Show that a complex number is on a line.












0












$begingroup$


Let $v$ and $w$ be two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t in mathbb{R}$. Prove that:



If $frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.



I have already tried using the hint as follows:



begin{equation}
frac{z - w}{z-v}=t
end{equation}
begin{equation}
z - w=t(z-v)
end{equation}
begin{equation}
z = frac{-v t +w}{1-t}
end{equation}



This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm:
begin{equation}
frac{z - w}{z-v}(w-v)=t(w-v)
end{equation}
begin{equation}
frac{z - w}{z-v}(w-v)+v=v+ t(w-v)
end{equation}
Which can be rewritten as:
begin{equation}
frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v)
end{equation}



I would have hoped it to simplify to $z$. Do you people have any pointers or tips?






geometry complex-numbers







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $v$ and $w$ be two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t in mathbb{R}$. Prove that:



    If $frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.



    I have already tried using the hint as follows:



    begin{equation}
    frac{z - w}{z-v}=t
    end{equation}
    begin{equation}
    z - w=t(z-v)
    end{equation}
    begin{equation}
    z = frac{-v t +w}{1-t}
    end{equation}



    This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm:
    begin{equation}
    frac{z - w}{z-v}(w-v)=t(w-v)
    end{equation}
    begin{equation}
    frac{z - w}{z-v}(w-v)+v=v+ t(w-v)
    end{equation}
    Which can be rewritten as:
    begin{equation}
    frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v)
    end{equation}



    I would have hoped it to simplify to $z$. Do you people have any pointers or tips?






    geometry complex-numbers







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $v$ and $w$ be two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t in mathbb{R}$. Prove that:



      If $frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.



      I have already tried using the hint as follows:



      begin{equation}
      frac{z - w}{z-v}=t
      end{equation}
      begin{equation}
      z - w=t(z-v)
      end{equation}
      begin{equation}
      z = frac{-v t +w}{1-t}
      end{equation}



      This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm:
      begin{equation}
      frac{z - w}{z-v}(w-v)=t(w-v)
      end{equation}
      begin{equation}
      frac{z - w}{z-v}(w-v)+v=v+ t(w-v)
      end{equation}
      Which can be rewritten as:
      begin{equation}
      frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v)
      end{equation}



      I would have hoped it to simplify to $z$. Do you people have any pointers or tips?






      geometry complex-numbers







      share|cite|improve this question











      $endgroup$




      Let $v$ and $w$ be two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t in mathbb{R}$. Prove that:



      If $frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.



      I have already tried using the hint as follows:



      begin{equation}
      frac{z - w}{z-v}=t
      end{equation}
      begin{equation}
      z - w=t(z-v)
      end{equation}
      begin{equation}
      z = frac{-v t +w}{1-t}
      end{equation}



      This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm:
      begin{equation}
      frac{z - w}{z-v}(w-v)=t(w-v)
      end{equation}
      begin{equation}
      frac{z - w}{z-v}(w-v)+v=v+ t(w-v)
      end{equation}
      Which can be rewritten as:
      begin{equation}
      frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v)
      end{equation}



      I would have hoped it to simplify to $z$. Do you people have any pointers or tips?






      geometry complex-numbers




      geometry complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 21:02









      Eric Wofsey

      184k13211338




      184k13211338










      asked Sep 9 '18 at 14:23









      Wesley StrikWesley Strik

      1,741423




      1,741423






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Your relation
          $$
          z = frac{-v t +w}{1-t}
          $$
          can be written as
          $$
          z=v+tau(w-v),
          $$
          where $displaystyletau={1over1-t}={z-vover w-v}$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
            $endgroup$
            – Aretino
            Sep 9 '18 at 14:35





















          1












          $begingroup$

          Hint for alternate solution: Think geometrically. We have that $l$ is the line going through the points $v$ and $w$ in the complex plane. What can you say about two complex numbers (interpreted as vectors in the plane) if their ratio (as complex numbers) is real?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This means that they are parallel.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:35










          • $begingroup$
            Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:39








          • 1




            $begingroup$
            Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
            $endgroup$
            – Arthur
            Sep 9 '18 at 15:42










          • $begingroup$
            Yeah, I had already used it, without knowing why it was true haha :P
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 16:09



















          1












          $begingroup$

          Just one more step.



          $$z=frac{-vt+w}{1-t}=v+frac{w-v}{1-t}=v+t'(w-v)$$



          with $t'=frac{1}{1-t}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks :) that did the trick!
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:43











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2910831%2fshow-that-a-complex-number-is-on-a-line%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Your relation
          $$
          z = frac{-v t +w}{1-t}
          $$
          can be written as
          $$
          z=v+tau(w-v),
          $$
          where $displaystyletau={1over1-t}={z-vover w-v}$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
            $endgroup$
            – Aretino
            Sep 9 '18 at 14:35


















          1












          $begingroup$

          Your relation
          $$
          z = frac{-v t +w}{1-t}
          $$
          can be written as
          $$
          z=v+tau(w-v),
          $$
          where $displaystyletau={1over1-t}={z-vover w-v}$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
            $endgroup$
            – Aretino
            Sep 9 '18 at 14:35
















          1












          1








          1





          $begingroup$

          Your relation
          $$
          z = frac{-v t +w}{1-t}
          $$
          can be written as
          $$
          z=v+tau(w-v),
          $$
          where $displaystyletau={1over1-t}={z-vover w-v}$.






