Sequence of Complex Stuff
x 1 = 0 , and x n + 1 = x n 2 − i for n > 1
Above is the sequence x 1 , x 2 , x 3 , … of the complex numbers.. Find the square of the distance between x 2 0 0 0 and x 1 9 9 7 in the complex plane.
Details and Assumptions:
i
=
−
1
is the imaginary number.
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The answer is 5.000.
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2 solutions
is there any specific approach or observation technique for solving sequence problems like this (or any Real number sequence problem) ?
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I am afraid there is none that I know of. Just try and error to solve it.
its written in the condition that n should be greater than 1,i guess question needs modification.
For n > 1 we notice ,
x 2 n = i
x 2 n − 1 = − 1 − i .
Hence. x 2 0 0 0 = z 1 = i ; x 1 9 9 7 = z 2 = − 1 − i
Hence distance = ∣ z 1 − z 2 ∣ = 5 .
Therefore answer is 5
x 1 = 0 and x n + 1 = x n 2 − i for n > 1
⇒ x 1 = 0 x 2 = − i x 3 = − 1 − i x 4 = 1 + 2 i − 1 − i = i x 5 = − 1 − i x 6 = i . . .
⇒ For n > 2 , x n = { − 1 − i i if n is odd if n is even
⇒ x 1 9 9 7 = − 1 − i , x 2 0 0 0 = i and the square of the distance between them = ( − 1 − 0 ) 2 + ( − 1 − 1 ) 2 = 1 + 4 = 5