Recent content by 53Mark53

Thanks I got the right answer by using conjugate pairs

Does that mean that I have to square the imaginary roots?

how would I make it real?

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...

Thanks that works :)

Why would it matter if it is positive or negative?

Homework Statement [/B] Suppose q(z) = z^3 − z^2 + rz + s, is a complex polynomial with 1 + i and i as zeros. Find r and s and the third complex zero. The Attempt at a Solution [/B] (z-(1+i)(z-i) = Z^2-z-1-2iz+i (Z^2-z-1-2iz+i)(z+d) = Z^3+z^2(d-1-zi)-z(d+1+2di-i)-d(1-i) Z^2 term...

does this meant that -x^2=(d-2)x^2 -1=d-2 d=1

Sum of roots taken 1 at a time = -(coeff of x^2)/(coeff of x^3) =-1/1=1 Sum of roots taken 2 at a time= (coeff of x)/(coeff of x^3) = a/1=a Sum of roots taken 3 at a time = -(constant)/(coeff of x^3) = b/1=b What would I do now?

I'm not sure that I understand would I just use C as a root and expand (1+i)(1-i)C

using constant term b=2d x terms a=2d+2 x^2 term -1=d-2 Would this be the equations that I need to use?

It should say x^3+dx^2-2x^2-2xd+2x+2d

I can Expand it which would give me x^3+dx^-2x^2-2xde+2x+2d What would I do now? Would I sub in a (1+i) in for x which would give me d2i-2d+2id+2d

That is the third root which I have to find How else would I go about solving it?

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 so (x-(1+i))(x-(1-i))=x^2-2x+2 let X^3-x^2+ax+b=x^2-2x+2(ax+d)...