Complex Numbers Simplified?

Algebra Level 5

Let x , y , z x,y,z be complex numbers such that the following equations hold. { x + y + z = 2 x 2 + y 2 + z 2 = 3 x y z = 4 \begin{cases} x+y+z &= ~~~2 \\ x^2+y^2+z^2 &=~~~3 \\ xyz &=~~~4 \end{cases} Find the value of 1 x y + z 1 + 1 y z + x 1 + 1 z x + y 1 . \dfrac{1}{xy+z-1}+\dfrac{1}{yz+x-1}+\dfrac{1}{zx+y-1}.


The answer is -0.222.

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4 solutions

Chew-Seong Cheong
Oct 16, 2014

Since x + y + z = 2 x+y+z = 2 , then z = 2 x y z = 2 - x -y .

x y + z 1 = x y + 2 x y 1 = x y x y + 1 = ( x 1 ) ( y 1 ) \Rightarrow xy + z - 1 = xy + 2 - x -y -1 = xy - x - y + 1 = (x-1)(y-1)

1 x y + z 1 + 1 y z + x 1 + 1 z x + y 1 \Rightarrow \dfrac {1}{xy+z-1} + \dfrac {1}{yz+x-1} + \dfrac {1}{zx+y-1}

= 1 ( x 1 ) ( y 1 ) + 1 ( y 1 ) ( z 1 ) + 1 ( z 1 ) ( x 1 ) \quad = \dfrac {1}{(x-1)(y-1)} + \dfrac {1}{(y-1)(z-1)} + \dfrac {1}{(z-1)(x-1)}

= x 1 + y 1 + z 1 ( x 1 ) ( y 1 ) ( z 1 ) = 1 ( x 1 ) ( y 1 ) ( z 1 ) \quad = \dfrac {x-1+y-1+z-1}{(x-1)(y-1)(z-1)} = \dfrac {-1}{(x-1)(y-1)(z-1)}

Now ( x 1 ) ( y 1 ) ( z 1 ) = x y z ( x y + y z + z x ) + ( x + y + z ) 1 (x-1)(y-1)(z-1) = xyz - (xy+yz+zx)+(x+y+z)-1 .

And x y + y z + z x = 1 2 [ ( x + y + z ) 2 ( x 2 + y 2 + z 2 ) ] = 1 2 ( 4 3 ) = 1 2 xy+yz+zx = \frac {1}{2} [(x+y+z)^2 - (x^2+y^2+z^2)] = \frac {1}{2} (4 - 3) = \frac {1}{2}

Therefore,

1 x y + z 1 + 1 y z + x 1 + 1 z x + y 1 \dfrac {1}{xy+z-1} + \dfrac {1}{yz+x-1} + \dfrac {1}{zx+y-1}

= 1 x y z ( x y + y z + z x ) + ( x + y + z ) 1 = \dfrac {-1}{xyz - (xy+yz+zx)+(x+y+z)-1}

= 1 4 1 2 + 2 1 = 1 9 2 = 2 9 0.222 = \dfrac {-1}{4 - \frac {1}{2} +2-1} = \dfrac {-1}{\frac {9}{2}} = -\dfrac {2}{9} \approx \boxed {-0.222}

Very nice solution and very intuitive

Trevor Arashiro - 6 years, 7 months ago

Nice solution ¨ \ddot\smile

Karthik Kannan - 6 years, 7 months ago

Nice solution :-)

Bhargav Upadhyay - 6 years, 5 months ago

Nice solution

Amartya Prakash - 5 years, 4 months ago
Ronak Agarwal
Oct 16, 2014

x y = 4 z xy=\frac { 4 }{ z }

Hence first term becomes :

1 4 z + z 1 = z z 2 z + 4 \large \frac { 1 }{ \frac { 4 }{ z } +z-1 } =\frac { z }{ { z }^{ 2 }-z+4 }

Hence our desired sum becomes :

S = c y c z z 2 z + 4 S=\displaystyle \sum _{ cyc }^{ }{ \frac { z }{ { z }^{ 2 }-z+4 } }

