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Roots of a third degree polynomal equation (complex numbers)

  1. Mar 10, 2012 #1
    Hey everyone! Here I have a problem I don't know how to solve so help would be greatly appreciated!

    1. The problem statement, all variables and given/known data
    Here is an equation z^3+az^2+bz+c where a, b and c are real numbers. If the roots are drawn in the complex plane they form a triangle with area of 9 units. One root of the equation is 1+3i.

    (a) Find other roots of the equation.
    (b) Find a, b and c


    2. The attempt at a solution
    The only thing I know is that if 1+3i is a root than 1-3i is also a root but more I don't know, I have no idea how to solve this without a, b and c i don't know how to find them.

    -Thanks in advance!

    EDIT: I think I finally figured out how to solve this. I tried to insert (1+3i) and (1-3i) into the equation and got a=1, b=4 and c=30 then I put it in the original equation and get the third root, z=-3. Is that correct? What is confusing me is that the area has to be 9, I don't get that.
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    tiny-tim

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    hi hrappur2! :smile:
    how? :confused:

    it might be easier to start with the triangle …

    you know the position of one of its sides …

    so what line does the opposite point have to be on if the area is 9 ? :wink:
     
  4. Mar 10, 2012 #3

    HallsofIvy

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    Unfortunately, I can't see the equation itself but yes, if the equation has only real coefficients and 1+ 3i is a root then 1- 3i is another root and the third root must be real. The fact that it is real means it is at the point (x, 0) in the complex plane for some real number x. Can you determine the area of such a triangle, in terms of x?

    Note tiny-tim's suggestion. The area of a triangle is "base times height" and we can take the line through 1+ 3i and 1- 3i ((1, 3) and (1, -3)) as base. What is its length? Knowing that and the fact that the area is 9, we can find the height of the triangle. The third vertex must lie on a line parallel to the line through (1, 3) and (1, -3) and a distance equal to the height of the triangle from it.
     
  5. Mar 10, 2012 #4
    Thanks a lot for your help!

    So if I'm understanding this correctly I draw 1-3i and 1+3i into the plane and get that the base is 6 so I have to have height 3 so the area is 9. So the third root must be either -2 or 4. Is that correct?
     
  6. Mar 10, 2012 #5

    tiny-tim

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    yes, the third root must be real, so it's -2 or 4 :smile:
     
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