polynomials (are GREAT) (1 Viewer)

[243_robbo]

hi i have another (!!!) question -

JB Fitzpatrick ex 36(c) q 41

write out an equation of the lowest possible degree with (i) comples coefficients, (ii) rational coefficients and having the folloring among it's roots:
sqrt(3) + 1, 2 - i


i can get part 1 by multiplying out (z - (sqrt(3) + 1)(z - (2 - i))

but how do i find the real coefficient one?

is it something to do with multiplying by the conjuages of the original complex roots so that the ratioanlise them?

please help?

 

[Riviet]

is it something to do with multiplying by the conjuages of the original complex roots so that the ratioanlise them?

Yep, remember the complex conjugate root theorem? If 2-i is a root of a polynomial equation with only real coefficients, then 2+i, it's conjugate must also be a root. Also, for the coefficients to be all rational, you also need the conjugate of sqrt3 + 1.

So your equation should be:

(z-(sqrt3 + 1))(z-(sqrt3 - 1)(z-(2+i))(z-(2-i))=0

 

[243_robbo]

would it be applicable just to multiply the answer for part 1 by the conjugate of the complex root.

also is the complex conjugate of (z - (x + iy))

(z - (x - iy)) or (z + (x + iy))


pretty sure its first

 

[243_robbo]

i just had a look at the answer - it is polynomial of degree 4 ???, should it be 3 if were only multiplying by one more z?

 

[Riviet]

Oh sorry, I forgot that you also need the conjugate of sqrt3 + 1 so that you get rational coefficients.

 

[243_robbo]

ok so in general to make a polynomial real multiply by conjugates of the complex roots, to make it rational, multiply by the conjugates of the irattional roots, so for that question the reason the answer has z^4 is that to make it both real and ratioanl you have to multiply by two conjugates, is that right? (so multiply by the jonjugate of an irrational and it'll just ratioanlise itself, cool thats easy)

 

[Mountain.Dew]

ok so in general to make a polynomial real multiply by conjugates of the complex roots, to make it rational, multiply by the conjugates of the irattional roots, so for that question the reason the answer has z^4 is that to make it both real and ratioanl you have to multiply by two conjugates, is that right? (so multiply by the jonjugate of an irrational and it'll just ratioanlise itself, cool thats easy)

thats pretty much it. because of the mentioned property of the poly in the question, there HAS to be 4 roots --> 4 linear factors --> hence you have polynomial with degree 4.

oh, and one last thing. the conjugate of sqrt3 + 1 isnt sqrt3 -1 - instead, its -sqrt 3 + 1, OR 1 - sqrt 3...remember, like complex numbers, it is always ideal to write the 'whole', real part first, THEN the part that defines what type of number it is (surds or complex) IE conjugate of a + sqrt b = a + - sqrt b, conjugate of x+iy = x-iy.

hope it clear things up, M.D.

 

[243_robbo]

oh, and one last thing. the conjugate of sqrt3 + 1 isnt sqrt3 -1 - instead, its -sqrt 3 + 1, OR 1 - sqrt 3...remember, like complex numbers, it is always ideal to write the 'whole', real part first, THEN the part that defines what type of number it is (surds or complex) IE conjugate of a + sqrt b = a + - sqrt b, conjugate of x+iy = x-iy.

hope it clear things up, M.D.

wow thats so helpful, i just dropped a bit of cheese as i read it ! i tried multiplying it out and gave up cos it is some massive numberthat will simplify out really well but it looked to huge to continue