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If-z-1-z-2-and-z-3-are-distinct-complex-numbers-such-that-z-1-z-2-z-3-1-and-z-1-2-z-2-z-3-z-2-2-z-3-z-1-z-3-2-z-1-z-2-1-then-the-value-o

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Question Number 139700 by EnterUsername last updated on 30/Apr/21
If z_1 , z_2 and z_3 are distinct complex numbers such that ∣z_1 ∣=∣z_2 ∣=∣z_3 ∣=1 and (z_1 ^2 /(z_2 z_3 ))+(z_2 ^2 /(z_3 z_1 ))+(z_3 ^2 /(z_1 z_2 ))=−1 then the value of ∣z_1 +z_2 +z_3 ∣ can be (A) 1/2 (B) 3 (C) 3/2 (D) 2
$$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{distinct}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{3}} \mid=\mathrm{1}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{z}_{\mathrm{1}} ^{\mathrm{2}} }{{z}_{\mathrm{2}} {z}_{\mathrm{3}} }+\frac{{z}_{\mathrm{2}} ^{\mathrm{2}} }{{z}_{\mathrm{3}} {z}_{\mathrm{1}} }+\frac{{z}_{\mathrm{3}} ^{\mathrm{2}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} }=−\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \mid\:\mathrm{can}\:\mathrm{be} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{3}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{2} \\ $$
Commented by hawa last updated on 30/Apr/21
B
$${B} \\ $$
Commented by EnterUsername last updated on 30/Apr/21
Welcome to the forum Hawa ! Right answer according to author is D. Would you mind verifying your working ? Thanks for your attention though!
$$\:\:\:\mathrm{Welcome}\:\mathrm{to}\:\mathrm{the}\:\mathrm{forum}\:\mathrm{Hawa}\:! \\ $$$$\mathrm{Right}\:\mathrm{answer}\:\mathrm{according}\:\mathrm{to}\:\mathrm{author}\:\mathrm{is}\:\mathrm{D}. \\ $$$$\mathrm{Would}\:\mathrm{you}\:\mathrm{mind}\:\mathrm{verifying}\:\mathrm{your}\:\mathrm{working}\:? \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{attention}\:\mathrm{though}! \\ $$
Answered by Ar Brandon last updated on 02/May/21
Let z=z_1 +z_2 +z_3 ⇒z^� =z_1 ^� +z_2 ^� +z_3 ^� =(1/z_1 )+(1/z_2 )+(1/z_3 )=((z_2 z_3 +z_3 z_1 +z_1 z_2 )/(z_1 z_2 z_3 )) ⇒z^� =((z_2 z_3 +z_3 z_1 +z_1 z_2 )/b)⇒z_2 z_3 +z_3 z_1 +z_1 z_2 =bz^� , b=z_1 z_2 z_(3 ) ∧∣b∣=1 (z_1 ^2 /(z_2 z_3 ))+(z_2 ^2 /(z_3 z_1 ))+(z_3 ^2 /(z_1 z_2 ))=−1⇒((z_1 ^3 +z_2 ^3 +z_3 ^3 )/(z_1 z_2 z_3 ))=−1 ⇒z_1 ^3 +z_2 ^3 +z_3 ^3 −3z_1 z_2 z_3 =−4z_1 z_2 z_3 ⇒(z_1 +z_2 +z_3 )(z_1 ^2 +z_2 ^2 +z_3 ^2 −z_1 z_2 −z_2 z_3 −z_3 z_1 )=−4z_1 z_2 z_3 ⇒(z_1 +z_2 +z_3 )((z_1 +z_2 +z_3 )^2 −3(z_1 z_2 +z_2 z_3 +z_3 z_1 ))=−4z_1 z_2 z_3 ⇒z^3 −3bz^� z=−4 ⇒z^3 =3b∣z∣^2 −4 ⇒∣z^3 ∣=∣3b∣z∣^2 −4∣=3∣z∣^2 −4 if 3∣z∣^2 −4>0, (∣b∣=1) ⇒∣z∣^3 −3∣z∣^2 +4=0, ∣z∣=2 D If 3∣z∣^2 −4<0 then ∣z∣=1
$$\mathrm{Let}\:{z}={z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \:\Rightarrow\bar {{z}}=\bar {{z}}_{\mathrm{1}} +\bar {{z}}_{\mathrm{2}} +\bar {{z}}_{\mathrm{3}} =\frac{\mathrm{1}}{{z}_{\mathrm{1}} }+\frac{\mathrm{1}}{{z}_{\mathrm{2}} }+\frac{\mathrm{1}}{{z}_{\mathrm{3}} }=\frac{{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} } \\ $$$$\Rightarrow\bar {{z}}=\frac{{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{{b}}\Rightarrow{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} ={b}\bar {{z}},\:{b}={z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}\:} \wedge\mid{b}\mid=\mathrm{1} \\ $$$$\frac{{z}_{\mathrm{1}} ^{\mathrm{2}} }{{z}_{\mathrm{2}} {z}_{\mathrm{3}} }+\frac{{z}_{\mathrm{2}} ^{\mathrm{2}} }{{z}_{\mathrm{3}} {z}_{\mathrm{1}} }+\frac{{z}_{\mathrm{3}} ^{\mathrm{2}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} }=−\mathrm{1}\Rightarrow\frac{{z}_{\mathrm{1}} ^{\mathrm{3}} +{z}_{\mathrm{2}} ^{\mathrm{3}} +{z}_{\mathrm{3}} ^{\mathrm{3}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} }=−\mathrm{1} \\ $$$$\Rightarrow{z}_{\mathrm{1}} ^{\mathrm{3}} +{z}_{\mathrm{2}} ^{\mathrm{3}} +{z}_{\mathrm{3}} ^{\mathrm{3}} −\mathrm{3}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} =−\mathrm{4}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{3}} ^{\mathrm{2}} −{z}_{\mathrm{1}} {z}_{\mathrm{2}} −{z}_{\mathrm{2}} {z}_{\mathrm{3}} −{z}_{\mathrm{3}} {z}_{\mathrm{1}} \right)=−\mathrm{4}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)\left(\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{3}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} +{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} \right)\right)=−\mathrm{4}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow{z}^{\mathrm{3}} −\mathrm{3}{b}\bar {{z}z}=−\mathrm{4}\:\Rightarrow{z}^{\mathrm{3}} =\mathrm{3}{b}\mid{z}\mid^{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow\mid{z}^{\mathrm{3}} \mid=\mid\mathrm{3}{b}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}\mid=\mathrm{3}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}\:\mathrm{if}\:\mathrm{3}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}>\mathrm{0},\:\left(\mid{b}\mid=\mathrm{1}\right) \\ $$$$\Rightarrow\mid{z}\mid^{\mathrm{3}} −\mathrm{3}\mid{z}\mid^{\mathrm{2}} +\mathrm{4}=\mathrm{0},\:\:\mid{z}\mid=\mathrm{2}\:\mathrm{D} \\ $$$$\mathrm{If}\:\mathrm{3}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}<\mathrm{0}\:\mathrm{then}\:\mid{z}\mid=\mathrm{1} \\ $$

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