Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{Let}\:{z}={z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \:\Rightarrow\bar {{z}}=\bar {{z}}_{\mathrm{1}} +\bar {{z}}_{\mathrm{2}} +\bar {{z}}_{\mathrm{3}} =\frac{\mathrm{1}}{{z}_{\mathrm{1}} }+\frac{\mathrm{1}}{{z}_{\mathrm{2}} }+\frac{\mathrm{1}}{{z}_{\mathrm{3}} }=\frac{{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} } \\ $$$$\Rightarrow\bar {{z}}=\frac{{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{{b}}\Rightarrow{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} ={b}\bar {{z}},\:{b}={z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}\:} \wedge\mid{b}\mid=\mathrm{1} \\ $$$$\frac{{z}_{\mathrm{1}} ^{\mathrm{2}} }{{z}_{\mathrm{2}} {z}_{\mathrm{3}} }+\frac{{z}_{\mathrm{2}} ^{\mathrm{2}} }{{z}_{\mathrm{3}} {z}_{\mathrm{1}} }+\frac{{z}_{\mathrm{3}} ^{\mathrm{2}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} }=−\mathrm{1}\Rightarrow\frac{{z}_{\mathrm{1}} ^{\mathrm{3}} +{z}_{\mathrm{2}} ^{\mathrm{3}} +{z}_{\mathrm{3}} ^{\mathrm{3}} }{{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} }=−\mathrm{1} \\ $$$$\Rightarrow{z}_{\mathrm{1}} ^{\mathrm{3}} +{z}_{\mathrm{2}} ^{\mathrm{3}} +{z}_{\mathrm{3}} ^{\mathrm{3}} −\mathrm{3}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} =−\mathrm{4}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{3}} ^{\mathrm{2}} −{z}_{\mathrm{1}} {z}_{\mathrm{2}} −{z}_{\mathrm{2}} {z}_{\mathrm{3}} −{z}_{\mathrm{3}} {z}_{\mathrm{1}} \right)=−\mathrm{4}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)\left(\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{3}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} +{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} \right)\right)=−\mathrm{4}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow{z}^{\mathrm{3}} −\mathrm{3}{b}\bar {{z}z}=−\mathrm{4}\:\Rightarrow{z}^{\mathrm{3}} =\mathrm{3}{b}\mid{z}\mid^{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow\mid{z}^{\mathrm{3}} \mid=\mid\mathrm{3}{b}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}\mid=\mathrm{3}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}\:\mathrm{if}\:\mathrm{3}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}>\mathrm{0},\:\left(\mid{b}\mid=\mathrm{1}\right) \\ $$$$\Rightarrow\mid{z}\mid^{\mathrm{3}} −\mathrm{3}\mid{z}\mid^{\mathrm{2}} +\mathrm{4}=\mathrm{0},\:\:\mid{z}\mid=\mathrm{2}\:\mathrm{D} \\ $$$$\mathrm{If}\:\mathrm{3}\mid{z}\mid^{\mathrm{2}} −\mathrm{4}<\mathrm{0}\:\mathrm{then}\:\mid{z}\mid=\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com