          share|cite|improve this answer











          $endgroup$



          Your relation
          $$
          z = frac{-v t +w}{1-t}
          $$
          can be written as
          $$
          z=v+tau(w-v),
          $$
          where $displaystyletau={1over1-t}={z-vover w-v}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 9 '18 at 14:34

























          answered Sep 9 '18 at 14:31









          AretinoAretino

          23.1k21443




          23.1k21443








          • 1




            $begingroup$
            $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
            $endgroup$
            – Aretino
            Sep 9 '18 at 14:35
















          • 1




            $begingroup$
            $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
            $endgroup$
            – Aretino
            Sep 9 '18 at 14:35










          1




          1




          $begingroup$
          $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
          $endgroup$
          – Aretino
          Sep 9 '18 at 14:35






          $begingroup$
          $v+t(w-v)=wt+v(1-t)$. I edited my answer to show that more clearly.
          $endgroup$
          – Aretino
          Sep 9 '18 at 14:35













          1












          $begingroup$

          Hint for alternate solution: Think geometrically. We have that $l$ is the line going through the points $v$ and $w$ in the complex plane. What can you say about two complex numbers (interpreted as vectors in the plane) if their ratio (as complex numbers) is real?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This means that they are parallel.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:35










          • $begingroup$
            Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:39








          • 1




            $begingroup$
            Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
            $endgroup$
            – Arthur
            Sep 9 '18 at 15:42










          • $begingroup$
            Yeah, I had already used it, without knowing why it was true haha :P
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 16:09
















          1












          $begingroup$

          Hint for alternate solution: Think geometrically. We have that $l$ is the line going through the points $v$ and $w$ in the complex plane. What can you say about two complex numbers (interpreted as vectors in the plane) if their ratio (as complex numbers) is real?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This means that they are parallel.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:35










          • $begingroup$
            Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:39








          • 1




            $begingroup$
            Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
            $endgroup$
            – Arthur
            Sep 9 '18 at 15:42










          • $begingroup$
            Yeah, I had already used it, without knowing why it was true haha :P
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 16:09














          1












          1








          1





          $begingroup$

          Hint for alternate solution: Think geometrically. We have that $l$ is the line going through the points $v$ and $w$ in the complex plane. What can you say about two complex numbers (interpreted as vectors in the plane) if their ratio (as complex numbers) is real?






          share|cite|improve this answer









          $endgroup$



          Hint for alternate solution: Think geometrically. We have that $l$ is the line going through the points $v$ and $w$ in the complex plane. What can you say about two complex numbers (interpreted as vectors in the plane) if their ratio (as complex numbers) is real?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 9 '18 at 14:32









          ArthurArthur

          113k7110193




          113k7110193












          • $begingroup$
            This means that they are parallel.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:35










          • $begingroup$
            Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:39








          • 1




            $begingroup$
            Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
            $endgroup$
            – Arthur
            Sep 9 '18 at 15:42










          • $begingroup$
            Yeah, I had already used it, without knowing why it was true haha :P
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 16:09


















          • $begingroup$
            This means that they are parallel.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:35










          • $begingroup$
            Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:39








          • 1




            $begingroup$
            Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
            $endgroup$
            – Arthur
            Sep 9 '18 at 15:42










          • $begingroup$
            Yeah, I had already used it, without knowing why it was true haha :P
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 16:09
















          $begingroup$
          This means that they are parallel.
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 14:35




          $begingroup$
          This means that they are parallel.
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 14:35












          $begingroup$
          Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 14:39






          $begingroup$
          Later in the same exercise I can use this fact to prove that $(z-w)(bar{z}-bar{v})= (bar{z} - bar{w})(z-v)$ if and only if $v, w, z$ are collinear.
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 14:39






          1




          1




          $begingroup$
          Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
          $endgroup$
          – Arthur
          Sep 9 '18 at 15:42




          $begingroup$
          Divide that relation by $(bar z-bar w)(z-w)$, and you get that one side is the conjugate of the other, meaning they are real, and thus you can use the result from this problem.
          $endgroup$
          – Arthur
          Sep 9 '18 at 15:42












          $begingroup$
          Yeah, I had already used it, without knowing why it was true haha :P
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 16:09




          $begingroup$
          Yeah, I had already used it, without knowing why it was true haha :P
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 16:09











          1












          $begingroup$

          Just one more step.



          $$z=frac{-vt+w}{1-t}=v+frac{w-v}{1-t}=v+t'(w-v)$$



          with $t'=frac{1}{1-t}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks :) that did the trick!
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:43
















          1












          $begingroup$

          Just one more step.



          $$z=frac{-vt+w}{1-t}=v+frac{w-v}{1-t}=v+t'(w-v)$$



          with $t'=frac{1}{1-t}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks :) that did the trick!
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:43














          1












          1








          1





          $begingroup$

          Just one more step.



          $$z=frac{-vt+w}{1-t}=v+frac{w-v}{1-t}=v+t'(w-v)$$



          with $t'=frac{1}{1-t}$.






          share|cite|improve this answer









          $endgroup$



          Just one more step.



          $$z=frac{-vt+w}{1-t}=v+frac{w-v}{1-t}=v+t'(w-v)$$



          with $t'=frac{1}{1-t}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 9 '18 at 14:33









          YutaYuta

          62429




          62429












          • $begingroup$
            Thanks :) that did the trick!
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:43


















          • $begingroup$
            Thanks :) that did the trick!
            $endgroup$
            – Wesley Strik
            Sep 9 '18 at 14:43
















          $begingroup$
          Thanks :) that did the trick!
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 14:43




          $begingroup$
          Thanks :) that did the trick!
          $endgroup$
          – Wesley Strik
          Sep 9 '18 at 14:43


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2910831%2fshow-that-a-complex-number-is-on-a-line%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more