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





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          3 Answers
          3






          active

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          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


















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          .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 ...
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