Complex slope of line $abar{z}+bar{a}z+b = 0$












7












$begingroup$


How can we prove.........



[1] The Complex slope of the line $abar{z}+bar{a}z+b = 0$ is $displaystyle omega = -frac{a}{bar{a}}$



[2] Complex slope of line joining the points $z_{1}$ and $z_{2}$ is $displaystyle omega = frac{z_{1}-z_{2}}{bar{z_{1}}-bar{z_{2}}}$



My Try:: Let General equation of line is $Ax+By+C=0$ , Now Let $z = x+iy$ and $bar{z} = x-iy$



Then $displaystyle Aleft(frac{z+bar{z}}{2}right)+Bleft(frac{z-bar{z}}{2i}right)+C = 0$



So we get $(B+iA)z+(-B+iA)bar{z}+iC = 0$



Now I did not understand How can i proceed further so that I can prove [1] and [2]






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How is "complex slope" defined for you?
    $endgroup$
    – Cameron Buie
    Sep 5 '13 at 5:26
















7












$begingroup$


How can we prove.........



[1] The Complex slope of the line $abar{z}+bar{a}z+b = 0$ is $displaystyle omega = -frac{a}{bar{a}}$



[2] Complex slope of line joining the points $z_{1}$ and $z_{2}$ is $displaystyle omega = frac{z_{1}-z_{2}}{bar{z_{1}}-bar{z_{2}}}$



My Try:: Let General equation of line is $Ax+By+C=0$ , Now Let $z = x+iy$ and $bar{z} = x-iy$



Then $displaystyle Aleft(frac{z+bar{z}}{2}right)+Bleft(frac{z-bar{z}}{2i}right)+C = 0$



So we get $(B+iA)z+(-B+iA)bar{z}+iC = 0$



Now I did not understand How can i proceed further so that I can prove [1] and [2]






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How is "complex slope" defined for you?
    $endgroup$
    – Cameron Buie
    Sep 5 '13 at 5:26














7












7








7


3



$begingroup$


How can we prove.........



[1] The Complex slope of the line $abar{z}+bar{a}z+b = 0$ is $displaystyle omega = -frac{a}{bar{a}}$



[2] Complex slope of line joining the points $z_{1}$ and $z_{2}$ is $displaystyle omega = frac{z_{1}-z_{2}}{bar{z_{1}}-bar{z_{2}}}$



My Try:: Let General equation of line is $Ax+By+C=0$ , Now Let $z = x+iy$ and $bar{z} = x-iy$



Then $displaystyle Aleft(frac{z+bar{z}}{2}right)+Bleft(frac{z-bar{z}}{2i}right)+C = 0$



So we get $(B+iA)z+(-B+iA)bar{z}+iC = 0$



Now I did not understand How can i proceed further so that I can prove [1] and [2]






algebra-precalculus







share|cite|improve this question











$endgroup$




How can we prove.........



[1] The Complex slope of the line $abar{z}+bar{a}z+b = 0$ is $displaystyle omega = -frac{a}{bar{a}}$



[2] Complex slope of line joining the points $z_{1}$ and $z_{2}$ is $displaystyle omega = frac{z_{1}-z_{2}}{bar{z_{1}}-bar{z_{2}}}$



My Try:: Let General equation of line is $Ax+By+C=0$ , Now Let $z = x+iy$ and $bar{z} = x-iy$



Then $displaystyle Aleft(frac{z+bar{z}}{2}right)+Bleft(frac{z-bar{z}}{2i}right)+C = 0$



So we get $(B+iA)z+(-B+iA)bar{z}+iC = 0$



Now I did not understand How can i proceed further so that I can prove [1] and [2]






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 15:27









Abcd

3,02321135




3,02321135










asked Sep 5 '13 at 4:45









juantheronjuantheron

34.1k1147142




34.1k1147142








  • 2




    $begingroup$
    How is "complex slope" defined for you?
    $endgroup$
    – Cameron Buie
    Sep 5 '13 at 5:26














  • 2




    $begingroup$
    How is "complex slope" defined for you?
    $endgroup$
    – Cameron Buie
    Sep 5 '13 at 5:26








2




2




$begingroup$
How is "complex slope" defined for you?
$endgroup$
– Cameron Buie
Sep 5 '13 at 5:26




$begingroup$
How is "complex slope" defined for you?
$endgroup$
– Cameron Buie
Sep 5 '13 at 5:26










1 Answer
1






active

oldest

votes


















4












$begingroup$

To answer question your $1^{st}$ question, I would have to answer your $2^{nd}$ question before that.



