Complex no. question help (1 Viewer)

[cssftw]

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.

Ok so here's what I did:

Using quadraticformula:

x= [-p +/- sqrt(p^2 - 4)]/2

But p^2-4 (<) 0 due the -2 < p < 2
So, x= [-p +/- i*sqrt(4-p^2)]/2

x = -p/2p +/- [i*sqrt(4-p^2)]/2p
So I tried to find the modulus

|z| = SQRT[(p^2 / 4p^2) + (4-p^2)/4p^2]
= SQRT[4/4p^2]
= SQRT[1/p^2]
= 1/p

Why did I not get a modulus of 1? What am I supposed to do?

Thanks guys, appreciate the help.

 

[ohexploitable]

Since all coefficients are real, the roots are complex conjugates, i.e.



But,



So,



By comparing coefficients we can see that,

 

[deterministic]

Also note that since |p|< 2, the discrimminant < 0 and hence roots must be complex and hence conjugates

 

[cssftw]

Thanks for the reply brah, but where on earth did this come from, I've nvr seen it before :S

 

[hscishard]

-P +/- i(root p^2 - 4) / 2
Then just use root(x^2 + y^2)
You'll get one

 

[hscishard]

Thanks for the reply brah, but where on earth did this come from, I've nvr seen it before :S

Pretty sure he meant = x^2 + xRe(z) + |z|^2
Comes from conjugate roots theorem.

 

[cssftw]

-P +/- i(root p^2 - 4) / 2
Then just use root(x^2 + y^2)
You'll get one

nah i got 1/p... check my working out if you can be bothered, its on the original post

 

[cssftw]

Pretty sure he meant = x^2 + xRe(z) + |z|^2
Comes from conjugate roots theorem.

Which textbook is that quadratic equation from? I don't think I've seen it in Cambridge or Terry Lee...

 

[hscishard]

You magically introduced p in the denominator. Lol

You're meant to think of that up really...I don't think any textbook has that labelled as a formula to remember

 

[ohexploitable]

Thanks for the reply brah, but where on earth did this come from, I've nvr seen it before :S

 

[boredoffail6]

think you mean a^2 + b^2 in the working above noob

 

[boredoffail6]

ie x^2 -2ax +(a^2 +b^2) =0

 

[hscishard]

Oh yea. z + z_ is 2Re(z)
I hate recalling stuff, but I hate having to calculate it

 

[ohexploitable]

think you mean a^2 + b^2 in the working above noob

Fixed.

 

[cssftw]

You magically introduced p in the denominator. Lol

:| fml

It was the bloody p... All good guys, I ended up getting 1.

 

[cssftw]

Since all coefficients are real, the roots are complex conjugates, i.e.



But,



So,



By comparing coefficients we can see that,

By the way, that is a ridiculously ninja solution, braah, you really are a BEAST at maths