Not Quite the Unity I'm Looking For
1 2 1 n = 1 ∑ 2 4 ( z n + z n 1 ) 2
What is the value of the expression above if we are given that z 2 − z + 1 = 0 ?
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5 solutions
Nice solution sir but, i think it would have been a bit easier if you would have noticed that the roots of the given equation are − w , − w 2 . w h e r e w & w 2 a r e t h e c u b e r o o t s o f u n i t y .
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I actually tried that but I didn't find an easy way. I actually used a spreadsheet to calculate the values which is very fast. It just that presenting the solution is lengthy.
Nice solution sir :) I guess this one involved a lot of hard work for you in typing out .
α n 2 i s c y c l i c w i t h n ≡ 0 ( m o d 3 ) , α n 2 = 1 , n ≡ 1 ( m o d 3 ) , α n 2 = 1 , n ≡ 2 ( m o d 3 ) , α n 2 = 4 , S o t h e g r o u p o f t h r e e a d d u p t o 6 . T h e r e a r e 3 2 4 = 8 g r o u p s , T o t a l = 8 ∗ 6 = 4 8 . A n s w e r 1 2 4 8 = 4
We see that z 6 k = 1 , z 6 k + 1 = z , z 6 k + 2 = z − 1 , z 6 k + 3 = − 1 , z 6 k + 4 = − z , z 6 k + 5 = 1 − z
We have to find
1 2 1 n = 1 ∑ 2 4 z 2 n + z 2 n 1 + 2
Now we can 'sum out' 2 out and use GP.
z 2 − 1 z 2 ( z 4 8 − 1 ) − z 4 8 ( z 2 − 1 ) − z 4 8 + 1
= z 4 8 ( z 2 − 1 ) z 9 8 − z 5 0 − 1 + z 4 8
From above we get the values of powers of z ,
0
And now getting the trivial things back,
we get the answer as - 1 2 1 ( 0 + 4 8 ) = 4
Were you in a hurry in typing out this solution , I felt so after reading your solution .
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Yeah! A lot! Two "exactly different solutions"! Lol!
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Please no more oxymorons please ! Just gave up on English :P
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@A Former Brilliant Member – Okay, lol! :P I just love them though.
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@Kartik Sharma – So did you solve your doubt in Magnetism in Various circumstances ?
Z= Cos 60+i sin 60 And (e°iA)°n = e°inA
I had written full solution. It has vanished!!!! It often happens. I wait for reaction from Challenge Master.
Let z=-w (omega) and then it's a cakewalk!
z 2 − z + 1 = 0 ⇒ z − 1 + z 1 = 0 ⇒ z + z 1 = 1
Then z 2 + z 2 1 = ( z + z 1 ) 2 − 2 ( z ˙ z 1 ) = 1 − 2 = − 1
For n > 2 ,
\(\begin{array} {} z^n + \dfrac {1}{z^n} & = \left( z + \dfrac {1}{z}\right) \left( z^{n-1} + \dfrac {1}{z^{n-1}}\right)-\left( z \dot{} \dfrac {1}{z}\right) \left( z^{n-2} + \dfrac {1}{z^{n-2}}\right) \\ & = \left( z^{n-1} + \dfrac {1}{z^{n-1}}\right)-\left( z^{n-2} + \dfrac {1}{z^{n-2}}\right) \end{array} \)
Let α n = z n + z n 1 , for n > 2 ⇒ α n = α n − 1 − α n − 2
Therefore,
α 1 = 1 α 7 = 1 α 1 3 = 1 α 1 9 = 1 α 2 = − 1 α 8 = − 1 α 1 4 = − 1 α 2 0 = − 1 α 3 = − 2 α 9 = − 2 α 1 5 = − 2 α 2 1 = − 2 α 4 = − 1 α 1 0 = − 1 α 1 6 = − 1 α 2 2 = − 1 α 5 = 1 α 1 1 = 1 α 1 7 = 1 α 2 3 = 1 α 6 = 2 α 1 2 = 2 α 1 8 = 2 α 2 4 = 2
Therefore,
\(\begin{array} {} \displaystyle \frac {1}{12} \sum_{n=1}^{24} { \left( z^{n} + \dfrac {1}{z^{n}}\right)^2} & = \dfrac {1}{12} ( 1+1+4+1+1+4+...+1+1+4) \\ & = \dfrac {8}{12}(1+1+4) \\ & = \boxed{4} \end{array} \)