MedVision ad

complex numbers prob (1 Viewer)

king.rafa

Member
Joined
Oct 7, 2007
Messages
48
Gender
Female
HSC
N/A
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

hey can anyone help me prove the modulus 1 part

thanks
 

ssglain

Member
Joined
Sep 18, 2006
Messages
445
Location
lost in a Calabi-Yau
Gender
Female
HSC
2007
As a hint, think about the relationship between the two roots and an expression that involves them and their moduli.

The full solution is attached.
 

kony

Member
Joined
Feb 10, 2006
Messages
322
Gender
Undisclosed
HSC
2007
because p is real, the roots are in conjugate pairs.

now conjugate pairs have the same moduli.

you also know that for roots R1 and R2, R1R2 = 1.

|R1R2| = 1
now |R1|=|R2|

therefore, |R1| = 1

hence, the roots of P(x)=0 have the same moduli 1.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top