The answer to your second question is that you cannot prove it as it is the term coined for the expression $dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$, where $z_1,z_2 in mathbb{C}$ . Now, why it had to be done can be answered by looking up the proof of either the condition that two lines be parallel or the two lines be perpendicular in the Argand plane. In these proofs, you can see that at the end that you always get a condition like $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=dfrac{z_3-z_4}{bar{z_3}-bar{z_4}}$$ $$or$$ $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=- left(dfrac{z_3-z_4}{bar{z_3}-bar{z_4}} right)$$.



Now just to make it cleaner we coined the term, complex slope, i.e. $w=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$



Now, onto the answer of first question.



What you did is the longer method. The shorter method would have been to take the equation of the line that you wrote in your question and let $z_1$ and $z_2$ lie on this line. Then these complex numbers satisfy the equation of the complex line. So, we get the following equations:-



$$begin{equation} label{eq:1} tag{1}
bar{a}z_1+abar{z_1}+b = 0
end{equation}$$
$$begin{equation} label{eq:2} tag{2}
bar{a}z_2+abar{z_2}+b = 0
end{equation}$$



Now, subtract $(2)$ from $(1)$, to get after rearrangement the result needed, i.e.



$$-dfrac{a}{bar{a}}=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=w$$



What you did is also correct and you have also almost reached to the conclusion of the proof. All you have to do is multiply the final equation obtained by $i$ and then you get the general form of a line in the Argand plane.






share|cite|improve this answer











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    1 Answer
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    active

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    4












    $begingroup$

    To answer question your $1^{st}$ question, I would have to answer your $2^{nd}$ question before that.



    The answer to your second question is that you cannot prove it as it is the term coined for the expression $dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$, where $z_1,z_2 in mathbb{C}$ . Now, why it had to be done can be answered by looking up the proof of either the condition that two lines be parallel or the two lines be perpendicular in the Argand plane. In these proofs, you can see that at the end that you always get a condition like $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=dfrac{z_3-z_4}{bar{z_3}-bar{z_4}}$$ $$or$$ $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=- left(dfrac{z_3-z_4}{bar{z_3}-bar{z_4}} right)$$.



    Now just to make it cleaner we coined the term, complex slope, i.e. $w=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$



    Now, onto the answer of first question.



    What you did is the longer method. The shorter method would have been to take the equation of the line that you wrote in your question and let $z_1$ and $z_2$ lie on this line. Then these complex numbers satisfy the equation of the complex line. So, we get the following equations:-



    $$begin{equation} label{eq:1} tag{1}
    bar{a}z_1+abar{z_1}+b = 0
    end{equation}$$
    $$begin{equation} label{eq:2} tag{2}
    bar{a}z_2+abar{z_2}+b = 0
    end{equation}$$



    Now, subtract $(2)$ from $(1)$, to get after rearrangement the result needed, i.e.



    $$-dfrac{a}{bar{a}}=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=w$$



    What you did is also correct and you have also almost reached to the conclusion of the proof. All you have to do is multiply the final equation obtained by $i$ and then you get the general form of a line in the Argand plane.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      To answer question your $1^{st}$ question, I would have to answer your $2^{nd}$ question before that.



      The answer to your second question is that you cannot prove it as it is the term coined for the expression $dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$, where $z_1,z_2 in mathbb{C}$ . Now, why it had to be done can be answered by looking up the proof of either the condition that two lines be parallel or the two lines be perpendicular in the Argand plane. In these proofs, you can see that at the end that you always get a condition like $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=dfrac{z_3-z_4}{bar{z_3}-bar{z_4}}$$ $$or$$ $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=- left(dfrac{z_3-z_4}{bar{z_3}-bar{z_4}} right)$$.



      Now just to make it cleaner we coined the term, complex slope, i.e. $w=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$



      Now, onto the answer of first question.



