Monic polynomial of the lowest possible degree

  • #1
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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


[/B]
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
 

Answers and Replies

  • #2
SammyS
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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


[/B]
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
Don't forget: It's a real polynomial.
 
  • #3
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Don't forget: It's a real polynomial.
how would I make it real?
 
  • #4
SteamKing
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how would I make it real?
Hint: what happened to the conjugates of the original roots?
 
  • #5
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Hint: what happened to the conjugates of the original roots?
Hint: what happened to the conjugates of the original roots?
Does that mean that I have to square the imaginary roots?
 
  • #6
SteamKing
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Does that mean that I have to square the imaginary roots?
What is the conjugate of the complex number, a + bi?
 
  • #7
SammyS
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@53Mark53 ,
Here's what you said in a previous thread:

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


[/B]
1-i is also a zero as it is the conjugate of 1+i
The same applies here, but now you have 3 complex roots, no two of which form a conjugate pair.
 
  • #8
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@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.
Thanks I got the right answer by using conjugate pairs
 

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