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Finding complex number with the lowest argument.

  1. Dec 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Of all complex numbers that fit requirement: ## |z-25i| \leq 15## find the one with the lowest argument.

    2. Relevant equations


    3. The attempt at a solution
    z=a + ib (a, b are real numbers)

    ## \sqrt{a^2 + (b-25)^2} \leq 15 \\ a^2 + (b-25)^2 \leq 225 ##

    The lowest possible argument is zero, in that case, complex number has only real part, which means that imaginary part is zero, it is obvious that none of the numbers with such argument fit this requirement, because, if b is zero then on the LHS we have a squared plus 625, which is way greater than 225 even without a^2.
     
  2. jcsd
  3. Dec 11, 2015 #2

    Samy_A

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    You have to bring the argument ##\phi## into play.
    You could transform your inequality by using:
    ##
    \begin{cases}
    a=r\cos \phi \\
    b=r\sin \phi
    \end{cases}
    ##
     
    Last edited: Dec 11, 2015
  4. Dec 11, 2015 #3
    Ok, let's see,

    ## r^2cos^2\phi + (rsin\phi - 25)^2 \leq 225 \\ r^2cos^2\phi + r^2sin^2\phi - 50rsin\phi + 625 \leq 225 \\ r^2 -r50sin\phi +400 \leq 0 ##

    So i ended up with quadratic equation by unknown r, but i don't know ##\phi## either.
     
  5. Dec 11, 2015 #4

    Samy_A

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    Rewrite the inequality with ##\phi## on one side and r on the other side:
    ##\frac{r²+400}{50r} \leq \sin \phi##
    Now remember you have to find the lowest possible ##\phi##.
    The LHS is clearly positive, so you are looking for the r-value that minimizes that LHS.
     
  6. Dec 11, 2015 #5

    Samy_A

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    A geometrical approach would also work, but I don't want to distract you if you don't need another approach.
    Take the tangent from the origin to the circle centered at ##25i## with radius 15 (the tangent with positive slope, as we want the lowest positive argument). The point you need is the intersection of the tangent and the circle.
    This will easily give you r, and, from there you get ##\phi##.
     
  7. Dec 11, 2015 #6
    But, how can i find the lowest possible value, is it by taking derivative?
     
  8. Dec 11, 2015 #7

    Samy_A

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    Yes, find a zero of the derivative and (if there is one) establish that it is a minimum.
     
  9. Dec 11, 2015 #8
    I think i solved it, lowest r = 20 so ## \phi = \arcsin\frac{4}{5}##.

    But, i would really like if you could explain me geometrical approach (i hope im not bothering you too much), it might help me understand what is really going on there.
     
  10. Dec 11, 2015 #9

    Samy_A

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    So ##z=12 + 16i##.

    Snapshot.jpg
    You are looking for the point ##z## that satisfies ##|z-25i| \leq 15## and with the lowest possible argument.
    ##z## has to be in (or on) the circle with centre at (0,25) and with radius 15.
    You will get the lowest argument with the tangent from the origin to the circle. That gives you the point S (or ##z## as complex number). As POS is a right triangle, you easily get that r=20.
    And then you can compute ##\arcsin \phi## from the same triangle POS.

    (Sorry for the clumsy drawing, never used that whiteboard before. P is supposed to be the centre of the circle.:smile:)
     
    Last edited: Dec 11, 2015
  11. Dec 11, 2015 #10

    Mark44

    Staff: Mentor

    Keep in mind that |z - 25i| represents the distance from a complex number z to the complex number 25i. You should be able to solve this problem simply by drawing a diagram.
     
  12. Dec 11, 2015 #11

    Samy_A

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    Deleted as irrelevant.
     
    Last edited: Dec 11, 2015
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