A Problem in Complex Numbers!

Algebra Level 4

Let z 1 , z 2 , z 3 z_1,z_2,z_3 be 3 complex numbers lying on a circle whose centre is at the origin such that z i + z j z k z_i+z_{j}z_k (where i , j , k ∈ { 1 , 2 , 3 } i,j,k\in \{1,2,3\} and i ≠ j ≠ k i \ne j \ne k ) are real numbers, then find z 1 × z 2 × z 3 z_1\times z_2\times z_3 .

Note

  • z 1 ≠ z 2 ≠ z 3 z_1\ne z_2\ne z_3 . In other words, z 1 , z 2 , z 3 z_1,z_2,z_3 are all distinct points on the circle.

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The answer is 1.

Miraj Shah by Miraj Shah , 23, India

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2 solutions

Rishi Sharma
May 17, 2016

Relevant wiki: Complex Numbers - Absolute Values

F i r s t w e n o t i c e t h a t a s z 1 , z 2 , z 3 l i e o n a c i r c l e c e n t e r d a t o r i g i n ∣ z 1 ∣ = ∣ z 2 ∣ = ∣ z 3 ∣ = r ( h e r e r i s t h e r a d i u s o f c i r c l e ) L e t z 1 = r e i θ 1 z 2 = r e i θ 2 z 3 = r e i θ 3 I t i s g i v e n t h a t z 1 + z 2 z 3 ∈ ℜ h e n c e z 1 = z 2 z 3 ‾ r e i θ 1 = r e − i θ 2 . r e − i θ 3 ⟹ r = r e i ( θ 3 + θ 2 + θ 1 ) F r o m h e r e w e c a n i n f e r t h a t r = 1 N o w z 1 × z 2 × z 3 = z 2 z 3 ‾ × z 3 z 2 = ( ∣ z 2 ∣ × ∣ z 3 ∣ ) 2 = r 4 = 1 4 = 1 First\quad we\quad notice\quad that\quad as\quad { z }_{ 1 },{ z }_{ 2 },{ z }_{ 3 }\quad lie\quad on\quad a\quad circle\quad centerd\quad at\quad origin\\ \left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =r\quad (here\quad r\quad is\quad the\quad radius\quad of\quad circle)\\ Let\quad { z }_{ 1 }=r{ e }^{ i{ \theta }_{ 1 } }\\ \quad \quad \quad { z }_{ 2 }=r{ e }^{ i{ \theta }_{ 2 } }\\ \quad \quad \quad { z }_{ 3 }=r{ e }^{ i{ \theta }_{ 3 } }\\ It\quad is\quad given\quad that\quad { z }_{ 1 }+{ z }_{ 2 }{ z }_{ 3 }\in \Re \\ hence\quad { z }_{ 1 }=\overline { { z }_{ 2 }{ z }_{ 3 } } \\ r{ e }^{ i{ \theta }_{ 1 } }=r{ e }^{ -i{ \theta }_{ 2 } }.r{ e }^{ -i{ \theta }_{ 3 } }\\ \Longrightarrow \quad r=r{ e }^{ i({ \theta }_{ 3 }+{ \theta }_{ 2 }{ +\theta }_{ 1 }) }\\ From\quad here\quad we\quad can\quad infer\quad that\quad r=1\\ Now\quad \\ { z }_{ 1 }\times { z }_{ 2 }\times { z }_{ 3 }=\overline { { z }_{ 2 }{ z }_{ 3 } } \times { z }_{ 3 }{ z }_{ 2 }={ \left( \left| { z }_{ 2 } \right| \times \left| { z }_{ 3 } \right| \right) }^{ 2 }={ r }^{ 4 }={ 1 }^{ 4 }=1\\

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Moderator note:

Good writeup using the various ideas in complex numbers. Here are some things to take note of:

  • The equation after the first implication sign should be r = r 2 r − i ( θ 1 + θ 2 + θ 3 ) r = r^2 r^{ -i ( \theta_1 + \theta_2 + \theta_3) } .
  • That equation has 2 solutions in r r , and we should explain why r = 0 r= 0 is rejected.
  • We have found a necessary condition for r r . However, is it sufficient? IE Does there exist 3 complex numbers that satisfy the conditions in the problem?

Nice Solution! Short and sweet!

Miraj Shah - 5 years ago

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Are you giving JEE this year or the next.

Rishi Sharma - 5 years ago

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Yes! Are you there on slack? Probably we can talk there

Miraj Shah - 5 years ago

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@Miraj Shah – I am not on slack and now i have to go and urgently meet someone. BTW good luck for you JEE P.S:-I'll giving it next year.

Rishi Sharma - 5 years ago

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@Rishi Sharma – Thank You very much! And good luck to you too! Just keep enjoying what you do!

Miraj Shah - 5 years ago

uh ! MOTHER of over-rated questions use euler form of complex numbers and some basic trigo , u get the answer :}

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