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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-

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Question Number 200087 by universe last updated on 13/Nov/23
if ω ≠ 1 is a root of unity aand z is a complex number such that ∣z∣ = 1 then ∣((2+3ω+4zω^2 )/(4ω+3ω^2 z+2z))∣= ?
$$\:\:\mathrm{if}\:\omega\:\neq\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{aand}\:\mathrm{z}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mid{z}\mid\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\:\:\mid\frac{\mathrm{2}+\mathrm{3}\omega+\mathrm{4}{z}\omega^{\mathrm{2}} }{\mathrm{4}\omega+\mathrm{3}\omega^{\mathrm{2}} {z}+\mathrm{2}{z}}\mid=\:? \\ $$
Commented by Frix last updated on 13/Nov/23
Btw. it′s also 1 with w=1
$$\mathrm{Btw}.\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{1}\:\mathrm{with}\:{w}=\mathrm{1} \\ $$
Commented by Frix last updated on 13/Nov/23
...now I think we can easily calculate this with w=1. We′re rotating the coordinate system by ±120°. We don′t have to rotate z=e^(iθ) since we′re only interested in ∣z∣=1. This leads to ∣((5+4z)/(4+5z))∣=((∣5+4cos θ +4i sin θ∣)/(∣4+5cos θ +5i sin θ∣))= =((√(41+40cos θ))/( (√(41+40cos θ))))=1
$$…\mathrm{now}\:\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{calculate}\:\mathrm{this} \\ $$$$\mathrm{with}\:{w}=\mathrm{1}.\:\mathrm{We}'\mathrm{re}\:\mathrm{rotating}\:\mathrm{the}\:\mathrm{coordinate} \\ $$$$\mathrm{system}\:\mathrm{by}\:\pm\mathrm{120}°.\:\mathrm{We}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to}\:\mathrm{rotate} \\ $$$${z}=\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{since}\:\mathrm{we}'\mathrm{re}\:\mathrm{only}\:\mathrm{interested}\:\mathrm{in}\:\mid{z}\mid=\mathrm{1}. \\ $$$$\mathrm{This}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mid\frac{\mathrm{5}+\mathrm{4}{z}}{\mathrm{4}+\mathrm{5}{z}}\mid=\frac{\mid\mathrm{5}+\mathrm{4cos}\:\theta\:+\mathrm{4i}\:\mathrm{sin}\:\theta\mid}{\mid\mathrm{4}+\mathrm{5cos}\:\theta\:+\mathrm{5i}\:\mathrm{sin}\:\theta\mid}= \\ $$$$=\frac{\sqrt{\mathrm{41}+\mathrm{40cos}\:\theta}}{\:\sqrt{\mathrm{41}+\mathrm{40cos}\:\theta}}=\mathrm{1} \\ $$
Answered by Frix last updated on 13/Nov/23
p=re^(iα) ∧q=se^(iβ) ∣(p/q)∣=∣(r/s)e^(i(α−β)) ∣=(r/s)=((∣re^(iα) ∣)/(∣se^(iβ) ∣))=((∣p∣)/(∣q∣)) w=−(1/2)±((√3)/2)i=−(1/2)+(((√3)σ)/2)i; σ=±1⇒σ^2 =1 z=cos θ +i sin θ =c+si ⇒ c^2 +s^2 =1 2+3w+4zw^2 = =−2c+2(√3)σs+(1/2)−(2(√3)σc+2s−((3(√3)σ)/3))i ∣2+3w+4zw^2 ∣=(√(23−4(5c+(√3)σs))) 4w+3w^2 z+2z= =(c/2)+((3(√3)σs)/2)−2−(((3(√3)σc)/2)−(s/2)−2(√3)σ)i ∣4w+3w^2 z+2z∣=(√(23−4(5c+(√3)σs))) ⇒ Answer is 1
$${p}={r}\mathrm{e}^{\mathrm{i}\alpha} \wedge{q}={s}\mathrm{e}^{\mathrm{i}\beta} \\ $$$$\mid\frac{{p}}{{q}}\mid=\mid\frac{{r}}{{s}}\mathrm{e}^{\mathrm{i}\left(\alpha−\beta\right)} \mid=\frac{{r}}{{s}}=\frac{\mid{r}\mathrm{e}^{\mathrm{i}\alpha} \mid}{\mid{s}\mathrm{e}^{\mathrm{i}\beta} \mid}=\frac{\mid{p}\mid}{\mid{q}\mid} \\ $$$$ \\ $$$${w}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}\sigma}{\mathrm{2}}\mathrm{i};\:\sigma=\pm\mathrm{1}\Rightarrow\sigma^{\mathrm{2}} =\mathrm{1} \\ $$$${z}=\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\:={c}+{s}\mathrm{i}\:\Rightarrow\:{c}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$$\mathrm{2}+\mathrm{3}{w}+\mathrm{4}{zw}^{\mathrm{2}} = \\ $$$$=−\mathrm{2c}+\mathrm{2}\sqrt{\mathrm{3}}\sigma{s}+\frac{\mathrm{1}}{\mathrm{2}}−\left(\mathrm{2}\sqrt{\mathrm{3}}\sigma{c}+\mathrm{2}{s}−\frac{\mathrm{3}\sqrt{\mathrm{3}}\sigma}{\mathrm{3}}\right)\mathrm{i} \\ $$$$\mid\mathrm{2}+\mathrm{3}{w}+\mathrm{4}{zw}^{\mathrm{2}} \mid=\sqrt{\mathrm{23}−\mathrm{4}\left(\mathrm{5}{c}+\sqrt{\mathrm{3}}\sigma{s}\right)} \\ $$$$ \\ $$$$\mathrm{4}{w}+\mathrm{3}{w}^{\mathrm{2}} {z}+\mathrm{2}{z}= \\ $$$$=\frac{{c}}{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}\sigma{s}}{\mathrm{2}}−\mathrm{2}−\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\sigma{c}}{\mathrm{2}}−\frac{{s}}{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\sigma\right)\mathrm{i} \\ $$$$\mid\mathrm{4}{w}+\mathrm{3}{w}^{\mathrm{2}} {z}+\mathrm{2}{z}\mid=\sqrt{\mathrm{23}−\mathrm{4}\left(\mathrm{5}{c}+\sqrt{\mathrm{3}}\sigma{s}\right)} \\ $$$$\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{1} \\ $$

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