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Argument of a random complex no. lying on given line segment.

  1. Dec 16, 2015 #1
    1. The problem statement, all variables and given/known data
    In the argand plane z lies on the line segment joining # z_1 = -3 + 5i # and # z_2 = -5 - 3i # . Find the most suitable answer from the following options .

    A) -3∏/4

    B) ∏/4

    C) 5∏/6

    D) ∏/6

    2. MY ATTEMPT AT THE SOLUTION

    We get two points ( -3 , 5 ) & ( -5 , -3 ) => The line segment must intersect x - axis and lie in the 2nd and 3rd quadrant .
    => #z# lies on x-axis or 2nd or 3rd qudrant .

    But , since there is no option ∏ , so z must be lying in 2nd or 3rd quadrant.

    => -3∏/4 or 5∏/6 should be the solution .

    I cant proceed further from here so as to differentiate between the above two choices.

    The answer in the book is : 5∏/6 .
     
    Last edited: Dec 16, 2015
  2. jcsd
  3. Dec 16, 2015 #2

    haruspex

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    The set of choices is only defined as far as directions from the origin. So I would think the approach is to find the arguments of z1 and z2.
     
    Last edited: Dec 17, 2015
  4. Dec 17, 2015 #3

    LCKurtz

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    How is your line situated with respect to the line ##y = x##, which is where a ##z## with argument ##-\frac{3\pi}{4}## would be? Draw a graph.
     
  5. Dec 18, 2015 #4
    You could also compute the argument : z=end-start and arg z or atan dy/dx
     
  6. Dec 18, 2015 #5

    BvU

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    Hello Rick, :welcome:

    If you pick random points on the line between those two points on your drawing, the average postion of those points should end up where, do you think ? You calculated the intersection with the negative x-axis, but you should have calculated the midpoint of the line segment. The angle that is closest to the argument of that point is your best answer.

    Oh, and: you did draw a graph, I hope ?

    :smile:
     
  7. Dec 18, 2015 #6

    SammyS

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    Is that a complete statement of the problem - word for word?

    If not, please give a complete statement of the problem - word for word. Also, if a portion of your question is in the thread title, please include it in the body of the thread as well.
     
  8. Dec 18, 2015 #7

    LCKurtz

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    Anyone want to give odds on the OP ever returning to this thread?
     
  9. Dec 18, 2015 #8

    SammyS

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    ##\displaystyle \ \left(\frac{1}{e}\right)^\pi \ ##
     
  10. Dec 18, 2015 #9

    haruspex

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    That's completely irrational. Probably.
     
  11. Dec 18, 2015 #10
    I am extremely sorry . I was in a hurry . Thats not the complete question . The question asked for " find the suitable solution for arg(z) " .
     
  12. Dec 18, 2015 #11

    haruspex

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    Have you tried my suggestion in post #2?
     
  13. Dec 18, 2015 #12
    Yup that solved it , but for that I needed a calculator to find tan inverse 3/5 & 5/3 .
     
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