find equation for parabola in complex plane












0












$begingroup$


Let $z_0,z_1,z_2inmathbb{C}$. Find equation for a parabola with $z_0$ the focus and directrix $|z-z_1|=|z-z_2|$.



By definition, the equation for parabola should be
$$
|z-z_0|=text{distance between focus and the directrix.}
$$



However, how to express the distance between $z_0$ and the directrix now?






complex-analysis







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $z_0,z_1,z_2inmathbb{C}$. Find equation for a parabola with $z_0$ the focus and directrix $|z-z_1|=|z-z_2|$.



    By definition, the equation for parabola should be
    $$
    |z-z_0|=text{distance between focus and the directrix.}
    $$



    However, how to express the distance between $z_0$ and the directrix now?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $z_0,z_1,z_2inmathbb{C}$. Find equation for a parabola with $z_0$ the focus and directrix $|z-z_1|=|z-z_2|$.



      By definition, the equation for parabola should be
      $$
      |z-z_0|=text{distance between focus and the directrix.}
      $$



      However, how to express the distance between $z_0$ and the directrix now?






      complex-analysis







      share|cite|improve this question









      $endgroup$




      Let $z_0,z_1,z_2inmathbb{C}$. Find equation for a parabola with $z_0$ the focus and directrix $|z-z_1|=|z-z_2|$.



      By definition, the equation for parabola should be
      $$
      |z-z_0|=text{distance between focus and the directrix.}
      $$



      However, how to express the distance between $z_0$ and the directrix now?






      complex-analysis




      complex-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 11 at 1:55









      whereamIwhereamI

      368115




      368115






















          1 Answer
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          1












          $begingroup$

          The distance of $z$ from the y axis is $|Re {z}|$



          Lets work in standard Cartesian, and then look for the complex analogue.



          The distance of a point $(x,y)$ from a line $ax + by + c = 0$



          $frac {|ax+by - c|}{sqrt {a^2+b^2}}$



          What are $a,b,c?$



          the point $frac {z_1+z_2}{2}$ is on the line.



          and $(z_2-z_1)$ is normal to the line.



          $a = Re (z_2-z_1)\ b = Im (z_2-z_1)\
          c = frac 12 (|z_2|^2 - |z_1|^2)\
          sqrt {a^2+b^2} = |z_2-z_1|$



          $ax + by = Re (bar zz_2-bar zz_1)$



          $|z - z_0| = frac {|Re (bar zz_2-bar zz_1) - frac 12 (|z_2|^2 - |z_1|^2)|}{|z_2-z_1|}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
            $endgroup$
            – whereamI
            Jan 11 at 4:11












          • $begingroup$
            You might be right about that... $(overline {z_2- z_1})$ would work, too.
            $endgroup$
            – Doug M
            Jan 11 at 21:05














          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The distance of $z$ from the y axis is $|Re {z}|$



          Lets work in standard Cartesian, and then look for the complex analogue.



          The distance of a point $(x,y)$ from a line $ax + by + c = 0$



          $frac {|ax+by - c|}{sqrt {a^2+b^2}}$



          What are $a,b,c?$



          the point $frac {z_1+z_2}{2}$ is on the line.



          and $(z_2-z_1)$ is normal to the line.



          $a = Re (z_2-z_1)\ b = Im (z_2-z_1)\
          c = frac 12 (|z_2|^2 - |z_1|^2)\
          sqrt {a^2+b^2} = |z_2-z_1|$



          $ax + by = Re (bar zz_2-bar zz_1)$



          $|z - z_0| = frac {|Re (bar zz_2-bar zz_1) - frac 12 (|z_2|^2 - |z_1|^2)|}{|z_2-z_1|}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
            $endgroup$
            – whereamI
            Jan 11 at 4:11












          • $begingroup$
            You might be right about that... $(overline {z_2- z_1})$ would work, too.
            $endgroup$
            – Doug M
            Jan 11 at 21:05


















          1












          $begingroup$

          The distance of $z$ from the y axis is $|Re {z}|$



          Lets work in standard Cartesian, and then look for the complex analogue.