      What you did is the longer method. The shorter method would have been to take the equation of the line that you wrote in your question and let $z_1$ and $z_2$ lie on this line. Then these complex numbers satisfy the equation of the complex line. So, we get the following equations:-



      $$begin{equation} label{eq:1} tag{1}
      bar{a}z_1+abar{z_1}+b = 0
      end{equation}$$
      $$begin{equation} label{eq:2} tag{2}
      bar{a}z_2+abar{z_2}+b = 0
      end{equation}$$



      Now, subtract $(2)$ from $(1)$, to get after rearrangement the result needed, i.e.



      $$-dfrac{a}{bar{a}}=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=w$$



      What you did is also correct and you have also almost reached to the conclusion of the proof. All you have to do is multiply the final equation obtained by $i$ and then you get the general form of a line in the Argand plane.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        To answer question your $1^{st}$ question, I would have to answer your $2^{nd}$ question before that.



        The answer to your second question is that you cannot prove it as it is the term coined for the expression $dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$, where $z_1,z_2 in mathbb{C}$ . Now, why it had to be done can be answered by looking up the proof of either the condition that two lines be parallel or the two lines be perpendicular in the Argand plane. In these proofs, you can see that at the end that you always get a condition like $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=dfrac{z_3-z_4}{bar{z_3}-bar{z_4}}$$ $$or$$ $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=- left(dfrac{z_3-z_4}{bar{z_3}-bar{z_4}} right)$$.



        Now just to make it cleaner we coined the term, complex slope, i.e. $w=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$



        Now, onto the answer of first question.



        What you did is the longer method. The shorter method would have been to take the equation of the line that you wrote in your question and let $z_1$ and $z_2$ lie on this line. Then these complex numbers satisfy the equation of the complex line. So, we get the following equations:-



        $$begin{equation} label{eq:1} tag{1}
        bar{a}z_1+abar{z_1}+b = 0
        end{equation}$$
        $$begin{equation} label{eq:2} tag{2}
        bar{a}z_2+abar{z_2}+b = 0
        end{equation}$$



        Now, subtract $(2)$ from $(1)$, to get after rearrangement the result needed, i.e.



        $$-dfrac{a}{bar{a}}=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=w$$



        What you did is also correct and you have also almost reached to the conclusion of the proof. All you have to do is multiply the final equation obtained by $i$ and then you get the general form of a line in the Argand plane.






        share|cite|improve this answer











        $endgroup$



        To answer question your $1^{st}$ question, I would have to answer your $2^{nd}$ question before that.



        The answer to your second question is that you cannot prove it as it is the term coined for the expression $dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$, where $z_1,z_2 in mathbb{C}$ . Now, why it had to be done can be answered by looking up the proof of either the condition that two lines be parallel or the two lines be perpendicular in the Argand plane. In these proofs, you can see that at the end that you always get a condition like $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=dfrac{z_3-z_4}{bar{z_3}-bar{z_4}}$$ $$or$$ $$dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=- left(dfrac{z_3-z_4}{bar{z_3}-bar{z_4}} right)$$.



        Now just to make it cleaner we coined the term, complex slope, i.e. $w=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}$



        Now, onto the answer of first question.



        What you did is the longer method. The shorter method would have been to take the equation of the line that you wrote in your question and let $z_1$ and $z_2$ lie on this line. Then these complex numbers satisfy the equation of the complex line. So, we get the following equations:-



        $$begin{equation} label{eq:1} tag{1}
        bar{a}z_1+abar{z_1}+b = 0
        end{equation}$$
        $$begin{equation} label{eq:2} tag{2}
        bar{a}z_2+abar{z_2}+b = 0
        end{equation}$$



        Now, subtract $(2)$ from $(1)$, to get after rearrangement the result needed, i.e.



        $$-dfrac{a}{bar{a}}=dfrac{z_1-z_2}{bar{z_1}-bar{z_2}}=w$$



        What you did is also correct and you have also almost reached to the conclusion of the proof. All you have to do is multiply the final equation obtained by $i$ and then you get the general form of a line in the Argand plane.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 7 '17 at 13:45

























        answered Jul 7 '16 at 10:17









        user350331user350331

        1,26511131




        1,26511131






























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