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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
letx,y,zbeacomplexnumbersasx∣=∣y∣=∣z∣=1{x+y+z=1xyz=1calcul1x+1y+1z=?x=?y=?z=?pleaseineedahelp
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
(x+y+z)2=x2+y2+z2+2(xy+xz+yz)x2+y2+z2=12(xy+xz+yz)1x+1y+1z=x+yxy+1z=xz+yz+xyxyz1x+1y+1z=xy+xz+yz=1(x2+y2+z2)2needmoreinformation
Commented by pticantor last updated on 10/Aug/20
yes sir ∣x∣=∣y∣=∣z∣=1
yessirx∣=∣y∣=∣z∣=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
ifx∣=∣y∣=∣z∣=11x+1y+1z=132=1
Commented by pticantor last updated on 10/Aug/20
how sir ???  we have x^2 # ∣x∣
howsir???You can't use 'macro parameter character #' in math mode
Commented by bemath last updated on 10/Aug/20
if ∣x∣ = 1 it does mean → { ((x=−1 or)),((x=1)) :}  or x^2 =1
ifx=1itdoesmean{x=1orx=1orx2=1
Commented by Her_Majesty last updated on 10/Aug/20
∣x∣ ⇒ x=cosθ+isinθ=e^(iθ)
xx=cosθ+isinθ=eiθ
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
ei(α+β+γ)=1α+β+γ=0γ=(α+β)eiα+eiβ+eiγ=1cosα+cosβ+cos(α+β)=1sinα+sinβsin(α+β)=0ifα<β<γ:α=π/2β=0γ=π/2x=iy=1z=i1x+1y+1z=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
howx∣=1x=i.impossiblex2=12x2=1butifx=ix2=i2=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
yourewronga+bi∣=a2+b20±1i∣=02+12=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!
whomarkedthisasinappropriate?thequestionsaysxyzarecomplexnumbersz=reiθ=rcosθ+irsinθ=a+biwithrR+z∣=r=a2+b2ifyoudonotknowthis,learnitrightnow!
Commented by bemath last updated on 10/Aug/20
wow .... I don't know who did that action.

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