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Math Help - What is complex slope of a line in complex numbers

  1. #1
    Senior Member pankaj's Avatar
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    What is complex slope of a line in complex numbers

    If equation of a line in complex plane is given by
     <br />
\bar az+a\bar z+b=0<br />
    where a is a complex constant and b is a real number
    Then complex slope of the line is defined as \omega =-\frac{a}{\bar a}

    Now,what I want to know is that what is the complex slope of the line,i.e. what is its geometrical meaning.Also,what do a and b signify geometrically
    Last edited by mash; March 4th 2012 at 12:09 PM. Reason: fixed latex
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  2. #2
    MHF Contributor

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    Let a= u+ iv, b= p+ iq, z= x+ iy. Then [tex]\overline{a}z+ a\overline{z}+ b= 0[tex] becomes (u- iv)(x+ iy)+ (u+ iv)(x- iy)+ p+ iq= 0

    (ux+ vy)+ i(uy- vx)+ (ux+ vy)+ i(vx- uy)+ p+ iq= 0
    Notice that i(uy- vx) and i(vx-uy) cancel. Separating real and imaginary parts, 2ux+ 2vy+ p= 0 and q= 0 (the latter is why b= p is a real number).
    We can rewrite the first equation as 2vy= -2ux+ b or y= -(u/v)x+ b/(2v) so that is, in fact, a straight line with slope -u/v and y-intercept b/(2v).

    Now, I honestly don't see what that has to do with the "complex slope", -\frac{a}{\overline{a}}! Where did you see that formula?
    Last edited by mash; March 4th 2012 at 12:09 PM. Reason: fixed latex
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  3. #3
    Senior Member pankaj's Avatar
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    It was given in a textbook .

    Among other things it was also given that complex slope of line joining z_{1} and z_{2} is given by \frac{z_{1}-z_{2}}{\bar z_{1}-\bar z_{2}} .

    also if complex slopes of two lines equal then the two lines are parallel and the two lines will be perpendicular if the sum of the slopes is 0.

    It is also given that complex slope of line making angle \theta with the real axis is given by \omega=e^{2i\theta}

    All this I have verified and they are true.

    But I want to understand the geometrical interpretation of this concept.
    Last edited by mash; March 4th 2012 at 12:10 PM. Reason: fixed latex
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