          The distance of a point $(x,y)$ from a line $ax + by + c = 0$



          $frac {|ax+by - c|}{sqrt {a^2+b^2}}$



          What are $a,b,c?$



          the point $frac {z_1+z_2}{2}$ is on the line.



          and $(z_2-z_1)$ is normal to the line.



          $a = Re (z_2-z_1)\ b = Im (z_2-z_1)\
          c = frac 12 (|z_2|^2 - |z_1|^2)\
          sqrt {a^2+b^2} = |z_2-z_1|$



          $ax + by = Re (bar zz_2-bar zz_1)$



          $|z - z_0| = frac {|Re (bar zz_2-bar zz_1) - frac 12 (|z_2|^2 - |z_1|^2)|}{|z_2-z_1|}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
            $endgroup$
            – whereamI
            Jan 11 at 4:11












          • $begingroup$
            You might be right about that... $(overline {z_2- z_1})$ would work, too.
            $endgroup$
            – Doug M
            Jan 11 at 21:05
















          1












          1








          1





          $begingroup$

          The distance of $z$ from the y axis is $|Re {z}|$



          Lets work in standard Cartesian, and then look for the complex analogue.



          The distance of a point $(x,y)$ from a line $ax + by + c = 0$



          $frac {|ax+by - c|}{sqrt {a^2+b^2}}$



          What are $a,b,c?$



          the point $frac {z_1+z_2}{2}$ is on the line.



          and $(z_2-z_1)$ is normal to the line.



          $a = Re (z_2-z_1)\ b = Im (z_2-z_1)\
          c = frac 12 (|z_2|^2 - |z_1|^2)\
          sqrt {a^2+b^2} = |z_2-z_1|$



          $ax + by = Re (bar zz_2-bar zz_1)$



          $|z - z_0| = frac {|Re (bar zz_2-bar zz_1) - frac 12 (|z_2|^2 - |z_1|^2)|}{|z_2-z_1|}$






          share|cite|improve this answer











          $endgroup$



          The distance of $z$ from the y axis is $|Re {z}|$



          Lets work in standard Cartesian, and then look for the complex analogue.



          The distance of a point $(x,y)$ from a line $ax + by + c = 0$



          $frac {|ax+by - c|}{sqrt {a^2+b^2}}$



          What are $a,b,c?$



          the point $frac {z_1+z_2}{2}$ is on the line.



          and $(z_2-z_1)$ is normal to the line.



          $a = Re (z_2-z_1)\ b = Im (z_2-z_1)\
          c = frac 12 (|z_2|^2 - |z_1|^2)\
          sqrt {a^2+b^2} = |z_2-z_1|$



          $ax + by = Re (bar zz_2-bar zz_1)$



          $|z - z_0| = frac {|Re (bar zz_2-bar zz_1) - frac 12 (|z_2|^2 - |z_1|^2)|}{|z_2-z_1|}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 21:06

























          answered Jan 11 at 3:02









          Doug MDoug M

          45.4k31954




          45.4k31954












          • $begingroup$
            Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
            $endgroup$
            – whereamI
            Jan 11 at 4:11












          • $begingroup$
            You might be right about that... $(overline {z_2- z_1})$ would work, too.
            $endgroup$
            – Doug M
            Jan 11 at 21:05




















          • $begingroup$
            Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
            $endgroup$
            – whereamI
            Jan 11 at 4:11












          • $begingroup$
            You might be right about that... $(overline {z_2- z_1})$ would work, too.
            $endgroup$
            – Doug M
            Jan 11 at 21:05


















          $begingroup$
          Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
          $endgroup$
          – whereamI
          Jan 11 at 4:11






          $begingroup$
          Why we can have $ax + by = Re (zz_2-zz_1)$ now? Since $x+yi=z$, then we have $z(z_2-z_1)=(x+yi)(a+bi)=xa-by+(ay+bx)i$ which gives us $ax - by = Re (zz_2-zz_1)$. Should we use the conjugate of $z$ instead?
          $endgroup$
          – whereamI
          Jan 11 at 4:11














          $begingroup$
          You might be right about that... $(overline {z_2- z_1})$ would work, too.
          $endgroup$
          – Doug M
          Jan 11 at 21:05






          $begingroup$
          You might be right about that... $(overline {z_2- z_1})$ would work, too.
          $endgroup$
          – Doug M
          Jan 11 at 21:05




















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