if-1-is-a-root-of-unity-aand-z-is-a-complex-number-such-that-z-1-then-2-3-4z-2-4-3-2-z-2z-

Question Number 200087 by universe last updated on 13/Nov/23

if ω \neq 1 is a root of unity aand z is a

complex number such that ∣ z ∣ = 1 then

∣ \frac{2 + 3 ω + 4 z ω^{2}}{4 ω + 3 ω^{2} z + 2 z} ∣= ?

Commented by Frix last updated on 13/Nov/23

Btw . {it}^{'} s also 1 with w = 1

Commented by Frix last updated on 13/Nov/23

\dots now I think we can easily calculate this

with w = 1 . {We}^{'} re rotating the coordinate

system by \pm 120 ° . We {don}^{'} t have to rotate

z = e^{i θ} since {we}^{'} re only interested in ∣ z ∣= 1 .

This leads to

∣ \frac{5 + 4 z}{4 + 5 z} ∣= \frac{∣ 5 + 4 \cos θ + 4 i \sin θ ∣}{∣ 4 + 5 \cos θ + 5 i \sin θ ∣} =

= \frac{\sqrt{41 + 40 \cos θ}}{\sqrt{41 + 40 \cos θ}} = 1

Answered by Frix last updated on 13/Nov/23

p = r e^{i α} \land q = s e^{i β}

∣ \frac{p}{q} ∣=∣ \frac{r}{s} e^{i (α - β)} ∣= \frac{r}{s} = \frac{∣ r e^{i α} ∣}{∣ s e^{i β} ∣} = \frac{∣ p ∣}{∣ q ∣}

w = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = - \frac{1}{2} + \frac{\sqrt{3} σ}{2} i; σ = \pm 1 \Rightarrow σ^{2} = 1

z = \cos θ + i \sin θ = c + s i \Rightarrow c^{2} + s^{2} = 1

2 + 3 w + 4 {z w}^{2} =

= - 2 c + 2 \sqrt{3} σ s + \frac{1}{2} - (2 \sqrt{3} σ c + 2 s - \frac{3 \sqrt{3} σ}{3}) i

∣ 2 + 3 w + 4 {z w}^{2} ∣= \sqrt{23 - 4 (5 c + \sqrt{3} σ s)}

4 w + 3 w^{2} z + 2 z =

= \frac{c}{2} + \frac{3 \sqrt{3} σ s}{2} - 2 - (\frac{3 \sqrt{3} σ c}{2} - \frac{s}{2} - 2 \sqrt{3} σ) i

∣ 4 w + 3 w^{2} z + 2 z ∣= \sqrt{23 - 4 (5 c + \sqrt{3} σ s)}

\Rightarrow

Answer is 1

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