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Prove-that-a-If-z-1-z-2-z-1-z-2-the-difference-of-the-arguements-of-z-1-and-z-2-is-pi-2-b-If-arg-z-1-z-2-z-1-z-2-pi-2-then-z-1-z-2-

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Question Number 51167 by Tawa1 last updated on 24/Dec/18
Prove that: (a) If ∣z_1 + z_2 ∣ = ∣z_1 − z_2 ∣, the difference of the arguements of z_1 and z_2 is (π/2) (b) If arg{((z_1 + z_2 )/(z_1 − z_2 ))} = (π/2) , then ∣z_1 ∣ = ∣z_2 ∣
Provethat:(a)If∣z1+z2∣=∣z1−z2∣,thedifferenceofthearguementsofz1andz2isπ2(b)Ifarg{z1+z2z1−z2}=π2,then∣z1∣=∣z2∣
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18
z_1 =a+ib z_2 =c+id a)∣(a+c)+i(b+d)∣=∣(a−c)+i(b−d)∣ (√((a+c)^2 +(b+d)^2 )) =(√((a−c)^2 +(b−d)^2 )) 2ac+2bd=−2ac−2bd ac+bd=0 ac=−bd (b/a)=−(c/d) tanθ_1 =−cotθ_2 tanθ_1 =tan((π/2)+θ_2 ) θ_1 −θ_2 =(π/2) b)((z_1 +z_2 )/(z_1 −z_2 )) (((a+c)+i(b+d))/((a−c)+i(b−d))) (((({(a+c)+i(b+d)}{(a−c)−i(b−d)})/((a−c)^2 +(b−d)^2 )) )/) =(({(a^2 −c^2 +b^2 −d^2 )+i(−ab+ad−bc+cd+ab−bc+ad−cd})/((a−c)^2 +(b−d)^2 )) tan(π/2)=((2ad−2bc)/((a−c)^2 +(b−d)^2 )) (a−c)^2 +(b−d)^2 =0 so a=c and b=d ∣z_1 ∣=(√(a^2 +b^2 )) =(√(c^2 +d^2 )) =∣z_2 ∣
z1=a+ibz2=c+ida)∣(a+c)+i(b+d)∣=∣(a−c)+i(b−d)∣(a+c)2+(b+d)2=(a−c)2+(b−d)22ac+2bd=−2ac−2bdac+bd=0ac=−bdba=−cdtanθ1=−cotθ2tanθ1=tan(π2+θ2)θ1−θ2=π2b)z1+z2z1−z2(a+c)+i(b+d)(a−c)+i(b−d){(a+c)+i(b+d)}{(a−c)−i(b−d)}(a−c)2+(b−d)2={(a2−c2+b2−d2)+i(−ab+ad−bc+cd+ab−bc+ad−cd}(a−c)2+(b−d)2tanπ2=2ad−2bc(a−c)2+(b−d)2(a−c)2+(b−d)2=0soa=candb=d∣z1∣=a2+b2=c2+d2=∣z2∣
Commented by Tawa1 last updated on 24/Dec/18
God bless you sir
Godblessyousir
Commented by Tawa1 last updated on 24/Dec/18
Sir, how is − cotθ_2 = tan((π/2) + θ_2 ) ??
Sir,howis−cotθ2=tan(π2+θ2)??
