Australian HSC mathematics extension 2 exam Polynomial

[Mentallic]

Homework Statement

This problem is from the Australian HSC mathematics extension 2 exam. Q6b)

It states:

b) Let P(x)=x^3+qx^2+qx+1, where q is real. One zero of P(x) is -1

i) Show that if \alpha is a zero of P(x) then \frac{1}{\alpha} is a zero of P(x)

ii) Suppose that \alpha is a zero of P(x) and \alpha is not real.
(1) show that |\alpha|=1
(2) show that Re(\alpha)=\frac{1-q}{2}

The Attempt at a Solution

i) I was able to answer this one

ii) (1) I was uncertain about this one so I fudged the answer, but for a quick and unsatisfactory answer, I went with the idea that if \alpha and 1/\alpha are complex roots to a real polynomial, then the roots must occur in conjugate pairs and thus have the same modulus. Actually... also

P(x)=x^3+qx^2+qx+1\equiv (x+1)(x^2+(q-1)x+1)

The quadratic factor: x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}
Thought I had it... guess not...

ii) (2) I scabbed the answer from the quadratic (without really understanding why it works) that Re(z)=Re(\alpha)=\frac{1-q}{2}
This is the part without the surd. Also, unsure of a correct approach to the answer.

Help for these 2 parts would be greatful.

 

[Office_Shredder]

For part ii, the idea that they have to come in conjugate pairs sounds good to me. What did you find wrong with it?

For the last part notice <br /> x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}<br /> is real unless the part inside the square root is negative. If that's negative, what's the real part of the number?

 

[Mentallic]

for ii) I thought that while I was able to show that|\alpha|=|\frac{1}{\alpha}|, I wasn't able to show they're both equal to 1. But now that I look at it again, for a complex number \alpha and its reciprocal to be equal, they must have a mod of 1.

for iii) well yes the real part is ofcourse obvious, but this assumes that (q-3)(q+1)&lt;0 thus -1&lt;q&lt;3
How do I explain/show that q is in fact in this range? Or can I just assume it because the question specifically stated that the roots are imaginary?

 

[Office_Shredder]

If it's not in that range, then you've found all the roots, and they are real. But one of them has to be imaginary, so that's a contradiction

 

[Mentallic]

I understand now. Thanks for your help Office_Shredder