The discussion focuses on finding the real, monic polynomial of the lowest possible degree with given zeros: -1-2i, -2i, and i. The correct approach involves recognizing that complex roots must occur in conjugate pairs to ensure the polynomial remains real. Therefore, the conjugates of the complex roots, specifically 1+2i and 2i, must also be included. The final polynomial is derived by expanding the product of the factors corresponding to these roots, resulting in a polynomial of degree four.
PREREQUISITESMathematics students, educators, and anyone interested in polynomial theory, particularly those dealing with complex roots and their implications in real polynomials.
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.