Roots of a third degree polynomal equation (complex numbers)

hrappur2
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Hey everyone! Here I have a problem I don't know how to solve so help would be greatly appreciated!

Homework Statement


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 c2. 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.
 
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hi hrappur2! :smile:
hrappur2 said:
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 …

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:
 
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.
 
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?
 
yes, the third root must be real, so it's -2 or 4 :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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