1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angle made by the line

  1. Oct 26, 2012 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Consider a line [itex]\alpha\overline{z}+\overline{\alpha}z+i\beta [/itex] such that [itex]\frac{-\alpha}{\overline{\alpha}} = \lambda(1+i), \lambda \in R^{+}[/itex], then the angle made by line with real axis is [itex]\dfrac{\pi}{k},[/itex] then k is


    2. Relevant equations

    3. The attempt at a solution
    I have tried nearly every possible method but still did not get the answer. Suppose I substitute α in the equation of given line I get

    [itex] (-\overline{\alpha}\lambda(1+i) )\overline{z}+\overline{\alpha}z+i\beta[/itex].

    But I don't see any point in doing these things as it won't help me.

    Second Method
    The equation of line perpendicular to this line is given by
    [itex]z\overline{\alpha}-\overline{z}\alpha+b[/itex] (for some 'b')
    If somehow I could get the slope of this line I would get my answer. But the problem is how?
     
  2. jcsd
  3. Oct 26, 2012 #2

    Mark44

    Staff: Mentor

    ... then k is what? Is some information missing here?
    I haven't worked this all the way through, but I started by writing ##\alpha = \alpha_1 + \alpha_2 i## and z = z1 + z2 i to write the given complex number in rectangular form.

    I'm assuming that ##\beta## is also complex.
     
  4. Oct 26, 2012 #3

    utkarshakash

    User Avatar
    Gold Member

    In this question I have to find k.
    Here β is not complex. It is real.
     
  5. Oct 27, 2012 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    How does that define a line? Do you mean this expression = 0?
    If so, I think there must be something wrong. Substitute z=x+iy and similarly for alpha and notice what happens to the imaginary part.
     
  6. Oct 27, 2012 #5

    utkarshakash

    User Avatar
    Gold Member

    Of course, that expression is equal to 0.
    Doing so I get the following

    [itex]2(x_{1}x_{2}+y_1y_2) + i\beta=0 [/itex]
     
  7. Oct 27, 2012 #6

    Mark44

    Staff: Mentor

    IMO the problem was stated unclearly. I interpret the problem this way: Consider the [STRIKE]line[/STRIKE] complex number ...

    Graphing the complex number in the plane produces a line segment that makes a certain angle with the positive real axis.
     
  8. Oct 27, 2012 #7
    The equation relating α and α* (conjugate) places a constraint on what values α can take on as a function of (real) lambda. For example, suppose α is written in polar form, what does that constraint say about α? You'll find that λ is fixed and α is arbitrary length at a fixed angle.

    But there is something wrong with that equation. You can see there is trouble because αz* and α*z are complex conjugates of each other so that their sum is real. So your final equation is a+jβ = 0 for some real a, which can only be satisfied if a and β are both zero. ie any solutions (if there is more than one) will lie on the real axis.
     
    Last edited: Oct 27, 2012
  9. Oct 27, 2012 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I thought of that, but I still don't see it. What is being varied to generate the complex number? Presumably it's z, which I take to be a complex number, so it could generate the whole plane, not just a line. As it happens, the imaginary parts of the first two terms cancel, so it ends up generating the line x+iβ. But that makes nonsense of the rest of the question.
     
  10. Oct 28, 2012 #9

    utkarshakash

    User Avatar
    Gold Member

    So does that mean that the question is incorrect?
     
  11. Oct 28, 2012 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I've thought of another interpretation: it's the lambda that's the independent variable. Varying that, keeping z and beta constant, looks like it should make the expression generate a line. But that runs into another problem.
    [itex]|\frac{-\alpha}{\overline{\alpha}}| = 1[/itex], so |λ| = 1/√2, leaving only two possible values for λ.
    I'm not certain the question is wrong, but I can't make sense of it. Have you triple-checked it's copied out correctly?
     
  12. Oct 29, 2012 #11

    utkarshakash

    User Avatar
    Gold Member


    My teacher says its incorrect
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook