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Show-that-if-z-1-z-2-z-3-z-4-0-and-z-1-z-2-0-then-the-complex-numbers-z-1-z-2-z-3-z-4-are-concyclic-

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Question Number 20549 by Tinkutara last updated on 28/Aug/17
Show that if z_1 z_2 + z_3 z_4 = 0 and z_1 + z_2 = 0, then the complex numbers z_1 , z_2 , z_3 , z_4 are concyclic.
$${Show}\:{that}\:{if}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} {z}_{\mathrm{4}} \:=\:\mathrm{0}\:{and}\:{z}_{\mathrm{1}} \:+ \\ $$$${z}_{\mathrm{2}} \:=\:\mathrm{0},\:{then}\:{the}\:{complex}\:{numbers}\:{z}_{\mathrm{1}} , \\ $$$${z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{z}_{\mathrm{4}} \:{are}\:{concyclic}. \\ $$
Commented by ajfour last updated on 28/Aug/17
Answered by ajfour last updated on 28/Aug/17
Origin is the midpoint of the join of z_1 and z_2 . (as z_1 +z_2 =0 ) z_3 z_4 +z_1 z_2 =0 ⇒ z_3 z_4 =z_1 ^2 ⇒ ((z_4 /z_1 ))((z_3 /z_1 ))=1 ⇒ arg((z_4 /z_1 ))=−arg((z_3 /z_1 )) also ∣z_3 ∣∣z_4 ∣=∣z_1 ∣^2 or ∣z_5 ∣∣z_4 ∣=∣z_1 ∣^2 ⇒ xy=R^2 power of point (here origin) is the same for chord joining z_1 ,z_2 and chord joining z_4 , z_5 . and as ∣z_3 ∣=∣z_5 ∣ ; z_1 , z_2 , z_3 , z_4 are concyclic.
$${Origin}\:{is}\:{the}\:{midpoint}\:{of}\:{the}\:{join} \\ $$$${of}\:{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} .\:\:\:\:\left({as}\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} =\mathrm{0}\:\right) \\ $$$${z}_{\mathrm{3}} {z}_{\mathrm{4}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} =\mathrm{0}\:\:\:\Rightarrow\:\:\:\:{z}_{\mathrm{3}} {z}_{\mathrm{4}} ={z}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left(\frac{{z}_{\mathrm{4}} }{{z}_{\mathrm{1}} }\right)\left(\frac{{z}_{\mathrm{3}} }{{z}_{\mathrm{1}} }\right)=\mathrm{1} \\ $$$$\Rightarrow\:{arg}\left(\frac{{z}_{\mathrm{4}} }{{z}_{\mathrm{1}} }\right)=−{arg}\left(\frac{{z}_{\mathrm{3}} }{{z}_{\mathrm{1}} }\right) \\ $$$${also}\:\:\:\mid{z}_{\mathrm{3}} \mid\mid{z}_{\mathrm{4}} \mid=\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \: \\ $$$${or}\:\:\:\:\mid{z}_{\mathrm{5}} \mid\mid{z}_{\mathrm{4}} \mid=\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{xy}={R}^{\mathrm{2}} \\ $$$${power}\:{of}\:{point}\:\left({here}\:{origin}\right)\:{is} \\ $$$${the}\:{same}\:{for}\:{chord}\:{joining}\:{z}_{\mathrm{1}} ,{z}_{\mathrm{2}} \\ $$$${and}\:{chord}\:{joining}\:{z}_{\mathrm{4}} ,\:{z}_{\mathrm{5}} . \\ $$$${and}\:{as}\:\mid{z}_{\mathrm{3}} \mid=\mid{z}_{\mathrm{5}} \mid\:;\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{z}_{\mathrm{4}} \:{are} \\ $$$${concyclic}. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by Tinkutara last updated on 28/Aug/17
Why ∣z_3 ∣ = ∣z_5 ∣? How you have defined z_5 ?
$$\mathrm{Why}\:\mid{z}_{\mathrm{3}} \mid\:=\:\mid{z}_{\mathrm{5}} \mid?\:\mathrm{How}\:\mathrm{you}\:\mathrm{have}\:\mathrm{defined} \\ $$$${z}_{\mathrm{5}} ? \\ $$
Commented by ajfour last updated on 28/Aug/17
it is reflection of z_3 about the perpendicular bisector of the join of z_1 and z_2 . Hence ∣z_5 ∣=∣z_3 ∣ .
$${it}\:{is}\:{reflection}\:{of}\:{z}_{\mathrm{3}} \:{about}\:{the} \\ $$$${perpendicular}\:{bisector}\:{of}\:{the}\:{join} \\ $$$${of}\:{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} .\:{Hence}\:\mid{z}_{\mathrm{5}} \mid=\mid{z}_{\mathrm{3}} \mid\:. \\ $$

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