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18
tan((π/2)−θ)=cotθ because (π/2)−θ=1st quadrant but tan((π/2)+θ)=−cotθ because ((π/2)+θ) 2nd quadrant
tan(π2−θ)=cotθbecauseπ2−θ=1stquadrantbuttan(π2+θ)=−cotθbecause(π2+θ)2ndquadrant
Commented by Tawa1 last updated on 24/Dec/18
God bless you sir
Godblessyousir
Answered by peter frank last updated on 24/Dec/18
z_1 =x_1 +iy_1 z_2 =x_2 +iy_2 z_(1 ) +z_2 =(x_1 +x_(2 ) )+i(y_(1 ) +y_2 ) ∣z_(1 ) +z_(2 ) ∣=∣(x_1 +x_(2 ) )+i(y_(1 ) +y_2 )∣ (√((x_1 +x_2 )^2 +(y_1 +y_(2 ) )^2 )) ....(i) ∣z_(1 ) −z_(2 ) ∣=(√((x_(1 ) −x_(2 ) )^2 +(y_1 −y_2 )^2 )) .....(ii) ∣z_(1 ) −z_2 ∣=∣z_1 +z_2 ∣ ∣(√((x_(1 ) −x_(2 ) )^2 +(y_1 −y_2 )^2 )) =(√((x_(1 ) +x_2 )^2 +(y_1 +y_2 )^2 )) 2x_(1 ) x_2 −2y_1 y_2 =−2x_1 x_2 −2y_1 y_2 x_(1 ) x_(2 ) =−y_(1 ) y_2 1=−((y_1 /x_1 ) ).((y_2 /x_2 )) 1=−tan θ_(1 ) .tan θ_2 −tan^(−1) (π/4)=tan θ_1 tan^(−1) (π/4)=tan θ_(2 ) −2tan^(−1) (π/4)=tan (θ_1 −θ_(2 ) ) 2.(π/4)=θ_1 −θ_2 (π/2)=θ_1 −θ_2 right or wrong?
z1=x1+iy1z2=x2+iy2z1+z2=(x1+x2)+i(y1+y2)∣z1+z2∣=∣(x1+x2)+i(y1+y2)∣(x1+x2)2+(y1+y2)2….(i)∣z1−z2∣=(x1−x2)2+(y1−y2)2…..(ii)∣z1−z2∣=∣z1+z2∣∣(x1−x2)2+(y1−y2)2=(x1+x2)2+(y1+y2)22x1x2−2y1y2=−2x1x2−2y1y2x1x2=−y1y21=−(y1x1).(y2x2)1=−tanθ1.tanθ2−tan−1π4=tanθ1tan−1π4=tanθ2−2tan−1π4=tan(θ1−θ2)2.π4=θ1−θ2π2=θ1−θ2rightorwrong?
Commented by Tawa1 last updated on 24/Dec/18
God bless you sir
Godblessyousir
Answered by peter frank last updated on 24/Dec/18
[arg(z_1 +z_(2 ) )−arg(z_1 −z_2 )=(π/2) arg[(x_(1 ) +x_2 )+i(y_1 +y_2 )]−[arg(x_1 −x_2 +i(y_1 −y_2 )] tan^(−1) (((y_(1 ) +y_2 )/(x_1 +x_2 )))−tan^(−1) (((y_1 −y_2 )/(x_1 −x_2 )))=(π/2) (((y_(1 ) +y_2 )/(x_1 +x_2 )))+(((y_1 −y_2 )/(x_1 −x_2 )))÷[1−(((y_1 +y_2 )/(x_1 +x_2 ))).(((y_1 −y_2 )/(x_(1 ) −x_2 )))=(π/2) (((y_(1 ) +y_2 )/(x_1 +x_2 )))+(((y_1 −y_2 )/(x_1 −x_2 )))÷[1−(((y_1 +y_2 )/(x_1 +x_2 ))).(((y_1 −y_2 )/(x_(1 ) −x_2 )))=(1/0) x_1 ^2 −x_2 ^2 +y_2 ^2 −y_1 ^2 =0 x_1 ^2 +y_1 ^2 −x_2 ^2 −y_2 ^2 =0 x_1 ^2 +y_1 ^2 =x_2 ^2 +y_(2 ) ^2 ∣z_(1 ) ∣=∣z_2 ∣
[arg(z1+z2)−arg(z1−z2)=π2arg[(x1+x2)+i(y1+y2)]−[arg(x1−x2+i(y1−y2)]tan−1(y1+y2x1+x2)−tan−1(y1−y2x1−x2)=π2(y1+y2x1+x2)+(y1−y2x1−x2)÷[1−(y1+y2x1+x2).(y1−y2x1−x2)=π2(y1+y2x1+x2)+(y1−y2x1−x2)÷[1−(y1+y2x1+x2).(y1−y2x1−x2)=10x12−x22+y22−y12=0x12+y12−x22−y22=0x12+y12=x22+y22∣z1∣=∣z2∣
Commented by Tawa1 last updated on 24/Dec/18
God bless you sir
Godblessyousir

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