Complex Numbers (1 Viewer)

[askit]

Just a quick question with this, I divided by z^2 to get the conjugates, but how do yk that the modulus of the roots are both 1? The product of roots = 9/9 = 1 = |w1|^2|w2|^2 --> is this enough to warrent that the modulus' are 1?

 

[liamkk112]

View attachment 43711
Just a quick question with this, I divided by z^2 to get the conjugates, but how do yk that the modulus of the roots are both 1? The product of roots = 9/9 = 1 = |w1|^2|w2|^2 --> is this enough to warrent that the modulus' are 1?

dividing by 9;
z^4 - 2z^3 + 5/9 z^2 -2z + 1 = 0
again dividing by z^2 we get
(z^2 + 1/z^2) -2(z + 1/z) + 5/9 = 0
now we want this to be a quadratic that we can solve. so (z+ 1/z)^2 = z^2 + 1/z^2 + 2; so we will add and subtract two on the LHS;
(z+1/z)^2 -2(z+1/z) -13/9 = 0
now consider the discriminant, which is 88/9 > 0. thus, the solutions are both real; that is, z + 1/z is real for the solutions to the equations.
we can write z + 1/z = (|z|^2 z + z bar) /|z|^2. ill save the algebra but because z + 1/z is real, it must be the case that the imaginary part is zero;
that is y(x^2+y^2-1) = 0 writing z = x + iy. now either y = 0, in which case there is no imaginary part to z, which means it's real, but because there were no real solutions this can't be true. thus, x^2 +y^2 = 1; so, |z| = sqrt(x^2+y^2) = 1

the product of roots condition doesn't seem sufficient to me. eg, |w1|^2 = 1/2 , |w2|^2 = 2, then the modulus isn't 1.

 

[askit]

dividing by 9;
z^4 - 2z^3 + 5/9 z^2 -2z + 1 = 0
again dividing by z^2 we get
(z^2 + 1/z^2) -2(z + 1/z) + 5/9 = 0
now we want this to be a quadratic that we can solve. so (z+ 1/z)^2 = z^2 + 1/z^2 + 2; so we will add and subtract two on the LHS;
(z+1/z)^2 -2(z+1/z) -13/9 = 0
now consider the discriminant, which is 88/9 > 0. thus, the solutions are both real; that is, z + 1/z is real for the solutions to the equations.
we can write z + 1/z = (|z|^2 z + z bar) /|z|^2. ill save the algebra but because z + 1/z is real, it must be the case that the imaginary part is zero;
that is y(x^2+y^2-1) = 0 writing z = x + iy. now either y = 0, in which case there is no imaginary part to z, which means it's real, but because there were no real solutions this can't be true. thus, x^2 +y^2 = 1; so, |z| = sqrt(x^2+y^2) = 1

the product of roots condition doesn't seem sufficient to me. eg, |w1|^2 = 1/2 , |w2|^2 = 2, then the modulus isn't 1.

Hmm, the answers did a different sort of thing - From this can you see where they found out that the modulus = 1?

 

[liamkk112]

View attachment 43712
Hmm, the answers did a different sort of thing - From this can you see where they found out that the modulus = 1?

yeah they just assumed it to be 1. i suppose you can assume it's 1, and then check your answers at the end, but the proof i gave above would justify assuming it to be 1 in the first place

is this from 2023 stuff btw? i think i had this on my trials lol

 

[askit]

yeah they just assumed it to be 1. i suppose you can assume it's 1, and then check your answers at the end, but the proof i gave above would justify assuming it to be 1 in the first place

is this from 2023 stuff btw? i think i had this on my trials lol

2023 NEAP - I found it to be pretty hard (I got 73 lmao) - good paper doe

 

[liamkk112]

2023 NEAP - I found it to be pretty hard (I got 73 lmao) - good paper doe

yeah thats the paper i had for trials lol
i think i got similar to what u got, pretty sure i got this question wrong too because i wasn't sure if we could just assume the modulus was one either

 

[askit]

yeah thats the paper i had for trials lol
i think i got similar to what u got, pretty sure i got this question wrong too because i wasn't sure if we could just assume the modulus was one either

Thats reassuring coming from someone who got 95 in e2 haha