FIBONACCI SEQUENCE AND COMPLEX NUMBERS



 

Again the Fibonacci sequence can be represented as

1, 1, 2, 3, 5, 8, 13, 21, 34, 5, 89, 144, 23, 377, 610, 987, .
 

Interesting connections as between successive Fibonacci terms can be demonstrated with inclusion of imaginary numbers.

For example if we subtract i (the square root of - 1) from each term,

then for example (5 - i)/(3 - i) = (8 + i)/5

When we replace 5 by 8 and 3 by 5 (i.e. the next Fibonacci terms), then a similar relation results (where i is replaced by 1.

Thus (8 - 1)/(5 - 1) = (13 + 1)/8.

When we move on to the next Fibonacci terms, the relationship once again reverts back to that involving imaginary numbers.

Thus (13 - i)/(8 - i) = (21 + i)/13

Therefore we can generalize the relationship as follows

When n is odd

(Tn - i)/( Tn-1 - i) = (Tn+1 + i)/ Tn

When n is even

(Tn - 1)/( Tn-1 - 1) = (Tn+1 + 1)/ Tn

So we have a fascinating relationship here involving Fibonacci terms which alternately replacing an imaginary number (i) with its real counterpart (1).
 
 

Similar relationships can be demonstrated when we subtract imaginary values from real Fibonacci terms which themselves represent Fibonacci values

Thus for example (8 - 2i)/(5 - i) = (21 - i)/13

Again when we move on to the next terms in the sequence (i.e. 13 and 8), once again i is replaced by 1.

Thus (13 - 2)/8 - 1) = (34 - 1)/21

Thus when n is odd

(Tn+1 - 2i)/( Tn - i) = (Tn+3 + i)/Tn+2
 
 

When n is even

Then (Tn+1 - 2)/( Tn - 1) = (Tn+3 + 1)/Tn+2

We can always establish similar type relationship where successive Fibonacci term values (real and imaginary) are subtracted from original Fibonacci terms.

Also when the use the next set of consecutive terms (real and imaginary) to subtract from the original Fibonacci terms we alternate from imaginary to real (and imaginary to real).

Thus whereas (8 - 2i)/(5 - i) = (21 - i)/13

(8 - 3)/(5 - 2) = (34 + 1)/21
 

and then

(13 - 3i)/(8 - 2i) = (55 + i)/34
 

Thus when n is even

(Tn+1 - 3i)/(Tn - 2i) = (Tn+4 + i)/Tn+3
 
 

However when n is odd

(Tn+1 - 3)/(Tn - 2) = (Tn+4 + 1)/Tn+3