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


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Question Number 51167 by Tawa1 last updated on 24/Dec/18
$Prove that: (a) If ∣z_1 + z_2 ∣ = ∣z_1 − z_2 ∣, the difference of the arguements of z_1 and z_2 is (π/2) (b) If arg{((z_1 + z_2 )/(z_1 − z_2 ))} = (π/2) , then ∣z_1 ∣ = ∣z_2 ∣$
$Prove that :$ $(a) If ∣ z_{1} + z_{2} ∣ = ∣ z_{1} - z_{2} ∣, the difference of the arguements of z_{1}$ $and z_{2} is \frac{π}{2}$ $(b) If \arg {\frac{z_{1} + z_{2}}{z_{1} - z_{2}}} = \frac{π}{2}, then ∣ z_{1} ∣ = ∣ z_{2} ∣$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18
	$z_1 =a+ib z_2 =c+id a)∣(a+c)+i(b+d)∣=∣(a−c)+i(b−d)∣ (√((a+c)^2 +(b+d)^2 )) =(√((a−c)^2 +(b−d)^2 )) 2ac+2bd=−2ac−2bd ac+bd=0 ac=−bd (b/a)=−(c/d) tanθ_1 =−cotθ_2 tanθ_1 =tan((π/2)+θ_2 ) θ_1 −θ_2 =(π/2) b)((z_1 +z_2 )/(z_1 −z_2 )) (((a+c)+i(b+d))/((a−c)+i(b−d))) (((({(a+c)+i(b+d)}{(a−c)−i(b−d)})/((a−c)^2 +(b−d)^2 )) )/) =(({(a^2 −c^2 +b^2 −d^2 )+i(−ab+ad−bc+cd+ab−bc+ad−cd})/((a−c)^2 +(b−d)^2 )) tan(π/2)=((2ad−2bc)/((a−c)^2 +(b−d)^2 )) (a−c)^2 +(b−d)^2 =0 so a=c and b=d ∣z_1 ∣=(√(a^2 +b^2 )) =(√(c^2 +d^2 )) =∣z_2 ∣$
	$z_{1} = a + i b z_{2} = c + i d$ $a) ∣ (a + c) + i (b + d) ∣=∣ (a - c) + i (b - d) ∣$ $\sqrt{{(a + c)}^{2} + {(b + d)}^{2}} = \sqrt{{(a - c)}^{2} + {(b - d)}^{2}}$ $2 a c + 2 b d = - 2 a c - 2 b d$ $a c + b d = 0$ $a c = - b d$ $\frac{b}{a} = - \frac{c}{d}$ $t a n θ_{1} = - c o t θ_{2}$ $t a n θ_{1} = t a n (\frac{π}{2} + θ_{2})$ $θ_{1} - θ_{2} = \frac{π}{2}$ $b) \frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ $\frac{(a + c) + i (b + d)}{(a - c) + i (b - d)}$ $\frac{\frac{{(a + c) + i (b + d)} {(a - c) - i (b - d)}}{{(a - c)}^{2} + {(b - d)}^{2}}}{}$ $= \frac{{(a^{2} - c^{2} + b^{2} - d^{2}) + i (- a b + a d - b c + c d + a b - b c + a d - c d}}{{(a - c)}^{2} + {(b - d)}^{2}}$ $t a n \frac{π}{2} = \frac{2 a d - 2 b c}{{(a - c)}^{2} + {(b - d)}^{2}}$ ${(a - c)}^{2} + {(b - d)}^{2} = 0$ $s o a = c a n d b = d$ $∣ z_{1} ∣= \sqrt{a^{2} + b^{2}}$ $= \sqrt{c^{2} + d^{2}}$ $=∣ z_{2} ∣$
	Commented by Tawa1 last updated on 24/Dec/18

	$God bless you sir$
	Commented by Tawa1 last updated on 24/Dec/18

	$Sir, how is - \cot θ_{2} = \tan (\frac{π}{2} + θ_{2}) ? ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

	$t a n (\frac{π}{2} - θ) = c o t θ b e c a u s e \frac{π}{2} - θ = 1 s t q u a d r a n t$ $b u t t a n (\frac{π}{2} + θ) = - c o t θ b e c a u s e (\frac{π}{2} + θ) 2 n d q u a d r a n t$
	Commented by Tawa1 last updated on 24/Dec/18

	$God bless you sir$
	Answered by peter frank last updated on 24/Dec/18

	$z_{1} = x_{1} + {i y}_{1}$ $z_{2} = x_{2} + {i y}_{2}$ $z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2})$ $∣ z_{1} + z_{2} ∣=∣ (x_{1} + x_{2}) + i (y_{1} + y_{2}) ∣$ $\sqrt{{(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2}} . . . . (i)$ $∣ z_{1} - z_{2} ∣= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} . . . . . (i i)$ $∣ z_{1} - z_{2} ∣=∣ z_{1} + z_{2} ∣$ $∣ \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} = \sqrt{{(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2}}$ $2 x_{1} x_{2} - 2 y_{1} y_{2} = - 2 x_{1} x_{2} - 2 y_{1} y_{2}$ $x_{1} x_{2} = - y_{1} y_{2}$ $1 = - (\frac{y_{1}}{x_{1}}) . (\frac{y_{2}}{x_{2}})$ $1 = - \tan θ_{1} . \tan θ_{2}$ $- \tan^{- 1} \frac{π}{4} = \tan θ_{1}$ $\tan^{- 1} \frac{π}{4} = \tan θ_{2}$ $- {2 \tan}^{- 1} \frac{π}{4} = \tan (θ_{1} - θ_{2})$ $2 . \frac{π}{4} = θ_{1} - θ_{2}$ $\frac{π}{2} = θ_{1} - θ_{2}$ $r i g h t o r w r o n g ?$
	Commented by Tawa1 last updated on 24/Dec/18

	$God bless you sir$
	Answered by peter frank last updated on 24/Dec/18

	$[a r g (z_{1} + z_{2}) - a r g (z_{1} - z_{2}) = \frac{π}{2}$ $a r g [(x_{1} + x_{2}) + i (y_{1} + y_{2})] - [a r g (x_{1} - x_{2} + i (y_{1} - y_{2})]$ $\tan^{- 1} (\frac{y_{1} + y_{2}}{x_{1} + x_{2}}) - \tan^{- 1} (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) = \frac{π}{2}$ $(\frac{y_{1} + y_{2}}{x_{1} + x_{2}}) + (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) \div [1 - (\frac{y_{1} + y_{2}}{x_{1} + x_{2}}) . (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) = \frac{π}{2}$ $(\frac{y_{1} + y_{2}}{x_{1} + x_{2}}) + (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) \div [1 - (\frac{y_{1} + y_{2}}{x_{1} + x_{2}}) . (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) = \frac{1}{0}$ $x_{1}^{2} - x_{2}^{2} + y_{2}^{2} - y_{1}^{2} = 0$ $x_{1}^{2} + y_{1}^{2} - x_{2}^{2} - y_{2}^{2} = 0$ $x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2}$ $∣ z_{1} ∣=∣ z_{2} ∣$
	Commented by Tawa1 last updated on 24/Dec/18

	$God bless you sir$
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