2)\
                          (underbrace{z_2-bar z_2-z_1+bar z_1}_{-sqrt2i|z_1-z_2|})(underbrace{z_2-bar z_2+z_1-bar z_1}_{2ioperatorname{Im}(z_1+z_2)})&=4(underbrace{z_1+bar z_1-z_2-bar z_2}_{sqrt2|z_1-z_2|})\
                          |z_1-z_2|(operatorname{Im}(z_1+z_2)-2)&=0.
                          end{align}







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 8 at 12:48









                          A.Γ.A.Γ.

                          22.9k32656




                          22.9k32656






























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                              Marschland

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                              Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                              Dieringhausen

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                              Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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