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Algebra Level 5

( z + 1 z ) + ( z 2 + 1 z 2 ) 2 + ( z 3 + 1 z 3 ) 3 + ⋯ + ( z 10 + 1 z 10 ) 10 \left(z + \dfrac{1}{z}\right) + \left(z^2 + \dfrac{1}{z^2}\right)^2 + \left(z^3 + \dfrac{1}{z^3}\right)^3 + \cdots + \left(z^{10} + \dfrac{1}{z^{10}}\right)^{10}

If z z is a complex number satisfying z 2 + z + 1 = 0 z^2 + z + 1 = 0 , then what is the value of the expression above?

Original problem


The answer is 585.

Akhil Bansal by Akhil Bansal , 22, India

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2 solutions

Akhil Bansal
Nov 24, 2015

As De Moivre's Theorem will tell us, the two non-real cube roots of unity, ( ω \omega and ω 2 \omega^2 ), are the roots of z 2 + z + 1 = 0 , z^2 + z + 1 = 0, so it follows that 1 + ω + ω 2 = 0 1 + \omega + \omega^2 = 0 and that ω 3 = 1 \omega^3 = 1

⇒ ( ω + 1 ω ) + ( ω 2 + 1 ω 2 ) 2 + ( ω 3 + 1 ω 3 ) 3 + … + ( ω 10 + 1 ω 10 ) 10 \Rightarrow \left(\omega + \dfrac{1}{\omega}\right) + \left(\omega^2 + \dfrac{1}{\omega^2}\right)^2 + \left(\omega^3 + \dfrac{1}{\omega^3}\right)^3 + \ldots + \left(\omega^{10} + \dfrac{1}{\omega^{10}}\right)^{10} ⇒ ( ω 2 + 1 ω ) + ( ω 4 + 1 ω 2 ) 2 + ( 1 + 1 1 ) 3 + … + ( ω 20 + 1 ω 10 ) 10 \Rightarrow \left( \dfrac{\omega^2 + 1}{\omega}\right) + \left( \dfrac{\omega^4 + 1}{\omega^2}\right)^2 + \left(1+ \dfrac{1}{1}\right)^3 + \ldots + \left( \dfrac{\omega^{20} + 1}{\omega^{10}}\right)^{10} ⇒ ( − 1 ) + ( − 1 ) 2 + 2 3 + ( − 1 ) 4 + ( − 1 ) 5 + 2 6 + ( − 1 ) 7 + ( − 1 ) 8 + 2 9 + ( − 1 ) 10 \Rightarrow (-1) + (-1)^2 + 2^3 + (-1)^4 + (-1)^5 + 2^6 + (-1)^7 + (-1)^8 + 2^9 + (-1)^{10} ⇒ 585 \Rightarrow 585

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Moderator note:

Is this answer independent of which cube root of unity you chose? Why, or why not?

I love it! Great solution, Akhil.

Note that I've added the Brilliant wiki link.
You can do this too by typing this in your solution: [[wiki-

Andrew Ellinor - 5 years, 6 months ago

Standard approach

Aakash Khandelwal - 5 years, 6 months ago

In response to challenge master note : I choose ω \omega to solve above question but even if i use ω 2 \omega^2 then also answer will remain same..
What will happen is just value of first and second term in above expansion will swap and similarly like terms and multiple of third term will remain at its position.
Thus, answer is independent whether we choose ω \omega or ω 2 \omega^2 .

Akhil Bansal - 5 years, 6 months ago

From where u get to know about these theorems

A Former Brilliant Member - 5 years, 6 months ago

Log in to reply

From my teachers when i was in class xi

Akhil Bansal - 5 years, 6 months ago

Same method!!