Now 2 ( x y + y z + x z ) = ( x + y + z ) 2 ( x 2 + y 2 + z 2 ) 2(xy+yz+xz)={(x+y+z)}^{2}-({x}^{2}+{y}^{2}+{z}^{2})

x y + y z + x z = 1 2 \Rightarrow xy+yz+xz=\frac{1}{2}

We assume x,y,z to be roots of a cubic polynomial hence the polynomial is given by :

p 3 ( c y c x ) p 2 + ( c y c x y ) p x y z = 0 { p }^{ 3 }-(\displaystyle \sum _{ cyc }^{ }{ x } ){ p }^{ 2 }+(\displaystyle \sum _{ cyc }^{ }{ xy } )p-xyz=0

Putting the given values we get :

2 p 3 4 p 2 + p 8 = 0 2{p}^{3}-4{p}^{2}+p-8=0

Since z is a root hence we have :

2 z 3 4 z 2 + z 8 = 0 2{z}^{3}-4{z}^{2}+z-8=0

Factorising we get :

2 ( z 1 ) ( z 2 z + 4 ) = 9 z 2(z-1)({z}^{2}-z+4)=9z

Hence we have :

2 9 ( z 1 ) = z z 2 z + 4 \frac { 2 }{ 9 } (z-1)=\frac { z }{ { z }^{ 2 }-z+4 }

Our desired sum becomes :

S = 2 9 c y c ( z 1 ) S=\frac { 2 }{ 9 } \displaystyle \sum _{ cyc }^{ }{ (z-1) }

Which it turns out to be :

S = 2 9 ( x + y + z 3 ) = 2 9 S=\frac { 2 }{ 9 } (x+y+z-3)=\frac{-2}{9}

Can you pls explain how you factorized 2 z 3 4 z 2 + z 8 = 0 2{z}^{3}-4{z}^{2}+z-8=0 to 2 ( z 1 ) ( z 2 z + 4 ) = 9 z 2(z-1)({z}^{2}-z+4)=9z , I mean how did you know that you had to isolate 9z and keep it on RHS .

otherwise I liked your solution .

Pls help me : @SanjeetRaria ,@CalvinLin,@SandeepBharadwaj,@PiHanGoh,@MeghChoksi,@ArronKau,@MichaelMendrin,@Chew-SeongCheong .

This is a good question .

Kudou Shinichi - 6 years, 5 months ago

i did the exact same!!!!

Ashutosh Sharma - 3 years, 4 months ago

Nice solution.

Sandeep Bhardwaj - 6 years, 8 months ago

Thanks for appreciating the solution.

Ronak Agarwal - 6 years, 7 months ago

Nice way to use polynomial... I solved it similar to chew's. But i like your solution.

Sanjeet Raria - 6 years, 7 months ago

Nice solution Ronak! ¨ \ddot\smile

Karthik Kannan - 6 years, 7 months ago

can u please explain transition from : eq--> p^3-sigma(x)p^2+sigma(xy)p-xyz=0 TO 2p^-4p^2+p-8

abhideep singh - 6 years, 6 months ago

Thats way more innovative, i expressed denominators as (x-1)(y-1) and so on then took LCM and added and inputted values after grouping terms accordingly,,,

Mvs Saketh - 6 years, 7 months ago
Sharky Kesa
Oct 16, 2014

This is my problem, Complicated Complex Fractions !!!

Welllll I think it'd be a better move to remove that link or your problem's rating would go down since anyone who clicked 'View Solution' here can get your problem right. @Sharky Kesa

Vishnu Bhagyanath - 5 years, 9 months ago

@Sandeep Bhardwaj

Sharky Kesa - 6 years, 7 months ago

Log in to reply

Yeah, I find it a little while ago. I'm sorry for posting it again. But I was not aware of this problem has been posted on Brilliant. @Sharky Kesa

Sandeep Bhardwaj - 6 years, 7 months ago
Lu Chee Ket
Oct 18, 2015

This answer is checked to have no cyclic order differences.

x = -0.229220113606684-1.25479322434906i

y = -0.229220113606684+1.25479322434906i

z = 2.45844022721337

2

3.00000000000003

3.99999999999999

All checked correct.

0.324097828269638

-0.27316002524593-0.278842938744237i

-0.27316002524593+0.278842938744237i

Sum = -0.222222222222222 such that -2/ 9

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