Another complex numbers questions (roots)?

Question

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 modulus 1.

Answer 1

x^2 + px + 1 = 0

use  the quadratic formula

a = 1　b = p　c = 1

-b ± sqrt ( b^2 - 4ac ) / 2a

= -p ± sqrt (p^2 - 4) / 2

if the roots are non-real

p^2 - 4 &lt;0

p^2 &lt; 4

-2 &lt; p &lt; 2

Answer 2

as p is real and - 2 &lt; p &lt; 2 we can put

p = 2 cos t and t != 0 t

now b^2 - 4ac = 4 cos^2 t - 4 = - 4 sin ^2 t

roots - (-b +/-sqrt(b^2-4ac))/ 2 = -2 cos t +/- 2 i sin t)/2

= - cos t +/- i sin t

as cos t != +/-1 and so sin t != 0

so roots are non real and complex

and cearly |r| = | - cos t +/-i sin t| = 1

hence proved