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Homework Statement
This problem is from the Australian HSC mathematics extension 2 exam. Q6b)
It states:
b) Let [tex]P(x)=x^3+qx^2+qx+1[/tex], where q is real. One zero of P(x) is -1
i) Show that if [itex]\alpha[/itex] is a zero of P(x) then [tex]\frac{1}{\alpha}[/tex] is a zero of P(x)
ii) Suppose that [itex]\alpha[/itex] is a zero of P(x) and [itex]\alpha[/itex] is not real.
(1) show that [tex]|\alpha|=1[/tex]
(2) show that [tex]Re(\alpha)=\frac{1-q}{2}[/tex]
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 [itex]\alpha[/itex] and [itex]1/\alpha[/itex] are complex roots to a real polynomial, then the roots must occur in conjugate pairs and thus have the same modulus. Actually... also
[tex]P(x)=x^3+qx^2+qx+1\equiv (x+1)(x^2+(q-1)x+1)[/tex]
The quadratic factor: [tex]x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}[/tex]
Thought I had it... guess not...
ii) (2) I scabbed the answer from the quadratic (without really understanding why it works) that [tex]Re(z)=Re(\alpha)=\frac{1-q}{2}[/tex]
This is the part without the surd. Also, unsure of a correct approach to the answer.
Help for these 2 parts would be greatful.