Divyansh Chaturvedi - 5 years, 6 months ago

L e t f ( n ) = z n + 1 z n . w h e r e n i s a n i n t e g e r . z 2 + z + 1 = 0 d i v i d i n g b o t h s i d e s b y z ∵ z ≠ 0 , ⟹ z + 1 z + 1 = 0 ∴ f ( 1 ) = − 1 f ( 2 ) + 2 = f ( 1 ) 2 = 1 , ∴ f ( 2 ) = − 1 f ( 3 ) + 3 ∗ 1 ∗ f ( 1 ) = f ( 1 ) 3 = − 1 , ∴ f ( 3 ) = 2 f ( 4 ) + 2 ∗ 1 ∗ f ( 2 ) = f ( 1 ) 4 = 1 , ∴ f ( 4 ) = − 1 f ( 5 ) + f ( 1 ) = f ( 2 ) ∗ f ( 3 ) = − 1 ∗ 2 , ∴ f ( 5 ) = − 1 A n d s o w e w i l l g e t f ( 1 ) = f ( 2 ) = f ( 4 ) = f ( 5 ) = f ( 7 ) = f ( 8 ) = f ( 10 ) = − 1 a n d f ( 3 ) = f ( 6 ) = f ( 9 ) = 2. ∴ f ( 1 ) + f ( 2 ) 2 = f ( 4 ) 4 + f ( 5 ) 5 = f ( 7 ) 7 + f ( 8 ) 8 = 0 , a n d f ( 10 ) 10 = 1 . a n d f ( 3 ) 3 = 2 3 , f ( 6 ) 6 = 2 6 . f ( 9 ) 9 = 2 9 . ∴ t h e g i v e n e x p r e s s i o n = 0 + 2 3 + 0 + 2 6 + 0 + 2 9 + 1 = 585 Let \ f(n)=z^n+\dfrac 1 {z^n}.\ \ \ where\ n\ is\ an\ integer.\\ z^2+z+1=0\ \ \ \ \ \ dividing\ \ both\ sides\ by\ z \ \ \because\ z\neq\ 0,\ \ \ \implies\ \ z+\dfrac1 z + 1=0\ \ \ \ \ \ \therefore\ \color{#3D99F6}{f(1)=\ -1}\\ f(2) +2=f(1)^2=1, \ \ \ \ \ \ \therefore \ \color{#3D99F6}{f(2)=\ -1} \\ f(3) +3*1*f(1)=f(1)^3= - 1, \ \ \ \ \ \ \therefore \ \color{#3D99F6}{f(3)=\ 2} \\ f(4) +2*1*f(2)=f(1)^4= 1, \ \ \ \ \ \ \therefore \ \color{#3D99F6}{f(4)=\ - 1} \\ f(5) +f(1)=f(2)*f(3)= - 1*2, \ \ \ \ \ \ \therefore \ \color{#3D99F6}{f(5)=\ - 1} \\ And \ so\ we\ will\ get \ \ \ \ \ \ f(1)=f(2)=f(4)=f(5)= f(7)=f(8)= f(10)= - 1\\ and\ f(3)=f(6)=f(9)=2.\\ \therefore\ \color{#3D99F6}{f(1)+f(2)^2=f(4)^4+f(5)^5= f(7)^7+f(8)^8= 0,\ \ \ and\ \ \ f(10)^{10}= 1}.\\ and \ \color{#3D99F6}{f(3)^3=2^3,\ \ \ \ \ f(6)^6=2^6.\ \ \ \ \ \ f(9)^9=2^9}.\\ \therefore\ the\ given\ expression =0+2^3+0+2^6+0+2^9+1=\Large\ \ \ \ \ \ \color{#D61F06}{585}\\ \ \ \ \ \\

Note:- I have not used Mod for those who might not be familiar with Mod.
However f ( n ≡ 1 ( m o d 3 ) ) = f ( n ≡ 2 ( m o d 3 ) ) = − 1 , f ( n ≡ 0 ( m o d 3 ) ) = 2. f ( n ≡ 1 ( m o d 3 ) ) o d d + f ( n ≡ 2 ( m o d 3 ) ) e v e n = 0 , f ( n ≡ 0 ( m o d 3 ) ) n = 2 n . ∴ ∑ i = 1 3 2 3 ∗ i + f ( 10 ) 10 = 584 + 1 = 585. f(n \equiv 1 \pmod {3})=f(n \equiv 2 \pmod {3}) = - 1,\ \ \ \ \ \ \ f(n \equiv 0 \pmod {3})=2.\\ f(n \equiv 1 \pmod {3})^{odd} + f(n \equiv 2 \pmod {3})^{even} =0,\ \ \ \ \ f(n \equiv 0 \pmod {3})^n=2^n.\\ \therefore\ \displaystyle \sum_{i=1}^3 2^{3*i} + f(10)^{10}=584+1=585.

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