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Question Number 107320 by pticantor last updated on 10/Aug/20
let x,y,z be a complex numbers as ∣x∣=∣y∣=∣z∣=1 { ((x+y+z=1)),((xyz=1)) :} calcul (1/x)+(1/y)+(1/z)=? x=? y=? z=? please i need a help
$${let}\:{x},{y},{z}\:{be}\:{a}\:{complex}\:{numbers} \\ $$$${as}\:\mid{x}\mid=\mid{y}\mid=\mid{z}\mid=\mathrm{1} \\ $$$$\begin{cases}{{x}+{y}+{z}=\mathrm{1}}\\{{xyz}=\mathrm{1}}\end{cases} \\ $$$${calcul}\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=? \\ $$$$\:\:{x}=?\:{y}=?\:{z}=? \\ $$$${please}\:{i}\:{need}\:{a}\:{help} \\ $$
Commented by bemath last updated on 10/Aug/20
(x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+xz+yz) x^2 +y^2 +z^2 =1−2(xy+xz+yz) (1/x)+(1/y)+(1/z)=((x+y)/(xy))+(1/z)=((xz+yz+xy)/(xyz)) (1/x)+(1/y)+(1/z)= xy+xz+yz = ((1−(x^2 +y^2 +z^2 ))/2) need more information
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{xz}+{yz}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}\left({xy}+{xz}+{yz}\right) \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{{x}+{y}}{{xy}}+\frac{\mathrm{1}}{{z}}=\frac{{xz}+{yz}+{xy}}{{xyz}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\:{xy}+{xz}+{yz} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$${need}\:{more}\:{information} \\ $$
Commented by pticantor last updated on 10/Aug/20
yes sir ∣x∣=∣y∣=∣z∣=1
$${yes}\:{sir}\:\mid{x}\mid=\mid{y}\mid=\mid{z}\mid=\mathrm{1} \\ $$
Commented by bemath last updated on 10/Aug/20
if ∣x∣=∣y∣=∣z∣=1 (1/x)+(1/y)+(1/z)=((1−3)/2)=−1
$${if}\:\mid{x}\mid=\mid{y}\mid=\mid{z}\mid=\mathrm{1}\: \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}}=−\mathrm{1} \\ $$
Commented by pticantor last updated on 10/Aug/20
how sir ??? we have x^2 # ∣x∣
$${how}\:{sir}\:??? \\ $$$${we}\:{have}\:{x}^{\mathrm{2}} #\:\mid{x}\mid \\ $$
Commented by bemath last updated on 10/Aug/20
if ∣x∣ = 1 it does mean → { ((x=−1 or)),((x=1)) :} or x^2 =1
$${if}\:\mid{x}\mid\:=\:\mathrm{1}\:{it}\:{does}\:{mean}\:\rightarrow\begin{cases}{{x}=−\mathrm{1}\:{or}}\\{{x}=\mathrm{1}}\end{cases} \\ $$$${or}\:{x}^{\mathrm{2}} =\mathrm{1} \\ $$
Commented by Her_Majesty last updated on 10/Aug/20
∣x∣ ⇒ x=cosθ+isinθ=e^(iθ)
$$\mid{x}\mid\:\Rightarrow\:{x}={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$
Answered by Her_Majesty last updated on 10/Aug/20
e^(i(α+β+γ)) =1 ⇒ α+β+γ=0 ⇔ γ=−(α+β) e^(iα) +e^(iβ) +e^(iγ) =1 ⇒ cosα+cosβ+cos(α+β)=1 sinα+sinβ−sin(α+β)=0 ⇒ if α<β<γ: α=−π/2 β=0 γ=π/2 ⇒ x=−i y=1 z=i (1/x)+(1/y)+(1/z)=1
$${e}^{{i}\left(\alpha+\beta+\gamma\right)} =\mathrm{1}\:\Rightarrow\:\alpha+\beta+\gamma=\mathrm{0}\:\Leftrightarrow\:\gamma=−\left(\alpha+\beta\right) \\ $$$${e}^{{i}\alpha} +{e}^{{i}\beta} +{e}^{{i}\gamma} =\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}\alpha+{cos}\beta+{cos}\left(\alpha+\beta\right)=\mathrm{1} \\ $$$${sin}\alpha+{sin}\beta−{sin}\left(\alpha+\beta\right)=\mathrm{0} \\ $$$$\Rightarrow\:{if}\:\alpha<\beta<\gamma:\:\alpha=−\pi/\mathrm{2}\:\beta=\mathrm{0}\:\gamma=\pi/\mathrm{2} \\ $$$$\Rightarrow\:{x}=−{i}\:{y}=\mathrm{1}\:{z}={i} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{1} \\ $$
Commented by bemath last updated on 10/Aug/20
how ∣x∣=1 ⇒ x=−i . impossible ∣x∣^2 =1^2 ⇒ x^2 =1 but if x=−i ⇒x^2 =i^2 =−1
$${how}\:\mid{x}\mid=\mathrm{1}\:\Rightarrow\:{x}=−{i}\:.\:{impossible} \\ $$$$\mid{x}\mid^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \Rightarrow\:{x}^{\mathrm{2}} =\mathrm{1} \\ $$$${but}\:{if}\:{x}=−{i}\:\Rightarrow{x}^{\mathrm{2}} ={i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$ \\ $$
Commented by Her_Majesty last updated on 10/Aug/20
you′re wrong ∣a+bi∣=(√(a^2 +b^2 )) ∣0±1i∣=(√(0^2 +1^2 ))=1
$${you}'{re}\:{wrong} \\ $$$$\mid{a}+{bi}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mid\mathrm{0}\pm\mathrm{1}{i}\mid=\sqrt{\mathrm{0}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\mathrm{1} \\ $$
Commented by Her_Majesty last updated on 10/Aug/20
who marked this as inappropriate? the question says “x y z are complex numbers” z=re^(iθ) =rcosθ+irsinθ=a+bi with r∈R^+ ∣z∣=r=(√(a^2 +b^2 )) if you do not know this, learn it right now!
$${who}\:{marked}\:{this}\:{as}\:{inappropriate}? \\ $$$${the}\:{question}\:{says}\:“{x}\:{y}\:{z}\:{are}\:{complex}\:{numbers}'' \\ $$$${z}={re}^{{i}\theta} ={rcos}\theta+{irsin}\theta={a}+{bi}\:{with}\:{r}\in\mathbb{R}^{+} \\ $$$$\mid{z}\mid={r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${if}\:{you}\:{do}\:{not}\:{know}\:{this},\:{learn}\:{it}\:{right}\:{now}! \\ $$
Commented by bemath last updated on 10/Aug/20
wow .... I don't know who did that action.

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