Cambridge Q (1 Viewer)

[Tsylana]

Excercise 2.2

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.


Yes I've checked the worked solutions xP... doesn't seem to explain the 2nd part of the question..

 

[cyl123]

For the equation x^2 + px + 1 = 0 since coefficients are real, then the roots must occur in conjugate pairs. Thus letting the roots be z1 and z2 where z2 is the conjugate of z1.

From product of roots: (z1)z2 = 1
z1z2 = |z1|^2 since z1 and z2 are conjugates (one of the rules of multiplying conjugates)
so |z1|^2 =1 thus |z1| =1 and also |z2|=1 (uses the fact the modulus of the complex number is the same as the modulus of the conjugate of that complex number)

To prove the complex numbers are not real, using quad equation gives:
x = (-p +- sqrt(p^2-4))/2 and using -2 < p < 2, p^2<4 and hence x is complex

 

[Tsylana]

yeahh i just realised ><" i kept forgetting to square the imaginary component ><"