some cambridge complex Q's (1 Viewer)

[ordnaiv]

sorry to be bothersome, but I'm kinda stuck on these 2, so if anyone can help out that would be awesome.

1. If p is real, and -2<p<2, show that the roots of the equation x^2 + px + 1 = 0 are non real complex numbers with a modulus of 1

2. Let H and K be the points representing the roots of x^2 + 2px + q, where p and q are real and p^2 < q. Find the algebraic relation satisfied by p and q in each of the following cases:

when angle HOK is a right angle

when A, B, H and K are equidistant from O

thanks

 

[Riviet]

1. If p is real, and -2<p<2, show that the roots of the equation x^2 + px + 1 = 0 are non real complex numbers with a modulus of 1

discriminant=p²-4(1)(1)
=p²-4

Now given -2 < p < 2,

then p²<4

p²-4<0

Therefore the discriminant is less than zero. Hence the roots are non real complex numbers.

Now product of 2 roots of equation = c/a = 1

So that means that the modulus of the product of the two complex roots is 1, and the only possibility for this is when both complex numbers have a modulus of 1 since 1x1=1.

Note: I've assumed that the roots are complex conjugates, which is a theorem that you'll learn about in the polynomials topic. Complex conjugates have the same modulus, their arguments are equal but opposite in sign e.g @ and -@.

Let z=rcis@ so that its conjugate is rcis(-@).
Then rcis@.rcis(-@)=r²cis0
=r²

Now r²=1 (product of roots)

.'. r=+1

But r is a distance so we take r=1, giving us the modulus.

 

[ordnaiv]

yeah that's what I thought but there are no answers at the back of the book for that particular question so i was unsure.

thanks for the help!