The discussion revolves around finding a real, monic polynomial of the lowest possible degree that has specific complex zeros: −1−2i, −2i, and i. The variable used is z.
Participants are actively engaging with the problem, raising questions about the necessity of including conjugate pairs of the complex roots to achieve a real polynomial. Some hints have been provided regarding the treatment of the imaginary roots.
There is an emphasis on the requirement for the polynomial to be real, which raises questions about the handling of complex roots, particularly when no two roots form a conjugate pair.
Don't forget: It's a real polynomial.53Mark53 said:Homework Statement
A monic polynomial is a polynomial which has leading coefficient 1. Find the real, monic polynomial of the lowest possible degree which has zeros −1−2i,−2i and i. Use z as your variable.The Attempt at a Solution
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Would I just expand the zeros giving me
(z+1+2i)(z+2i)(z-i)
z^3+(1+3*I)*z^2+I*x+I*4+2
how would I make it real?SammyS said:Don't forget: It's a real polynomial.
Hint: what happened to the conjugates of the original roots?53Mark53 said:how would I make it real?
SteamKing said:Hint: what happened to the conjugates of the original roots?
SteamKing said:Hint: what happened to the conjugates of the original roots?
What is the conjugate of the complex number, a + bi?53Mark53 said:Does that mean that I have to square the imaginary roots?
The same applies here, but now you have 3 complex roots, no two of which form a conjugate pair.53Mark53 said:Homework Statement
p(x) = x^3 − x^2 + ax + b is a real polynomial with 1 + i as a zero, find a and b and find all of the real zeros of p(x).
The Attempt at a Solution
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1-i is also a zero as it is the conjugate of 1+i
Thanks I got the right answer by using conjugate pairsSammyS said:@53Mark53 ,
Here's what you said in a previous thread:
The same applies here, but now you have 3 complex roots, no two of which form a conjugate pair.